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Copyright © 2014 Pearson Education, Inc.
Chapter 6
Normal Probability Distributions
Section 6-2, Basic Skills and Concepts
1. The word “normal” has a special meaning in statistics. It refers to a specific bell-shaped distribution that can be
described by Formula 6-1.
2.
3. The mean and standard deviation have values of 0
μ = and 1
σ=
4. The notation α
z represents the z score that has an area of α to its right.
5. ( )
0.2 5 1.25 0.75
− = 6. ( )
0.2 0.75 0 0.15
− =
7. ( )
0.2 3 1 0.40
− = 8. ( )
0.2 4.5 1.5 0.60
− =
9. You will use Excel’s NORM.S.DIST function to solve the problem. Click Formulas at the top of the screen.
Select Insert Function. Select the Statistical category. Select the NORM.S.DIST function. Click OK. Enter
the following values: z = 0.44, Cumulative = TRUE. Click OK. The function returns 0.6700.
( )
0.44 0.6700
P z < =
the following values: z = –1.04, Cumulative = TRUE. Click OK. The function returns 0.1492. The area of the
shaded region = 1 – 0.1492 = 0.8508. ( )
1.04 0.8508
P z > − =
11. You will use Excel’s NORM.S.DIST function to solve the problem. You will first find the cumulative area to
the left of z = 1.28. Next you will find the cumulative area to the left of z = –0.84. Then you will subtract the
cumulative area to the left of z = –0.84 from the cumulative area to the left of z = 1.28. Click Formulas at the
top of the screen. Select Insert Function. Select the Statistical category. Select the NORM.S.DIST function.
Click OK. Enter the following values: z = 1.28, Cumulative = TRUE. Click OK. The function returns
0.8997. Select the NORM.S.DIST function again. Enter the following values: z = –0.84, Cumulative =
TRUE. Click OK. The function returns 0.2005.
( ) ( ) ( )
0.84 1.28 1.28 0.84 0.8997 0.2005 0.6992
P z P z P z
− < < = < − < − = − =
Click OK. Enter the following values: z = 0.67, Cumulative = TRUE. Click OK. The function returns
0.7486. Select the NORM.S.DIST function again. Enter the following values: z = –1.07, Cumulative =
TRUE. Click OK. The function returns 0.1423.
( ) ( ) ( )
1.07 0.67 0.67 1.07 0.7486 0.1423 0.6063
P z P z P z
− < < = < − < − = − =

94 Chapter 6: Normal Probability Distributions
13. You will use Excel’s NORM.S.INV function to solve the problem. Click Formulas at the top of the screen.
Select Insert Function. Select the Statistical category. Select the NORM.S.INV function. Enter the
following value: Probability = 0.8907. The function returns 1.23. 1.23
z=
following value: Probability = 0.3050. The function returns –0.51. 0.51
z=−
15. You will use Excel’s NORM.S.INV function to solve the problem. First subtract 0.9265 from 1. 1 – 0.9265 =
0.0735. Click Formulas at the top of the screen. Select Insert Function. Select the Statistical category.
Select the NORM.S.INV function. Enter the following value: Probability = 0.0735. The function returns –
01.45. 1.45
z=−
16. You will use Excel’s NORM.S.INV function to solve the problem. First subtract 0.2061 from 1. 1 – 0.2061 =
0.7939. Click Formulas at the top of the screen. Select Insert Function. Select the Statistical category.
Select the NORM.S.INV function. Enter the following value: Probability = 0.7939. The function returns
0.82. 0.82
z=
Select Insert Function. Select the Statistical category. Select the NORM.S.DIST function. Enter the
following values: z = –2.04, Cumulative = TRUE. The function returns 0.0207. ( )
2.04 0.0207
P z < − =
the following values: z = –0.19, Cumulative = TRUE. Click OK. The function returns 0.4247.
( )
0.19 0.4247
P z < − =
the following values: z = 2.33, Cumulative = TRUE. Click OK. The function returns 0.9901.
( )
2.33 0.9901
P z < =
following values: z = 1.96, Cumulative = TRUE. The function returns 0.9750. ( )
1.96 0.9750
P z < =
following values: z = 0.82, Cumulative = TRUE. The function returns 0.7939. 1 – 0.7939 = 0.2061.
( )
0.82 1 0.7939 0.2061
P z > = − =
following values: z = 1.82, Cumulative = TRUE. The function returns 0.9656. 1 – 0.9656 = 0.0344.
( )
1.82 1 0.9656 0.0344
P z > = − =
following values: z = –1.50, Cumulative = TRUE. The function returns 0.0668. 1 – 0.0668 = 0.9332.
( )
1.50 1 0.0668 0.9332
P z > − = − =
( )
0.84 1 0.2005 0.7995
P z > − = − =
the left of z = 1.25. Next you will find the cumulative area to the left of z = 0.25. Then you will subtract the
cumulative area to the left of z = 0.25 from the cumulative area to the left of z = 1.25. Click Formulas at the

Section 6-2, Basic Skills and Concepts 95
Enter the following values: z = 1.25, Cumulative = TRUE. The function returns 0.8944.
Select the NORM.S.DIST function again. Enter the following values: z = 0.25, Cumulative = TRUE. The
function returns 0.5987. ( ) ( ) ( )
0.25 1.25 1.25 0.25 0.8944 0.5987 0.2957
P z P z P z
< < = < − < = − =
the left of z = 2.37. Next you will find the cumulative area to the left of z = 1.23. Then you will subtract the
cumulative area to the left of z = 1.23 from the cumulative area to the left of z = 2.37. Click Formulas at the
1.23 2.37 2.37 1.23 0.9911 0.8907 0.1004
P z P z P z
< < = < − < = − =
the left of z = –2.00. Next you will find the cumulative area to the left of z = –2.75. Then you will subtract the
cumulative area to the left of z = –2.00 from the cumulative area to the left of z = –2.75. Click Formulas at the
Enter the following values: z = –2.00, Cumulative = TRUE. The function returns 0.0228.
Select the NORM.S.DIST function again. Enter the following values: z = –2.75, Cumulative = TRUE. The
2.75 2.00 2.00 2.75 0.0228 0.0030 0.0198
P z P z P z
− < < − = < − − < − = − =
the left of z = –0.45. Next you will find the cumulative area to the left of z = –1.93. Then you will subtract the
cumulative area to the left of z = –1.93 from the cumulative area to the left of z = –0.45. Click Formulas at the
Enter the following values: z = –0.45, Cumulative = TRUE. The function returns 0.3264.
1.93 0.45 0.45 1.93 0.3264 0.0268 0.2996
P z P z P z
− < < − = < − − < − = − =
2.20 2.50 2.50 2.20 0.9938 0.0139 0.9799
P z P z P z
− < < = < − < − = − =
Click OK. Enter the following values: z = 1.78, Cumulative = TRUE. The function returns 0.9625.
0.62 1.78 1.78 0.62 0.9625 0.2676 0.6949
P z P z P z
− < < = < − < − = − =
function returns 0.017429.
( ) ( ) ( )
2.11 4.00 4.00 2.11 0.999968 0.017429 0.9825
P z P z P z
− < < = < − < − = − =

( ) ( ) ( )
3.90 2.00 2.00 3.90 0.97725 0.000048 0.9772
P z P z P z
− < < = < − < − = − =
following values: z = 3.65, Cumulative = TRUE. The function returns 0.999869. ( )
3.65 0.9999
P z < =
( )
3.80 0.9999
P z > − =
following values: z = 0, Cumulative = TRUE. The function returns 0.5000. ( )
0 0.5000
P z < =
following values: z = 0, Cumulative = TRUE. The function returns 0.5000. ( )
0 0.5000
P z > =
following value: Probability = 0.90. The function returns 1.28. 90 1.28
P =
following value: Probability = 0.05. The function returns –1.645. 5 1.645
P = −
following value: Probability = 0.025. The function returns –1.96. Because the distribution is symmetrical,
2.5 1.96
P = − and 97.5 1.96
P =
following value: Probability = 0.005. The function returns –2.576. Because the distribution is symmetrical,
0.5 2.576
P = − and 99.5 2.576
P =
following value: Probability = 0.025. The function returns –1.96. 0.025 1.96
z =
z =
z =

z =
the left of z = 1. Next you will find the cumulative area to the left of z = –1. Then you will subtract the
cumulative area to the left of z = –1 from the cumulative area to the left of z = 1. Click Formulas at the top of
the screen. Select Insert Function. Select the Statistical category. Select the NORM.S.DIST function. Enter
the following values: z = 1, Cumulative = TRUE. The function returns 0.8413.
Select the NORM.S.DIST function again. Enter the following values: z = –1, Cumulative = TRUE. The
1 1 1 1 0.8413 0.1587 0.6826 68.26%
P z P z P z
− < < = < − < − = − = =
2 2 2 2 0.9772 0.0228 0.9544 95.44%
P z P z P z
− < < = < − < − = − = =
3 3 3 3 0.9987 0.0013 0.9974 99.74%
P z P z P z
− < < = < − < − = − = =
cumulative area to the left of z = –3.5 from the cumulative area to the left of z = 3.5. Click Formulas at the top
of the screen. Select Insert Function. Select the Statistical category. Select the NORM.S.DIST function.
( ) ( ) ( )
3.5 3.5 3.5 3.5 0.99977 0.00023 0.99954 99.95%
P z P z P z
− < < = < − < − = − = =
Section 6-2, Beyond the Basics
49. a. You will use Excel’s NORM.S.DIST function to solve the problem. You will first find the cumulative area
to the left of z = 1. Next you will find the cumulative area to the left of z = –1. Then you will subtract the
cumulative area to the left of z = –1 from the cumulative area to the left of z = 1. Click Formulas at the top
of the screen. Select Insert Function. Select the Statistical category. Select the NORM.S.DIST
function. Enter the following values: z = 1, Cumulative = TRUE. The function returns 0.8413.
1 1 1 1 0.8413 0.1587 0.6826 68.26%
P z P z P z
− < < = < − < − = − = =
b. You will use Excel’s NORM.S.DIST function to solve the problem. You will find the cumulative area to
the left of z = –2. Click Formulas at the top of the screen. Select Insert Function. Select the Statistical
category. Select the NORM.S.DIST function. Enter the following values: z = –2, Cumulative = TRUE.
The function returns 0.0228. Because the distribution is symmetrical,
( ) ( ) ( )
2 or 2 2 2 0.0228 0.0228 0.0456 4.56%
P z z P z P z
< − > = < − + > = + = =

c. You will use Excel’s NORM.S.DIST function to solve the problem. You will find the cumulative area to
the left of z = –1.96. Click Formulas at the top of the screen. Select Insert Function. Select the
Statistical category. Select the NORM.S.DIST function. Enter the following values: z = –1.96,
Cumulative = TRUE. The function returns 0.0250. Because the distribution is symmetrical,
( ) ( ) ( )
1.96 or 1.96 1.96 1.96 0.0250 0.0250 0.0500
P z z P z P z
< − > = < − + > = + =
( )
1.96 1.96 1 0.0500 0.9500 95%
P z
− < < = − = =
d. You will use Excel’s NORM.S.DIST function to solve the problem. You will first find the cumulative area
to the left of z = 2. Next you will find the cumulative area to the left of z = –2. Then you will subtract the
cumulative area to the left of z = –2 from the cumulative area to the left of z = 2. Click Formulas at the top
of the screen. Select Insert Function. Select the Statistical category.
Select the NORM.S.DIST function. Enter the following values: z = 2, Cumulative = TRUE. The function
returns 0.9772. Select the NORM.S.DIST function again. Enter the following values: z = –2, Cumulative
= TRUE. The function returns 0.0228.
( ) ( ) ( )
2 2 2 2 0.9772 0.0228 0.9544 95.44%
P z P z P z
− < < = < − < − = − = =
e. You will use Excel’s NORM.S.DIST function to solve the problem. You will first find the cumulative area
to the left of z = –3. Click Formulas at the top of the screen. Select Insert Function. Select the
Statistical category. Select the NORM.S.DIST function. Enter the following values: z = –3, Cumulative
= TRUE. The function returns 0.0013. Because the distribution is symmetrical,
( ) ( ) ( )
3 or 3 3 3 0.0013 0.0013 0.0026 0.26%
P z z P z P z
< − > = < − + > = + = =
50. a.
15
2.5
6
μ = = min. and
5
1.4
12
σ = = min.
b. The probability is
1
3
or 0.5774, and it is very different from the probability of 0.6826 that would be
obtained by incorrectly using the standard normal distribution. The distribution does affect the results very
much.
1. a. 0
μ = and 1
σ= .
b. The z scores are numbers without units of measurements.
2. a. The area equals the maximum probability value of 1.
b. The median is the middle value and for normally distributed scores that is also the mean, which is 100.
c. The mode is also 100.
d. The variance is the square of the standard deviation which is 225.
3. The standard normal distribution has a mean of 0 and a standard deviation of 1, but a nonstandard normal
distribution has a different value for one or both of those parameters.
4. No. Randomly generated digits have a uniform distribution, but not a normal distribution. The probability of a
digit less than 3 is
3
0.3
10
= .
5. You will use Excel’s STANDARDIZE function to find z. Click FORMULAS at the top of the screen. Select
Insert Function. Select the Statistical category. Select the STANDARDIZE function. Enter the following
values: x = 118, Mean = 100, Standard deviation = 15. Click OK. The function returns 1.2; 118 1.2
x
z = = .
Next, you will use the NORM.S.DIST function to find the area. Select the Statistical category. Select the
NORM.S.DIST function. Enter the following values: z = 1.2, Cumulative = TRUE. The function returns
0.8848. The area to the left of z = 1.2 is 0.8848.
6. You will use Excel’s STANDARDIZE function to find z. Click FORMULAS at the top of the screen. Select
Insert Function. Select the Statistical category. Select the STANDARDIZE function. Enter the following

values: x = 91, Mean = 100, Standard deviation = 15. The function returns –0.6; 91 0.6
x
z = = − . Next, you
will use the NORM.S.DIST function to find the area. Select the Statistical category. Select the
NORM.S.DIST function. Enter the following values: z = –0.6, Cumulative = TRUE. The function returns
0.2743. The area to the right of z = 1.2 is 1-0.2743 0.7257
= .
7. You will use Excel’s STANDARDIZE function to find the two z values. Click FORMULAS at the top of the
screen. Select Insert Function. Select the Statistical category. Select the STANDARDIZE function. Enter
the following values: x = 79, Mean = 100, Standard deviation = 15. The function returns –1.4; 79 1.4
x
z = = − .
Select the STANDARDIZE function again. Enter the following values: x = 133, Mean = 100, Standard
deviation = 15. The function returns –2.2; 133 2.2
x
z = = . Next, you will use the NORM.S.DIST function to
find the area. Select the Statistical category. Select the NORM.S.DIST function. Enter the following values:
z = –1.4, Cumulative = TRUE. The function returns 0.0808. The area to the left of z = –1.4 is 0.0808.
function returns 0.9861. The area between the two scores is 0.9861 0.0808 0.9053
− = .
8. You will use Excel’s STANDARDIZE function to find the two z values. Click FORMULAS at the top of the
screen. Select Insert Function. Select the Statistical category. Select the STANDARDIZE function. Enter
the following values: x = 112, Mean = 100, Standard deviation = 15. The function returns 0.8; 112 0.8
x
z = = .
Select the STANDARDIZE function again. Enter the following values: x = 124, Mean = 100, Standard
deviation = 15. The function returns 1.6; 124 1.6
x
z = = .
Next, you will use the NORM.S.DIST function to find the area. Select the Statistical category. Select the
NORM.S.DIST function. Enter the following values: z = 0.8, Cumulative = TRUE. The function returns
0.7881. The area to the left of z = 0.8 is 0.7881.
function returns 0.9452. The area between the two scores is 0.9452 0.7881 0.1571
− = .
9. You will use Excel’s NORM.INV function to find the IQ score. Click FORMULAS at the top of the screen.
Select Insert Function. Select the Statistical category. Select the NORM.INV function. Enter the following
values: Probability = 0.9918, Mean = 100, Standard deviation = 15. The function returns an IQ score of 136.
10. First find the area to the left of x: 1 0.1587 0.8413
− = . Now you will use Excel’s NORM.INV function to find the
IQ score. Click FORMULAS at the top of the screen. Select Insert Function. Select the Statistical category.
Select the NORM.INV function. Enter the following values: Probability = 0.8413, Mean = 100, Standard
deviation = 15. The function returns an IQ score of 115.
11. First find the area to the left of x: 1 0.9798 0.0202
− = . Now you will use Excel’s NORM.INV function to find the
IQ score. Click FORMULAS at the top of the screen. Select Insert Function. Select the Statistical category.
Select the NORM.INV function. Enter the following values: Probability = 0.0202, Mean = 100, Standard
deviation = 15. The function returns an IQ score of 69.
13. You will use Excel’s NORM.DIST function to find the probability. Click FORMULAS at the top of the
screen. Select Insert Function. Select the Statistical category. Select the NORM.DIST function. Enter the
following values: x = 85, Mean = 100, Standard deviation = 15, Cumulative = TRUE. The function returns a
probability of 0.1587.
probability of 0.0228. This is the area to the left of an IQ equal to 70. The area to the right of an IQ equal to 70
1 .0228 0.9772
= − = .

probability of 0.2525.
Select the NORM.DIST function again. Enter the following values: x = 110, Mean = 100, Standard deviation
= 15, Cumulative = TRUE. The function returns a probability of 0.7475. The area between the two IQ scores
0.7475 0.2525 0.4950
= − = .
16. You will use Excel’s NORM.DIST function to find the probability. Click FORMULAS at the top of the screen.
Select Insert Function. Select the Statistical category. Select the NORM.DIST function. Enter the following
values: x = 110, Mean = 100, Standard deviation = 15, Cumulative = TRUE. The function returns a probability
of 0.7475.
Select the NORM.DIST function again. Enter the following values: x = 120, Mean = 100, Standard deviation
= 15, Cumulative = TRUE. The function returns a probability of 0.9088. The area between the two IQ scores
0.9088 0.7475 0.1613
= − = .
21. a. You will use Excel’s NORM.DIST function to solve the problem. Click FORMULAS at the top of the
screen. Select Insert Function. Select the Statistical category. Select the NORM.DIST function. Enter
the following values: x = 62, Mean = 63.8, Standard deviation = 2.6, Cumulative = TRUE. The function
returns a probability of 0.2444.
Select the NORM.DIST function again. Enter the following values: x = 78, Mean = 63.8, Standard
deviation = 2.6, Cumulative = TRUE. The function returns a probability of 0.999999976. The area
between 62 and 78 in. 0.999999976 0.2444 0.7556
= − = or 75.56%.
b. You will use Excel’s NORM.DIST function to solve the problem. Click FORMULAS at the top of the
deviation = 2.4, Cumulative = TRUE. The function returns a probability of 0.9998. The area between 62
and 78 in. 0.9998 0.0009 0.9989
= − = or 99.89%. No, only about 0.1% of men are not qualified because of their
heights.
c. You will use Excel’s NORM.INV function to find the heights. Click FORMULAS at the top of the screen.
Select Insert Function. Select the Statistical category. Select the NORM.INV function. Enter the
following values: Probability = 0.02, Mean = 63.8, Standard deviation = 2.6. The function returns height
of 58.5 inches.
Select the NORM.INV function again. Enter the following values: Probability = 0.98, Mean = 63.8,
Standard deviation = 2.6. The function returns height of 69.1inches. The new height requirements are 58.5
inches to 69.1 inches.

d. You will use Excel’s NORM.INV function to find the heights. Click FORMULAS at the top of the screen.
of 63.9 inches.
deviation = 2.6, Cumulative = TRUE. The function returns a probability of 0.99999981. The area between
64 and 77 in. 0.99999981 0.5307 0.4693
= − = or 46.93%.
and 77 in. 0.9991 0.0110 0.9881
= − = or 98.81%.
of 58.9 inches.
deviation = 2.6, Cumulative = TRUE. The function returns a probability of 0.9999918. The area between
56 and 75 in. 0.9999918 0.0013 0.9987
= − = or 99.87%.
and 75 in. 0.9890 0.000000009 0.9890
= − = or 98.90%.
of 59.5 inches.

the following values: x = 51.6, Mean = 69.5, Standard deviation = 2.4, Cumulative = TRUE. The function
returns a probability of 0.0000. 0.00% of adult men can fit through the door without bending.
returns a probability of 0.0000. 0.00% of adult women can fit through the door without bending.
c. The door design is very inadequate, but the jet is relatively small and seats only six people. A much higher
door would require such major changes in the design and cost of the jet, that the greater height is not
practical.
d. You will use Excel’s NORM.INV function to find the height. Click FORMULAS at the top of the screen.
following values: Probability = .6, Mean = 69.5, Standard deviation = 2.4. The function returns height of
70.1inches.
the following values: x = 174, Mean = 182.9, Standard deviation = 40.8, Cumulative = TRUE. The
function returns a probability of 0.4137.
b.
3500
25
140
= people
c.
3500
19.14
182.9
= , so 19 people
d. The mean weight is increasing over time, so safety limits must be periodically updated to avoid an unsafe
condition.
26. a. You will use Excel’s NORM.INV function to find the clearance. Click FORMULAS at the top of the
screen. Select Insert Function. Select the Statistical category. Select the NORM.INV function. Enter
the following values: Probability = .95, Mean = 21.4, Standard deviation = 1.2. The function returns
height of 23.4inches. If there is clearance for 95% of males, there will certainly be clearance for all women
in the bottom 5%.
returns a probability of 0.9599. Men: 95.99%.
Select the NORM.DIST function again. Enter the following values: x = 23.5, Mean = 19.6, Standard
deviation = 1.1, Cumulative = TRUE. The function returns a probability of 0.9998. Women: 99.98%. The
table will fit almost everyone except about 4% of the men with the largest sitting knee heights.
the following values: x = 308 Mean = 268, Standard deviation = 15, Cumulative = TRUE. The function
returns a probability of 0.9962. The probability of a pregnancy lasting 308 days or longer is
1 0.9962 0.0038
− = . Either a very rare event occurred or the husband is not the father.
b. You will use Excel’s NORM.INV function to solve the problem. Click FORMULAS at the top of the

the following values: Probability = .03, Mean = 268, Standard deviation = 15. The function returns 239.8.
The length that separates premature from not premature is 240 days.
the following values: x = 100.6 Mean = 98.2, Standard deviation = 0.62, Cumulative = TRUE. The
function returns a probability of 0.9999. 1 0.9999 0.0001 0.01%
− = = would be considered to have a fever. Yes,
the cutoff is appropriate.
the following values: Probability = 0.95, Mean = 98.2, Standard deviation = 0.62. The function returns
99.22.
the following values: x = 2 Mean = 1.184, Standard deviation = 0.587, Cumulative = TRUE. The function
returns a probability of 0.9178. 91.78% of earthquakes are considered microearthquakes.
returns a probability of 0.999999. 1 0.999999 0.000001 0.00%
− = = of earthquakes fall into this category.
c. You will use Excel’s NORM.INV function to solve the problem. Click FORMULAS at the top of the
2.15. Not all earthquakes about the 95th percentile will cause items to shake.
30. You will use Excel’s NORM.INV function to solve the problem. Click FORMULAS at the top of the screen.
values: Probability = 0.99, Mean = 14.4, Standard deviation = 1. The function returns 16.7.
31. You will use Excel’s NORM.INV function to solve the problem. Click FORMULAS at the top of the screen.
values: Probability = 0.01, Mean = 24, Standard deviation = 2.6. The function returns 18.0; P1 = 18 chocolate
chips.
Select the NORM.INV function again. Enter the following values: Probability = 0.99, Mean = 24, Standard
deviation = 2.6. The function returns 30.0; P99 = 30 chocolate chips. The values can be used to identify cookies
with an unusually low number of chocolate chips or an unusually high number of chocolate chips, so those
numbers can be used to monitor the production process to ensure that the numbers of chocolate chips stay
within reasonable limits.
the following values: x = 5.64 Mean = 5.67, Standard deviation = 0.06, Cumulative = TRUE. The function
returns a probability of 0.3085. Select the NORM.DIST function again. Enter the following values: x =
5.7 Mean = 5.67, Standard deviation = 0.06, Cumulative = TRUE. The function returns a probability of
0.6915. 0.6915 0.3085 0.3830
− = ; 1 38.3% 61.7%
− = are rejected. That percentage is too high because most
quarters will be rejected.
5.55; P2.5 = 5.55 g. Select the NORM.INV function again. Enter the following values: Probability =
0.975, Mean = 5.67, Standard deviation = 0.6. The function returns 5.79; P97.5 = 5.79 g. Accept quarters
with weights between 5.55 g and 5.79 g.
33. a. Open the MBODY data file on your data disk. Select the XLSTAT add-in. Select Visualizing data.
Select Histograms. Enter the pulse data range inclusive of the label. Select Discrete. Select Sample

labels. Click the Outputs tab. Select Descriptive statistics. Click the Charts tab. Select Histograms
and Bars. Click OK. Mean = 67.25 beats per minute. Standard deviation = 10.335 beats per minute. The
histogram, shown below, confirms that the distribution is roughly normal.
the following values: Probability = 0.025, Mean = 67.25, Standard deviation = 10.335. The function
returns 46.99; P2.5 = 47.0 beats per minute g.
Standard deviation = 10.335. Click OK. The function returns 87.51; P97.5 = 87.5 beats per minute.
34. a. Open the COLA data file on your data disk. Select the XLSTAT add-in. Select Visualizing data. Select
Histograms. Enter the PPDIETWT data range inclusive of the label. Select Continuous. Select Sample
labels. Click the Outputs tab. Select Descriptive statistics. Click the Charts tab. Select Histograms
and Bars. Click OK. Mean = 0.783858 lb. Standard deviation = 0.004362 lb. The histogram, shown
below, confirms that the distribution is roughly normal.
the following values: Probability = 0.005, Mean = 0.783858, Standard deviation = 0.004362. The function
returns 0.7726; P0.5 = 0.7726 lbs.
Standard deviation = 0.004362. The function returns 0.7951; P99.5 = 0.7951 lbs.

Section 6-3, Beyond the Basics 105
35. a. The new mean is equal to the old one plus the new points which is 75. The standard deviation is
unchanged at 10 (since we added the same amount to each student.)
b. No, the conversion should also account for variation.
c. You will use Excel’s NORM.INV function to solve the problem. Click FORMULAS at the top of the
the following values: Probability = 0.70, Mean = 40, Standard deviation = 10. The function returns 45.2.
Select the NORM.INV function again. Enter the following values: Probability = 0.90, Mean = 40,
Standard deviation = 10. The function returns 52.8. B grade: 45.2 to 52.8.
d. Using a scheme like the one in part (c), because variation is included in the curving process.
the following values: x = 20 Mean = 24, Standard deviation = 2.6, Cumulative = TRUE. The function
Select the NORM.DIST function again. Enter the following values: x = 30 Mean = 24, Standard
deviation = 2.6, Cumulative = TRUE. The function returns a probability of 0.9895. The percentage
between 20 and 30 chocolate chips inclusive 0.9895 0.0620 0.9275 92.75%
= − = = .
the following values: x = 19.5 Mean = 24, Standard deviation = 2.6, Cumulative = TRUE. The function
Select the NORM.DIST function again. Enter the following values: x = 30.5 Mean = 24, Standard
deviation = 2.6, Cumulative = TRUE. The function returns a probability of 0.9938. The percentage
between 20 and 30 chocolate chips inclusive using the continuity correction 0.9938 0.0417 0.9521 95.21%
= − = = .
c. The use of the continuity correction changes the result by a relatively small but not insignificant amount.
37. The z score for Q1 is –0.67, and the z score for Q3 is 0.67. The IQR is ( )
0.67 0.67 1.34
− − = . 1.5 2.01
IQR
⋅ =
so 1 1.5 0.67 2.01 2.68
Q IQR
− ⋅ = − − = − and 3 1.5 0.67 2.01 2.68
Q IQR
+ ⋅ = + = .
The percentage to the left of –2.68 is 0.0037 and the percentage to the right of 2.68 is 0.0037. Therefore, the
probability of an outlier is 0.0074.
38. a. You will use Excel’s NORM.INV function to solve the problem. Click FORMULAS at the top of the
the following values: Probability = 0.95, Mean = 1511, Standard deviation = 312. The function returns an
SAT score of 2024.
Standard deviation = 5.1. The function returns an ACT score of 29.5.
b. You will use Excel’s NORM.DIST function to find the probability associated with the SAT score. Click
FORMULAS at the top of the screen. Select Insert Function. Select the Statistical category. Select the
NORM.DIST function. Enter the following values: x = 2100, Mean = 1511, Standard deviation = 312,
Cumulative = TRUE. The function returns a probability of 0.9705.
Select the NORM.INV function to find the ACT score associate with a cumulative probability of 0.9705.
Enter the following values: Probability = 0.9705, Mean = 21.1, Standard deviation = 5.1. The function
returns an ACT score of 30.7.

1. a. The sample mean will tend to center about the population parameter of 5.67 g.
b. The sample mean will tend to have a distribution that is approximately normal.
c. The sample proportions will tend to have a distribution that is approximately normal.
2. a. Without replacement
b. (1) When selecting a relatively small sample from a large population, it makes no significant difference
whether we sample with replacement or without replacement. (2) Sampling with replacement results in
independent events that are unaffected by previous outcomes, and independent events are easier to analyze
and they result in simpler calculations and formulas.
3. Sample mean, sample variance, sample proportion
4. No. The data set is only one sample, but the sampling distribution of the mean is the distribution of the means
from all samples, not the one sample mean obtained from this single sample.
5. No. The sample is not a simple random sample from the population of all college Statistics students. It is very
possible that the students at Broward College do not accurately reflect the behavior of all college Statistics
students.
6. a. Normal
b.
0 1 2 3 4 5 6 7 8 9
4.5
10
+ + + + + + + + +
=
c.
5
0.5
10
= which is the proportion of the five odd numbers {1, 3, 5, 7, 9} to the ten digits {0, 1, 2, 3, 4, 5, 6,
7, 8, 9}
7. a. The mean of the population is
4 5 9
6
3
μ
+ +
= = , and the variance is
2 2 2
2 (4 6) (5 6) (9 6)
4.7
3
σ
− + − + −
= =
b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which have
the following variances {0, 0.5, 12.5, 0.5, 0, 8, 12.5, 8, 0} respectively.
Sample Variance Probability
0 3/9
0.5 2/9
8 2/9
c. The sample variances’ mean is
3 0 2 0.5 2 8 2 12.5
4.7
9
⋅ + ⋅ + ⋅ + ⋅
=
d. Yes. The mean of the sampling distribution of the sample variances (4.7) is equal to the value of the
population variance (4.7) so the sample variances target the value of the population variance.
8. a. The population standard deviation (using the result from the previous problem) is 4.7 2.160
σ = =
b. By taking the square root of the sample variances from the previous problem we get

Sample Standard Deviation Probability
0.000 3/9
0.707 2/9
2.828 2/9
3.536 2/9
c. The mean of the sample standard deviations is
3 0 2 0.707 2 2.828 2 3.536
1.571
9
⋅ + ⋅ + ⋅ + ⋅
=
d. No. The mean of the sampling distribution of the sample standard deviations is 1.571, and it is not equal to
the value of the population standard deviation (2.160), so the sample standard deviations do not target the
value of the population standard deviation.
9. a. The population median is 5.
the following medians {4, 4.5, 6.5, 4.5, 5, 7, 6.5, 7, 9}
Sample Median Probability
4 1/9
4.5 2/9
5 1/9
6.5 2/9
7 2/9
9 1/9
c. The mean of the sampling distribution of the sampling median is
4 4.5 4.5 5 6.5 6.5 7 7 9
6
9
+ + + + + + + +
=
d. No. The mean of the sampling distribution of the sample medians is 6, and it is not equal to the value of
the population median of 5, so the sample medians do not target the value of the population median.
10. a. The proportion of odd numbers is 2/3 (there are two odd numbers from the population of 4, 5, and 9)
the following proportion of odd numbers {0, 0.5, 0.5, 0.5, 1, 1, 0.5, 1, 1}
Sample Proportion Probability
0 1/9
0.5 4/9
1 4/9
c. The mean of the sampling distribution of sample proportions is
0 0.5 0.5 0.5 0.5 1 1 1 1 2
0.67
9 3
+ + + + + + + +
= =
d. Yes. The mean of the sampling distribution of the sample proportion of odd numbers is 2/3, and it is equal
to the value of the population proportion of odd numbers of 2/3, so the sample proportions target the value
of the population proportion.

11. a. The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58), (49,
46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)}
Sample Mean Age Probability
46 1/16
47.5 2/16
49 1/16
51 2/16
52 2/16
52.5 2/16
53.5 2/16
56 1/16
57 2/16
58 1/16
b. The mean of the population is
56 49 58 46
52.25
4
+ + +
= and the mean of the sample means is
46 47.5 47.5 49 51 51 52 52 52.5 52.5 53.5 53.5 56 57 57 58
52.25
16
+ + + + + + + + + + + + + + +
=
c. The sample means target the population mean. Sample means make good estimators of population means
because they target the value of the population mean instead of systematically underestimating or
overestimating it.
46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)}
Sample Median Age Probability
46 1/16
47.5 2/16
49 1/16
51 2/16
52 2/16
52.5 2/16
53.5 2/16
56 1/16
57 2/16
58 1/16
b. The median of the population is
49 56
52.5
2
+
= and the median of the sample medians is
52 52.5
52.25
2
+
= . The two values are not equal.

c. The sample medians do not target the population median of 52.5, so the sample medians do not make good
estimators of the population medians.
46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)} which have the following
ranges and associated probabilities:
Sample Range Probability
0 4/16
2 2/16
3 2/16
7 2/16
9 2/16
10 2/16
12 2/16
b. The range of the population is 58 46 12
− = , the mean of the sample ranges is
4 0 2 2 2 3 2 7 2 9 2 10 2 12
5.375
16
⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅
= . The values are not equal.
c. The sample ranges do not target the population range of 12, so sample ranges do not make good estimators
of the population range.
46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)} which have the following
variances and associated probabilities:
Sample Variance Probability
0 4/16
2 2/16
4.5 2/16
24.5 2/16
40.5 2/16
50 2/16
72 2/16
b. The variance of the population is
2 2 2 2
(56 52.25) (49 52.25) (58 52.25) (46 52.25)
24.1875
4
− + − + − + −
= .
The mean of the sample variances is
4 0 2 2 2 4.5 2 24.5 2 40.5 2 50 2 72
24.1875
16
⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅
= .
The two values are equal.
c. The sample variances do target the population variance , so sample variances do make good estimators of
the population variance.
15. The possible birth samples are {(b, b), (b, g), (g, b), (g, g)}.

Proportion of Girls Probability
0 0.25
1 / 2 0.5
2/2 0.25
Yes. The proportion of girls in 2 births is 0.5, and the mean of the sample proportions is 0.5. The result
suggests that a sample proportion is an unbiased estimator of the population proportion.
16. The possible birth samples are {bbb, bbg, bgb, gbb, ggg, ggb, gbg, bgg}.
Proportion of Girls Probability
0 1/3
1/3 3/8
2/3 3/8
3/3 1/8
Yes. The proportion of girls in 3 births is 0.5 and the mean of the sample proportions is 0.5. The result
suggests that a sample proportion is an unbiased estimator of the population proportion.
17. The possibilities are: both questions incorrect, one question correct (two choices), both questions correct.
a.
Proportion Correct Probability
0
2
4 4 16
5 5 25
⋅ =
1
2
1 4 8
2
5 5 25
⎛ ⎞
⎟
⎜
⋅ ⋅ =
⎟
⎜ ⎟
⎜
⎝ ⎠
2
2
1 1 1
5 5 25
⎛ ⎞
⎟
⎜ ⋅ =
⎟
⎜ ⎟
⎜
⎝ ⎠
b. The mean is
16 0 8 0.5 1 1
0.2
25
⋅ + ⋅ + ⋅
= .
c. Yes. The sampling distribution of the sample proportions has a mean of 0.2 and the population proportion
is also 0.2 (because there is 1 correct answer among 5 choices.) Yes, the mean of the sampling distribution
of the sample proportions is always equal to the population proportion.
18. a. The proportions of 0, 0.5, and 1 have the following probabilities
Proportion of Defective Probability
0 9 / 25
0.5 12 / 25
1 4 / 25
b. The mean is
9 0 12 0.5 4 1
0.4
25
⋅ + ⋅ + ⋅
= .
c. Yes. The population proportion is 0.4 (2 out of 5) and the mean of the sampling proportions is also 0.4.
Yes, the mean of the sampling distribution of proportions is always equal to the population proportion.

Section 6-4, Beyond the Basics 111
19.
1 1
(0) 0.25
2(2 2 0)!(2 0)! 4
P = = =
− ⋅ ⋅
,
1
(0.5) 0.5
2(2 2 0.5)!(2 0.5)!
P = =
− ⋅ ⋅
,
1
(1) 0.25
2(2 2 1)!(2 1)!
P = =
− ⋅ ⋅
.
The formula yields values which describes the sampling distribution of the sample proportions. The formula is
just a different way of presenting the same information in the table that describes the sampling distribution.
20. Sample values of the mean absolute deviation (MAD) do not usually target the value of the population MAD, so
a MAD statistic is not good for estimating a population MAD. If the population of {4, 5, 9} from Example 5 is
used, the sample MAD values of 0, 0.5, 2, and 2.5 have corresponding probabilities of 3/9, 2/9, 2/9, and 2/9.
For these values, the population MAD is 2, but the sample MAD values have a mean of 1.1, so the mean of the
sample MAD values is not equal to the population MAD.
1. Because the sample size is greater than 30, the sampling distribution of the mean ages can be approximated by a
normal distribution with mean μ and standard deviation
40
σ
.
2. No. Because the original population is normally distributed, the sample means will be normally distributed for
any sample size, not just those greater than 30.
3. 60.5
x
μ = cm and it represents the mean of the population consisting of all sample means.
6.6
1.1
36
x
σ = = cm,
and it represents the standard deviation of the population consisting of all sample means.
4. Because the digits are equally likely to occur, they have a uniform distribution. Because the sample means are
based on samples of size 3 drawn from a population that does not have a normal distribution, we should not
treat the sample means as having a normal distribution.
b. 205.5
x
μ μ
= = , / 8.6 / 49 1.2286
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 207.0, Mean = 205.5, Standard
deviation = 1.2286, Cumulative = TRUE. The function returns a probability of 0.8889.
the following values: x = 196.9, Mean = 205.5, Standard deviation = 8.6, Cumulative = TRUE. The
b. 205.5
x
μ μ
= = , / 8.6 / 36 1.4333
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 205, Mean = 205.5, Standard deviation
= 1.4333, Cumulative = TRUE. The function returns a probability of 0.3636.
function returns a probability of 0.9332. 1 0.9332 0.0668
− =
b. 205.5
x
μ μ
= = , / 8.6 / 9 2.8667
x n
σ σ
= = =
= 0.9556, Cumulative = TRUE. The function returns a probability of 0.3004. 1 0.3004 0.6996
− =

c. Because the original population has a normal distribution, the distribution of sample means is normal for
any sample size.
returns a probability of 0.1111. 1 0.1111 0.8889
− =
b. 205.5
x
μ μ
= = , / 8.6 / 25 1.72
x n
σ σ
= = =
− =
c. Because the original population has a normal distribution, the distribution of sample means is normal for
any sample size.
Select the NORM.DIST function again. Enter the following values: x = 231.3, Mean = 205.5, Standard
deviation = 8.6, Cumulative = TRUE. The function returns a probability of 0.9987. 0.9987 0.0013 0.9974
− =
b. 205.5
x
μ μ
= = , / 8.6 / 40 1.3598
x n
σ σ
= = =
= 1.3598, Cumulative = TRUE. The function returns a probability of 0.1350.
Select the NORM.DIST function again. Enter the following values: x = 206 Mean = 205.5, Standard
deviation = 1.3598, Cumulative = TRUE. The function returns a probability of 0.6435.
0.6435 0.1350 0.5085
− =
10. a. Select the NORM.DIST function. Enter the following values: x = 180, Mean = 205.5, Standard deviation
= 8.6, Cumulative = TRUE. The function returns a probability of 0.0015. Select the NORM.DIST
function again. Enter the following values: x = 200 Mean = 205.5, Standard deviation = 8.6, Cumulative =
TRUE. The function returns a probability of 0.2612.
b. 205.5
x
μ μ
= = , / 8.6 / 50 1.2162
x n
σ σ
= = =
= 1.2162, Cumulative = TRUE. The function returns a probability of 0.00000. Select the NORM.DIST
function again. Enter the following values: x = 206 Mean = 205.5, Standard deviation = 1.2162,
Cumulative = TRUE. The function returns a probability of 0.6595. 0.6595 0.00000 0.6595
− =
11. 182.9
x
μ μ
= = , / 40.8 / 16 10.2
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 195.3, Mean = 182.9, Standard deviation =
10.2, Cumulative = TRUE. The function returns a probability of 0.8879; 1 0.8879 0.1121
− = . The elevator does
not appear to be safe because there is a reasonable chance (0.1121) that it will be overloaded with 16 male
passengers.
12. 174
x
μ μ
= = , / 40.8 / 16 10.2
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 195.3, Mean = 174, Standard deviation =
10.2, Cumulative = TRUE. The function returns a probability of 0.9816; 1 0.9816 0.0184
− = . The elevator
appears to be relatively safe because there is a very small chance (0.0184) of overloading. Using an outdated
mean that is too low has the effect of making the elevator appear to be much safer than it really is.

Select the NORM.DIST function again. Enter the following values: x = 25 Mean = 22.65, Standard
deviation = 0.8, Cumulative = TRUE. The function returns a probability of 0.9983. 0.9983 0.0196 0.9787
− =
the following values: Probability = 0.025, Mean = 22.65, Standard deviation = 0.8. The function returns a
head size of 21.08 inches.
Standard deviation = 0.8. The function returns a head size of 24.22 inches. 21.08 to 24.22 inches.
c. 22.65
x
μ μ
= = , / 0.8 / 64 0.1
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 22, Mean = 22.65, Standard deviation =
0.1, Cumulative = TRUE. The function returns a probability of 4.016E–11 or 0.00000.
deviation = 0.1, Cumulative = TRUE. The function returns a probability of 0.999767;
0.999767 .00000 0.9998
− = . No, the hats must fit individual women, not the mean from 64 women. If all hats
are made to fit head circumferences between 22.00 in. and 23.00 in., the hats won’t fit about half of those
women.
the following values: x = 22 Mean = 18.2, Standard deviation = 1, Cumulative = TRUE. The function
returns a probability of 0.999928. 99.99% of men will fit.
b. 18.2
x
μ μ
= = , / 1.0 / 36 0.1667
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 18.5, Mean = 18.2, Standard deviation
= 0.1667, Cumulative = TRUE. The function returns a probability of 0.9640. No, when considering the
diameters of manholes, we should use a design based on individual men, not samples of 36 men.
15. a. The mean weight of passengers is
3500
140
25
= lb.
b. 182.9
x
μ μ
= = , / 40.8 / 25 8.16
x n
σ σ
= = =
= 8.16, Cumulative = TRUE. The function returns a probability of 7.30718E–08 or 0.000000073.
1 0.000000073 1.0000
− = rounded to 4 decimal places.
c. / 40.8 / 20 9.1232
x n
σ σ
= = =
− =
d. Given that there is a 0.8067 probability of exceeding the 3500 lb. limit when the water taxi is loaded with
20 random men, the new capacity of 20 passengers does not appear to be safe enough because the
probability of overloading is too high.
16. a. Select the NORM.DIST function. Enter the following values: x = 0.8535, Mean = 0.8565, Standard
deviation 0.0518, Cumulative = TRUE. The function returns a probability of 0.4769. 1 0.4769 0.5231
− =
b. 0.8565
x
μ μ
= = , / 0.0518 / 465 0.0024
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 0.8535, Mean = 0.8565, Standard
deviation = 0.0024, Cumulative = TRUE. The function returns a probability of 0.1056. 1 0.1056 0.8944
− =

c. Instead of filling each bag with exactly 465 M&Ms, the company probably fills the bags so that the weight
is as stated. In any event, the company appears to be doing a good job of filling the bags.
17. a. Select the NORM.DIST function. Enter the following values: x = 167, Mean = 182.9, Standard deviation
40.8, Cumulative = TRUE. The function returns a probability of 0.3484. 1 0.3484 0.6516
− =
b. / 40.8 / 12 11.7779
x n
σ σ
= = =
11.7779, Cumulative = TRUE. The function returns a probability of 0.0885. 1 0.0885 0.9115
− =
c. There is a high probability that the gondola will be overloaded if it is occupied by 12 more people, so it
appears that the number of allowed passengers should be reduced.
18. a. Select the NORM.INV function. Enter the following values: Probability = 0.01, Mean = 77.5, Standard
deviation = 11.6. The function returns 50.5; 1 50.5
P = beats per minute. Select the NORM.INV function
again. Enter the following values: Probability = 0.99, Mean = 77.5, Standard deviation = 11.6. The
function returns 104.5; 99 104.5
P = beats per minute.
b. 77.5
x
μ μ
= = , / 11.6 / 25 2.32
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 70, Mean = 77.5, Standard deviation =
2.32, Cumulative = TRUE. The function returns a probability of 0.0006. Select the NORM.DIST function
again. Enter the following values: x = 85, Mean = 77.5, Standard deviation = 2.32, Cumulative = TRUE.
The function returns a probability of 0.9994. 0.9994 0.0006 0.9988
− =
c. Instead of the mean pulse rate from the patients in a day , the cutoff values should be based on individual
patients, so it would be better to use the pulse rates of 50.5 beats per minute and 104.5 beats per minute.
19. a. Select the NORM.DIST function. Enter the following values: x = 140, Mean = 165, Standard deviation =
45.6, Cumulative = TRUE. The function returns a probability of 0.2918. Select the NORM.DIST function
again. Enter the following values: x = 211, Mean = 165, Standard deviation = 45.6, Cumulative = TRUE.
The function returns a probability of 0.8435. 0.8435 0.2918 0.5517
− =
b. 165.0
x
μ μ
= = , / 45.6 / 36 7.6
x n
σ σ
= = =
Select the NORM.DIST function. Click OK. Enter the following values: x = 140, Mean = 165, Standard
deviation = 7.6, Cumulative = TRUE. Click OK. The function returns a probability of 0.0005. Select the
NORM.DIST function again. Click OK. Enter the following values: x = 211, Mean = 165, Standard
deviation = 7.6, Cumulative = TRUE. Click OK. The function returns a probability of 0.9999.
0.9990 0.0005 0.9994
− =
c. Part (a) because the ejection seats will be occupied by individual women, not groups of women.
20. a. 182.9
x
μ μ
= = , / 40.8 / 50 5.770
x n
σ σ
= = =
= 5.77, Cumulative = TRUE. The function returns a probability of 5.22806E–14, or 0+.
1 0.0000 1.0000
− = when rounded to 4 decimal places.
b. 182.9
x
μ μ
= = , / 40.8 / 14 10.904
x n
σ σ
= = =
= 10.904, Cumulative = TRUE. The function returns a probability of 0.2072. The probability that the boat
is overloaded because their mean weight is greater than 174 lbs. is 1 0.2072 0.7928
− = . Because there is a high
probability (0.7928) of overloading, the new ratings do not appear to be safe when the boat is loaded with
14 male passengers.
21. a. Select the NORM.DIST function. Enter the following values: x = 69.5, Mean = 72, Standard deviation =
2.4, Cumulative = TRUE. The function returns a probability of 0.1488. The probability the male
passenger can fit through the doorway without bending is 1 0.1488 0.8512
− = .

b. 69.5
x
μ μ
= = , / 2.4 / 100 0.24
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 69.5, Mean = 72, Standard deviation =
0.24, Cumulative = TRUE. The function returns a probability of 1.0406E–25 or 0+. The probability that
the mean height of the 100 men is less than 72 in. is 1 0.0000 1.0000
− = rounded to four decimal places.
c. The probability of part (a) is more relevant because it shows that 85.08% of male passengers will not need
to bend. The result from part (b) gives us useful information about the comfort and safety of individual
male passengers.
d. Because men are generally taller than women, a design that accommodates a suitable proportion of men
will necessarily accommodate a greater proportion of women.
22. 182.9
x
μ μ
= = , / 40.8 / 37 6.7075
x n
σ σ
= = =
Select the NORM.DIST function. Enter the following values: x = 167.6, Mean = 182.9, Standard deviation =
6.7075, Cumulative = TRUE. The function returns a probability of 0.0113. The probability that the mean
weight is greater than 167.6 lb. is 1 0.0113 0.9887
− = . Because the probability is so high, the pilot should take
action, such as removing excess fuel and/or requiring that some passengers disembark and take a later flight.
23. a. Yes. The sampling is without replacement and the sample size of 50 is greater than 5% of the finite
population size of 275. 16 275 50
2.0504584
275 1
50
x
σ
−
= =
−
.
b. Select the NORM.DIST function. Enter the following values: x = 95, Mean = 95.5, Standard deviation =
2.0504584, Cumulative = TRUE. The function returns a probability of 0.4037. Select the NORM.DIST
function again. Enter the following values: x = 105, Mean = 95.5, Standard deviation = 2.0504584,
Cumulative = TRUE. The function returns a probability of 1.0000 rounded to four decimal places.
1.0000 0.4037 0.5963
− =
24. a. Yes. The sampling is without replacement (because each sample of 16 elevator passengers consists of 16
different people) and the sample size of 16 is greater than 5% of the finite population size of 300.
40 300 16
9.7459365
300 1
16
x
σ
−
= =
−
b.
3000
187.5
16
= lb.
c. Select the NORM.DIST function. Enter the following values: x = 187.5, Mean = 177, Standard deviation
= 9.7459365, Cumulative = TRUE. The function returns a probability of 0.8593. The probability that the
total load exceeds the safe limit is 1 0.8593 0.1407
− = . The probability is not as low as it should be, since it
would be overloaded 14% of the time.
d. The z score for 0.999 is 3.1 which means that we need to solve for n in the equation
40 16
3.1
16 1
n
n
−
=
−
which has a solution of 14
n= passengers.
25. a.
4 5 9
6
3
μ
+ +
= = ,
2 2 2
(4 6) (5 6) (9 6)
2.160246899
3
σ
− + − + −
= =
b. The possible samples of size 2 are:{ (4, 5), (4, 9), (5, 4), (5, 9), (9, 4), (9, 5)} which have the following
means {4.5, 6.5, 4.5, 7, 6.5, 7} respectively.

c.
4.5 6.5 4.5 7 6.5 7
6
6
x
μ
+ + + + +
= = and
2 2 2 2 2 2
(4.5 6) (6.5 6) (4.5 6) (7 6) (6.5 6) (7 6)
1.08012345
6
x
σ
− + − + − + − + − + −
= =
d. It is clear that 6
x
μ μ
= = . 2.160246899 3 2
1.08012345
3 1
2
x
σ σ
−
= = =
−
1. The histogram should be approximately bell-shaped, and the normal quantile plot should have points that
approximate a straight line pattern.
2. Either the points are not reasonably close to a straight line pattern, or there is some systematic pattern that is not
a straight line pattern.
3. We must verify that the sample is from a population having a normal distribution. We can check for normality
using a histogram, identifying the number of outliers, and constructing a normal quantile plot.
4. Because the histogram is roughly bell-shaped, conclude that the data are from a population having a normal
distribution.
5. Not normal. The points show a systematic pattern that is not a straight line pattern.
6. Normal. The points are reasonably close to a straight line pattern, and there is no other pattern that is not a
straight line pattern.
7. Normal. The points are reasonably close to a straight line pattern, and there is no other pattern that is not a
straight line pattern.
8. Not normal. The points are not reasonably close to a straight line pattern, and there appears to be a pattern that
is not a straight line pattern.
9. Open the FLIGHTS data file on your data disk. Select the
XLSTAT add-in. Select Visualizing data. Select
Histograms. Enter the Arr Delay data range inclusive of the
label. Select Discrete. Select Sample labels. Click the
Charts tab. Select Histograms and Bars. Select Frequency
for ordinate of the histogram. Click OK. Not normal.
10. Open the POTUS data file on your data disk. Select the XLSTAT add-in. Select Visualizing data. Select
Histograms. Enter the Ht data range inclusive of the label. Select Discrete. Select Sample labels. Click the
Charts tab. Select Histograms and Bars. Select Frequency for ordinate of the histogram. Click OK.
Normal.

11. Open the MBODY data file on your data disk. Select the
Histograms. Enter the SYS data range inclusive of the label.
Select Discrete. Select Sample labels. Click the Charts tab.
Select Histograms and Bars. Select Frequency for ordinate
of the histogram. Click OK. Normal.
12. Open the COTININE data file on your data disk. Select the
Histograms. Enter the NOETS data range inclusive of the
label. Select Discrete. Select Sample labels. Click the
Charts tab. Select Histograms and Bars. Select Frequency
for ordinate of the histogram. Click OK. Not normal.
13. Open the FLIGHTS data file on your data disk. Select the XLSTAT add-in. Select Describing data. Select
Normality tests. Enter the Arr Delay data range inclusive of the label. Select Sample labels. Click the
Charts tab. Select Normal Q-Q plots. Click OK. Not normal.
Note that XLSTAT uses normalized scores rather than z-scores in the construction of a quantile plot.

14. Open the POTUS data file on your data disk. Select the
XLSTAT add-in. Select Describing data. Select Normality
tests. Enter the Ht data range inclusive of the label. Select
Sample labels. Click the Charts tab. Select Normal Q-Q
plots. Click OK. Normal.
Note that XLSTAT uses normalized scores rather than z-scores
in the construction of a quantile plot.
15. Open the MBODY data file on your data disk. Select the
XLSTAT add-in. Select Describing data. Select Normality
tests. Enter the SYS data range inclusive of the label. Select
Sample labels. Click the Charts tab. Select Normal Q-Q
plots. Click OK. Normal.
Note that XLSTAT uses normalized scores rather than z-scores
in the construction of a quantile plot.
16. Open the COTININE data file on your data disk. Select the XLSTAT add-in. Select Describing data. Select
Normality tests. Enter the NOETS data range inclusive of the label. Select Sample labels. Click the Charts
tab. Select Normal Q-Q plots. Click OK. Not normal.

17. Enter the label Feet in cell A1 of an Excel worksheet followed by the five data values. Sort the data values
from lowest to highest. With a sample size n = 5, each value represents a proportion of 1/5 of the sample.
Next, you identify the cumulative areas to the left of the corresponding sample values. The formulas are:
1 3 5 7 9
, , , ,
2 2 2 2 2
n n n n n
. The specific areas, then, are:
1 3 5 7 9
, , , ,
10 10 10 10 10
. Expressed as decimals, the specific
areas are: 0.1, 0.3, 0.5, 0.7, and 0.9. Enter the label z score in cell B1. You will now use Excel’s
NORM.S.INV function to find the corresponding z scores. Click in cell B2 where the z score corresponding to
an area of 0.1 will be placed. Click Formulas at the top of the screen. Select Insert Function. Select the
Statistical category. Select the NORM.S.INV function. Click OK. Enter the following value: Probability =
0.1. Click OK. The function returns z = –1.28. Continue in the same way to find the remaining z scores. The
feet and z-scores are displayed below in the table on the left.
The feet and z score pairs are (x,y)
coordinates that you will plot in a
scatter plot. Select the XLSTAT add-
in. Select Visualizing data. Select
Scatter plots. Feet is the x variable.
Enter the data range of the Feet
variable inclusive of the label. z score
is the y variable. Enter the data range
of the z score variable inclusive of the label. Select Variable
labels. Click OK. Right-click directly on one of the plotted
points in the graph. Select Add trendline from the menu that
appears. Select Linear from trendline options. The normal
quantile plot is displayed on the right. The data appear to be
from a population with a normal distribution.
18. Enter the label Minutes in cell A1 of an Excel worksheet followed by the six data values. Sort the data values
1 3 5 7 9 11
, , , , ,
2 2 2 2 2 2
n n n n n n
1 3 5 7 9 11
, , , , ,
12 12 12 12 12 12
.
Expressed as decimals, the specific areas are: 0.0833, 0.25, 0.4167, 0.5833, 0.75, and
0.9167 Enter the label z score in cell B1. You will now use Excel’s NORM.S.INV
function to find the corresponding z scores. Click in cell B2 where the z score
corresponding to an area of 0.0833 will be placed. Click Formulas at the top of the
screen. Select Insert Function. Select the Statistical category. Select the
NORM.S.INV function. Enter the following value: Probability = 0.0833. The
function returns z = –1.38. Continue in the same way to find the remaining z scores.

The minutes and z score pairs are (x,y) coordinates that you
will plot in a scatter plot. Select the XLSTAT add-in. Select
Visualizing data. Select Scatter plots. Minutes is the x
variable. Enter the data range of the Minutes variable
inclusive of the label. z score is the y variable. Enter the data
range of the z score variable inclusive of the label. Select
Variable labels. Click OK. Right-click directly on one of the
plotted points in the graph. Select Add trendline from the
menu that appears. Select Linear from trendline options. The
normal quantile plot is displayed below on the right. The data
appear to be from a population with a distribution that is not
normal.
19. Enter the label cm^3 in cell A1 of an Excel worksheet followed by the eight data values. Sort the data values
1 3 5 7 9 11 13 15
, , , , , , ,
2 2 2 2 2 2 2 2
n n n n n n n n
1 3 5 7 9 11 13 15
, , , , , , ,
16 16 16 16 16 16 16 16
. Expressed as
decimals, the specific areas are: 0.0625, 0.1875, 0.3125, 0.4375, 0.5625, 0.6875, 0.8125, and 0.9375 Enter the
label z score in cell B1. You will now use Excel’s NORM.S.INV function to find the corresponding z scores.
Click in cell B2 where the z score corresponding to an area of 0.0625 will be placed. Click Formulas at the top
of the screen. Select Insert Function. Select the Statistical category. Select the NORM.S.INV function.
Enter the following value: Probability = 0.0625. The function returns z = –1.53. Continue in the same way to
find the remaining z scores. The cm3
and z-scores are displayed below in the table on the left.
The cm3
and z score pairs are (x,y)
Scatter plots. Cm3
is the x variable.
Enter the data range of the cm3
variable inclusive of the label. z score
is the y variable. Enter the data range
of the z score variable inclusive of the
label. Select Variable labels. Click
OK. Right-click directly on one of
the plotted points in the graph. Select Add trendline from the
menu that appears. Select Linear from trendline options. The
normal quantile plot is displayed on the right. The data appear
to be from a population with a distribution that is not normal.
20. Enter the label g in cell A1 of an Excel worksheet followed by the nine data values. Sort the data values from
lowest to highest. With a sample size n = 9, each value represents a proportion of 1/9 of the sample. Next, you
identify the cumulative areas to the left of the corresponding sample values. The formulas are:
1 3 5 7 9 11 13 15 17
, , , , , , , ,
2 2 2 2 2 2 2 2 2
n n n n n n n n n
1 3 5 7 9 11 13 15 17
, , , , , , , ,
18 18 18 18 18 18 18 18 18
.
Expressed as decimals, the specific areas are: 0.0556, 0.1667, 0.2778, 0.3889, 0.5, 0.6111, 0.7222, 0.8333, and
0.9444. Enter the label z score in cell B1. You will now use Excel’s NORM.S.INV function to find the
corresponding z scores. Click in cell B2 where the z score corresponding to an area of 0.0556 will be placed.
Click Formulas at the top of the screen. Select Insert Function. Select the Statistical category. Select the
NORM.S.INV function. Click OK. Enter the following value: Probability = 0.0556. Click OK. The function
returns z = –1.59. Continue in the same way to find the remaining z scores. The g and z-scores are displayed
in the table on the left.

The g and z score pairs are (x,y)
Scatter plots. g is the x variable.
Enter the data range of the g variable
inclusive of the label. z score is the y
variable. Enter the data range of the z
score variable inclusive of the label.
Select Variable labels. Click OK.
Right-click directly on one of the
plotted points in the graph. Select
Add trendline from the menu that appears. Select Linear
from trendline options. The normal quantile plot is displayed
on the right. The data appear to be from a population with a
normal distribution.
21. a. Yes
b. Yes
c. No
22. a. Open the QUAKE data file on your data disk. Select the XLSTAT add-in. Select Describing data.
Select Normality tests. Enter the MAG data range inclusive of the label. Select Sample labels. Click the
Charts tab. Select Normal Q-Q plots. Click OK. Normal.
b. The original measurements have a lognormal distribution
c. We can reverse the process of taking values be an exponent of 10. The normal quantile plot indicates that
the original values are not from a population with a normal distribution.
23. Enter the label Thousands of dollars in cell A1 of an Excel worksheet followed by the data given in Exercise
23. Select the XLSTAT add-in. Select Describing data. Select Normality tests. Enter the Thousands of
dollars data range inclusive of the label. Select Sample labels. Click the Charts tab. Select Normal Q-Q
plots. Click OK. The original values are not from a normally distributed population.

Enter the label Logarithm in cell B1 of the worksheet. Then
use Excel’s LOG10 function to take the logarithm of each
value. Use XLSTAT to construct a normality Q-Q plot of the
logarithms.
After taking the logarithm of each value, the values appear to
be from a normally distributed population. The original
values are from a population with a lognormal distribution.
Note that XLSTAT uses normalized scores rather than z-
scores in the construction of a quantile plot.
1. The Minitab display shows that the region representing 235 wins is a rectangle. The result of 0.0068 is an
approximation, but the result of 0.0066 is better because it is based on an exact calculation. The approximation
differs from the exact result by a very small amount.
2. The continuity correction is used to compensate for the fact that a continuous distribution (normal) is used to
approximate a discrete distribution (binomial). The discrete number of 13 is represented by the interval from
12.5 to 13.5.
3.
1
0.2
5
p = = ,
4
0.8
5
q = = , 25 0.2 5
μ = ⋅ = , 25 0.2 0.8 2
σ = ⋅ ⋅ = . The value of 5 for the mean shows that
for many people who make random guesses for the 25 questions, the mean number of correct answers is 5. For
many people who make random guesses, the standard deviation of 2 is a measure of how much the numbers of
correct responses vary.
4. Yes. The circumstances correspond to 25 independent trials of a binomial experiment in which the probability
of success is 0.2. Also, with 25
n= , 0.2
p = , 0.8
q = , the requirements of 5
np ≥ and 5
nq ≥ are both
satisfied.
5. The requirements are satisfied with a mean of 13 0.4 5.2
⋅ = and the standard deviation of
13 0.4 0.6 1.7663.
⋅ ⋅ = Select Excel’s NORM.DIST function. Enter the following values: x = 2.5, Mean =
5.2, Standard deviation = 1.7663, Cumulative = TRUE. The function returns a probability of 0.0632.
6. The requirement of 5
nq ≥ is not satisfied. Normal approximation should not be used.

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Title: Le baptême de Pauline Ardel
roman
Author: Emile Baumann
Release date: August 23, 2023 [eBook #71473]
Language: French
Original publication: Paris: Bernard Grasset, 1913
Credits: Laurent Vogel and the Online Distributed Proofreading Team
at https://www.pgdp.net (This file was produced from
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Archive/Canadian Libraries)
*** START OF THE PROJECT GUTENBERG EBOOK LE BAPTÊME
DE PAULINE ARDEL ***

ÉMILE BAUMANN
LE BAPTÊME
DE
PAULINE ARDEL
— ROMAN —
PARIS
BERNARD GRASSET, ÉDITEUR
61, RUE DES SAINTS-PÈRES, 61
1913
Tous droits de reproduction, de traduction et d’adaptation réservés pour tous pays.
Copyright by Bernard Grasset, 1913.

DU MÊME AUTEUR
Les Grandes Formes de la musique : l’œuvre de Camille
Saint-Saëns (Ollendorff, éditeur).
L’Immolé, roman, ouvrage couronné par l’Académie française
(Bernard Grasset, éditeur).
La Fosse aux lions, roman (Bernard Grasset, éditeur).
Trois villes saintes : Ars-en-Dombes, Saint-Jacques de
Compostelle, le Mont Saint-Michel (Bernard Grasset,
éditeur).
Pour paraître :
Le Fer sur l’enclume, roman.

Il a été tiré de cet ouvrage :
cinq exemplaires sur Japon
des Manufactures Impériales de Tokio
numérotés de 1 à 5
et vingt exemplaires sur Hollande Van Gelder
numérotés de 6 à 25.
Une édition hors commerce, réimposée in-8o raisin, a été tirée pour la société
« Les XX ».

A
SŒUR MARIE DE LA CROIX
du Carmel de V…
ce roman qu’elle ne lira jamais.
E. B.

LE
BAPTÊME DE PAULINE ARDEL
I
La cathédrale avait l’air triste sous la brume. Ses deux tours
austères fixaient l’Occident où le soleil de décembre se coucherait
sans avoir lui. Soumises depuis sept cents ans aux hivers enfumés
et aux nuées pleurantes, elles se résignaient, jusqu’à ce que, pour
leur délivrance, le clairon de l’archange mît debout le Christ de gloire
assis entre elles au-dessus du porche et du vitrail. En bas, sur le
parvis, bien que ce fût un dimanche et que l’heure des vêpres
approchât, les passants étaient rares, et ils traversaient vivement
comme des provinciaux casaniers qu’attend une maison chaude.
La place se trouvait déserte, lorsque M. Victorien Ardel,
accompagné de sa fille Pauline, déboucha de la rue des Quatre-
Vents. Tous deux s’avancèrent du côté de la halle, puis, s’arrêtant,
se retournèrent vers la façade de l’église, à la façon d’étrangers qui,
pour la première fois, l’examinaient.
M. Ardel, agrégé d’histoire, venait d’être, un mois auparavant,
nommé professeur à Sens. Il n’avait pas encore pris le loisir
d’étudier la cathédrale ; ce monument le touchait peu ; car, étant un

esprit fort, il s’évitait ainsi qu’à sa fille la rencontre d’images
mystiques qu’il éliminait de leur commune vie. Néanmoins, par une
curiosité d’historien et d’esthète, il s’était décidé, ce dimanche, à ne
plus différer une visite au reste inoffensive pour son indifférence
éprouvée.
Il considéra donc, d’un œil critique, d’abord la tour des cloches,
depuis l’étroite et courte ogive de la porte jusqu’au campanile
octogonal que la Renaissance a vissé tout au sommet.
Pauline la regardait en même temps ; mais quelque chose, dans
cette masse hautaine, lui déplaisait : était-ce le relief rude et
perpendiculaire des contreforts, la noirceur des abat-son, l’orgueil
des pinacles qui surplombent solitairement l’autre tour décoiffée et
tronquée ? Le visage de ces pierres la rebutait comme celui d’un
justicier rébarbatif.
Son attention, une seconde, s’accrocha aux cinq statues
blanches logées à mi-hauteur sous des niches pointues ; mais ces
évêques, avec leur crosse, ne lui exprimaient rien. Ses yeux
s’infléchirent à gauche vers le Saint Étienne du porche, en robe de
clerc, mince et long comme une colonnette, doux et méditatif,
présentant entre ses mains le Livre mystérieux. Puis elle se
détourna, le nez au vague, et, d’un air de discrète impatience, fit
deux ou trois pas en avant.
— Il n’y a pas à dire, remarqua, la canne levée vers la façade,
son père qui la rejoignit, le moyen âge eut la tradition de la force !
M. Ardel n’articula point le mot : force ! sans une certaine
emphase. Il laissait voir en sa démarche ce je ne sais quoi
d’autoritaire et de gourmé où se ressemblent un pédagogue et un
magistrat. Sa façon de balancer les bras et de porter sa tête
accusaient le contentement acquis d’une supériorité. C’était
d’ailleurs un homme d’une figure encore belle, quoique fatiguée par
d’excessifs travaux. Si des bajoues alourdissaient le contour de son
menton, sa bouche restait fine et mordante sous une moustache
drue ; son nez aurait pu servir de modèle à un sculpteur romain ;

l’arc étrangement noir des sourcils se dessinait sur des yeux d’une
mobilité sombre dont on avait peine à soutenir le choc.
Dans les traits de sa fille, comme dans les siens, une rectitude
latine était inscrite : l’ovale des joues de Pauline se détachait
noblement d’un cou ferme et délicat ; à l’œil bien fendu répondaient
une bouche et des oreilles un peu grandes, mais régulières ; une
moue d’orgueil renflait sa lèvre inférieure, mais son regard s’en allait
imbibé de tendresse. Une voilette noire et la froidure excitaient sous
sa peau l’éclat d’un sang radieux. Sans dépasser de beaucoup son
père qui était de stature moyenne, elle semblait d’une venue
vigoureuse en ses dix-huit ans ; son port et ses mouvements
offraient une harmonie naturelle plutôt grave que vive. Le sérieux de
sa mise exprimait, soit l’insouciance de paraître, soit la discipline
d’économie qu’elle tenait de sa mère, morte il y avait six ans.
Ils se rapprochèrent des portails et, devant les saints sculptés
contre le soubassement ou le long des voussures, tous sans tête, et
qui vivent, gesticulent pourtant, comme des martyrs impossibles à
tuer :
— Les pauvres gens ! dit Pauline. Quels misérables se sont
amusés à ce jeu de massacre ?
M. Ardel se dispensa de lui répondre ; du haut de la tour le
premier coup des vêpres tinta ; ils l’écoutèrent. La cloche émettait le
son d’un glas ; chacun de ses battements descendait à larges
intervalles et les vibrations s’amplifiaient sur la ville engourdie,
pareilles aux cercles ondulatoires que forme une goutte d’eau
tombant du plafond d’une grotte dans un lac ténébreux.
Pauline, à cet appel, n’éprouva qu’une oppression confuse.
— Entrons-nous ? demanda-t-elle, comme si elle avait eu le désir
de ne pas entrer.
— Mais oui, répondit froidement son père.
Et il pénétra la premier à l’intérieur de la cathédrale dont le vide
étonna Pauline et la mit à l’aise.

M. Ardel s’arrêta au bas des nefs, près de l’un des maîtres
piliers, formidable en son épaisseur, et cependant allégé par l’élan
des sveltes colonnes comme un buffet d’orgue par ses tuyaux. Il fut
saisi d’admiration ; d’autant plus sensible à la vigueur naïve de cet
art qu’il était saturé de culture livresque. Mais, tout de suite, en face
de la grande nef, reparurent ses habitudes de sèche analyse :
— Voûte sexpartite, observa-t-il, un peu basse. Alternance de
grosses colonnes géminées et de piliers…
Il continua, prenant le bras de Pauline :
— Tu vois ces deux arcs plus aigus que les autres ; le pan avait
dû s’effondrer, on les a refaits au quinzième siècle. Et, dans les bas-
côtés, ces alvéoles romanes, elles datent de Viollet-Leduc. On
s’imagine visiter un édifice du moyen âge, et c’est du Louis-Philippe
que nous touchons. De l’ancienne église il reste à peine la carcasse.
Si Pauline avait eu des velléités d’enthousiasme, des réflexions
pareilles visaient à les annihiler. Nourrie dans le dédain de toutes les
religions, elle croyait pouvoir explorer une cathédrale avec le même
détachement qu’une pagode hindoue. Ses aïeux, sa mère et son
père en leur enfance n’en avaient pas moins adoré dans des
temples semblables à celui-là. Des persistances obscures, un sourd
émoi, à son insu, la troublaient. Mais elle n’avait passé le seuil des
églises qu’à l’occasion de mariages et surtout d’enterrements. De
sinistres idées mortuaires se liaient pour elle à l’office chrétien.
Le crépuscule où ce jour d’hiver déclinant enfonçait le vaste
vaisseau aggravait cette impression. L’ombre filait ses toiles
d’araignée contre les murailles ; des tentures noires se dépliaient
entre les vitraux. Dans le chœur, un sacristain voûté, à demi perclus,
allumait près de chaque stalle une bougie. Sur les dalles où il traînait
ses pantoufles des lueurs froides coulaient. Deux vieilles femmes,
ramassées sous leurs manteaux, entrèrent et se signèrent ; la porte
du tambour claqua lourdement sur leur dos.
Pauline se demanda si on n’allait pas amener un mort. Une
anxiété la prit de se trouver dans le lieu saint ; elle le jugeait
sournoisement hostile. Pour un peu elle eût dit à son père : Allons-

nous-en. Mais l’ennui d’expliquer son inquiétude la dissuada d’y
céder ; elle se raisonna : qu’avait-elle à craindre de ces autels
muets ? Non vraiment, était-elle assez puérile de subir une émotion
superstitieuse, comme s’il y avait eu là quelqu’un !
Elle suivit d’un pas délibéré le professeur auprès du retable de
Salazar, vis-à-vis la chaire ; ils y firent une courte halte. Sous le dais
en fuseau d’une niche amenuisée, ouvragée à la façon d’une
broderie, entre deux Saints une Vierge au visage finement rustique
et grave soutient sur son bras l’Enfant qui lève le doigt.
— Qu’ils sont vrais, murmura Pauline, cette femme et cet enfant !
Elle ne songeait qu’à la vérité des figures et des attitudes ; mais
le sens de sa parole dépassait ce qu’elle avait cru dire.
Plus loin, M. Ardel s’intéressa aux cintres et aux chapiteaux
d’une primitive chapelle voûtée en cul-de-four. Pauline voulut savoir
ce qu’on faisait d’un bénitier oblong drapé d’un voile, qu’elle toucha
d’une main curieuse.
— Les fonts baptismaux, répondit-il négligemment.
Ils s’étaient engagés derrière l’orgue du chœur, dans le profond
déambulatoire, et ils longeaient une suite de vitraux anciens dont
Pauline, plus que son père, fut émerveillée. Elle se souciait peu
d’abord des scènes qu’ils racontaient, n’y voyant qu’une imagerie
d’Épinal éblouissante et enfantine. Mais les médaillons sertis dans
des armatures noires sollicitaient son âme par un mystère semblable
à l’intimité d’une musique pleine de nostalgies. Le bleu qui les
trempe, céruléen, presque violet, lui offrait un crépuscule tel que
jamais, dans les plus beaux soirs, elle n’en avait contemplé. Sur ce
fond, l’émail vert d’une robe, la tête d’un palefroi, caparaçonnée d’or,
le profil d’un moine brûlaient d’une flamme inextinguible ; une sorte
de chaleur joyeuse en descendait, ils semblaient s’aviver de toutes
les ténèbres qui s’épaississaient alentour.
L’un de ces vitraux, découpé en losanges et en arcs de cercle,
était si net de dessin qu’à le fixer une minute elle apprit, sans le
vouloir, la légende de saint Eustache. Trois ou quatre épisodes du
moins, au premier coup d’œil, s’élucidèrent. Dans une clairière bleue

comme les songes elle voyait entre les cornes d’un cerf brun une
croix de feu ; un chasseur s’agenouillait devant elle, tandis que son
cheval argenté, paisible, pâturait.
Plus haut, le même personnage reparaissait, amaigri, à genoux
dans une cuve baptismale, et un évêque infondait de l’eau sur son
front cerné d’un nimbe rouge. Ailleurs, elle le retrouvait
s’embarquant sur une mer ensoleillée…
Pauline l’abandonna en chemin ; mais elle pensa aux félicités
naïves des hommes qui avaient assez cru à de telles fables pour les
peindre avec tant de ferveur et de patience.
En continuant le tour de l’abside, M. Ardel s’attarda derrière le
fastueux baldaquin du maître-autel soutenu par quatre colonnes de
marbre opulentes, jadis taillées pour figurer sur la place des
Victoires, autour de la statue du grand Roi. Pauline l’avait devancé
jusqu’au bas de la fenêtre grillée d’où les archevêques, sans sortir
de leur palais, assistaient aux offices. Là, pend à la muraille nue un
Christ en bois, d’un jaune bruni, coiffé de sa couronne lamentable.
Des cheveux confus se collent le long de ses joues et sur sa
poitrine ; chacune de ses côtes paraît dire : Comptez-moi ; ses bras
décharnés sont raidis ; les rotules de ses genoux et les os de ses
jambes incurvés comme des baguettes distendent sa peau. Tout ce
que peut souffrir la chair de l’homme s’est abrégé dans ce cadavre
et dans sa tête encline, indiciblement meurtrie. Pauline fut affectée
d’une pitié vague, mais plus encore d’une répulsion :
— Est-ce possible, se dit-elle, que d’un affreux supplicié on ait fait
un dieu !
L’horloge de la tour sonna trois heures moins un quart avec la
lenteur dolente des vieilles horloges qui ne semblent plus croire au
temps ; le second coup des vêpres se prolongea. La cathédrale
commençait à s’animer : deux chanoines enveloppés d’amples
manteaux, l’un, obèse et court, l’autre, sec, long et pâle, enfilèrent le
couloir sombre de la sacristie au même instant qu’en sortait un petit
abbé rond dans son surplis, rubicond, vif et trotte-menu, montrant
sur sa mine la jovialité spirituelle d’un bourguignon content de vivre.

M. Ardel avait rejoint Pauline devant un escalier dont il loua les
gracieuses arcades ; il aborda le vicaire au passage pour s’enquérir
si le Trésor était visible.
— Pas maintenant, monsieur ; après les vêpres, répondit l’abbé,
s’arrêtant à peine ; et, preste comme un moineau qui s’envole, il
s’élança vers le chœur.
Le professeur fronça les sourcils et grommela :
— Sont-ils malotrus, ces curés !
Pauline et lui gagnèrent le milieu du transept ; pendant qu’il
s’assimilait d’un regard synthétique l’harmonie de la cathédrale, la
structure de l’ensemble, robuste et froide, sa fille admirait, au-dessus
du portail d’Abraham, la rosace du Paradis enfermant dans les
torsions ardentes de ses nervures un azur vierge où des anges qui
tiennent des violes éploient leurs ailes, d’une blancheur translucide ;
au centre, dans l’épais brasier d’un soleil couchant, s’enclôt une
Face triomphale ; était-ce le même Christ qu’elle venait de voir si
douloureux ? Elle aurait eu peine à concilier toute l’humiliation avec
toute la gloire ; elle ne l’essaya point ; car sa pensée ressemblait à
ces eaux des lacs qui ne savent rien du ciel dont elles absorbent la
splendeur. De telles images y déposaient pourtant l’idée incertaine
d’une vie supra-sensible que jusqu’alors elle n’avait pas conçue.
Les vêpres allaient commencer ; les chanoines et les clercs
étaient montés à leurs stalles. M. Ardel constata, non sans ironie, le
nombre dérisoire des fidèles : peu ou point d’hommes, des femmes
âgées, des petites filles, quelques religieuses à longue coiffe. Le
suisse, plein de majesté, se cambrait devant les chaises vides
comme s’il avait eu des foules à contenir.
Aussitôt que résonna le Deus in adjutorium, M. Ardel battit en
retraite ; le chant des psaumes l’eut ennuyé. Pauline et lui sortirent
par le portail de Moïse ; un aveugle fit tinter inutilement sa sébile où
dansaient des sous rares. Le professeur, à respirer hors de l’église,
sentit une légère satisfaction.
— Leurs cathédrales, énonça-t-il, ne sont que des nécropoles :
tout y est bien mort…

Il n’appuya pas sur sa remarque, trouvant superflu d’affirmer que
le catholicisme, au dedans de lui-même, ne rendait plus aucun son.
— J’aime cette cour, dit Pauline, qu’un instinct juvénile poussait à
contredire l’aridité de l’incroyance paternelle.
Elle indiquait, au bout des grilles pompeuses déployées à droite
et à gauche, sur une porte voûtée, un pavillon en briques rehaussé
de moulures délicates, avec des croisées étroites à meneaux, telles
qu’on en voit aux châteaux français de la Renaissance ; et, derrière
eux, la rose magnifique qui, même en l’absence du soleil, flamboyait.
La cour n’en concentrait pas moins une mélancolie de cimetière
abandonné. Sous des thuyas et des sureaux moisissaient, dans
l’herbe jaunie, des feuilles mortes. D’un côté, les ardoises de
l’archevêché, ses fenêtres toujours closes depuis que les
archevêques ont été chassés de leur demeure ; de l’autre, les tuiles
vernissées du palais synodal, le flanc de la tour des cloches et le toit
des nefs l’enfermaient sévèrement. Des corneilles, parmi les
gargouilles, s’envolaient de leurs ailes pesantes ; elles se jetaient un
cri aigre-doux, analogue à celui d’une girouette usée. De l’intérieur,
les ronflements de l’orgue et la psalmodie des prêtres ne
s’épandaient qu’atténués, lointains.
Pauline se plut quelques instants à les entendre ; cette musique
sourde la captivait comme l’illusoire écho d’un monde fini.
L’éducation rigide, hautaine, qu’elle avait reçue auprès d’un père
despote et studieux la prédisposait à comprendre la solitude d’un
lieu vénérable ; de ces édifices, où six siècles s’étaient continués,
émanait une paix accueillante. C’était une influence dont M. Ardel
non plus ne cherchait pas à se défendre, tant il croyait défuntes
toutes ces choses, et il se taisait, induit à un attendrissement qu’il ne
voulait point laisser voir. Mais, soudain, il secoua les épaules, frappa
le pavé de ses pieds impatients.
— On gèle ici, dit-il ; marchons.
Ils descendirent d’un pas allègre le long et noir boyau de la
grande rue ; et, arrivés au quai de l’Yonne, ils franchirent le vieux
pont trapu que dominait une croix de fer.

Devant l’église Saint-Maurice, un passant coiffé d’un gibus,
portant une rosette à la boutonnière de son pardessus, les honora
d’un salut cérémonieux. M. Ardel reconnut un de ses collègues, M.
Lemerle, lequel, depuis vingt-neuf ans, professait au lycée la
rhétorique. M. Lemerle, outre son invariable gibus, se signalait par
des lunettes bleues, une barbe grisâtre et courte, des façons de
pasteur protestant ; sur sa figure probe, mais rogue, s’était durci un
masque de sévérité qu’il semblait devoir conserver jusqu’à sa mort
et au delà.
— Tu as vu cet homme, exposa M. Ardel à sa fille ; il est un des
derniers survivants d’une race qui va s’éteindre en France, comme
s’est éteinte celle des bons domestiques. C’est le professeur-né, le
cuistre à lunettes ! Il ne peut concevoir une existence ayant d’autre
but que d’expliquer du Bossuet et de corriger des versions. Il fait sa
classe à la manière dont Dandin jugeait. Et quel fonctionnaire ! Il ne
vous parle que d’inspections, de dossiers, d’avancement. A propos
d’une copie où un élève avait risqué cette phrase : « Nous
regardions avec indifférence défiler le cortège officiel » : — Monsieur,
s’est écrié Lemerle en courroux, au passage d’un cortège officiel on
ne doit jamais être indifférent !
Pauline, à ce trait, fit un éclat de rire. Son père n’appartenait
certes pas à la race des Lemerle, et elle en était fière, bien qu’ayant
pâti elle-même de son humeur intraitable. Envoyé fort jeune à
Bordeaux, Victorien Ardel donnait toutes les promesses d’un sujet,
selon la formule, « très distingué ». Mais, un jour que son Recteur
visitait sa classe, sur une critique qui lui fut faite, il s’emporta, eut
avec lui une altercation. Cinq jours après on l’expédiait à Roanne où
il resta seize ans, oublié, disait-il, « comme au jeu de l’oie, dans le
puits ». C’était à Roanne, parmi la laideur des fabriques, et en un
foyer pauvre, qu’avait grandi Pauline. Des séjours à Lyon, chez son
grand-oncle Jérôme, lui révélèrent, avec une existence plus large, la
majesté d’une ville insigne et antique.
La souche des Ardel était lyonnaise depuis près d’un siècle. Le
bisaïeul, Fabricio Ardello, natif de Turin, avait fondé, en 1812, sur la
place Bellecour, une maison d’armurerie. Son fils aîné, Octave, le

père de Victorien, tint boutique jusqu’aux dernières années du
second Empire ; une sotte affaire où l’ensorcela un aigrefin, la
construction, aux Brotteaux, d’un Alcazar fastueux, culbuta son
patrimoine et sa maison. Des trois fils d’Octave, le premier, Adolphe,
lieutenant de chasseurs alpins, avait péri, précipité dans un trou par
une avalanche ; le troisième, Jacques, avait pris la soutane au grand
séminaire de Saint-Just. Quant à Victorien, brillant écolier, il s’était
décidé pour l’enseignement ; car il apercevait là une position
prompte et la certitude de loisirs copieux.
Dans sa tâche un seul attrait l’excita, la part de vie pensante qu’il
savait maintenir au-dessus des rabâchages quotidiens. Un livre sur
le Duc de Saint-Simon lui avait acquis un renom d’originalité. Au
rebours des historiens révolutionnaires, il y justifiait l’aristocrate en
lutte avec les gens de robe. Le contact d’êtres asservis n’avait pu
qu’irriter son besoin d’indépendance. Les hommes « francs du
collier », rares dans l’histoire, avaient toute son estime autant que
les autres son aversion. C’est pourquoi, à la vue de M. Lemerle, il
s’était soulagé par quelques sarcasmes. Du haut de son orgueil
sauvage il considérait ce collègue à peu près comme un épervier
des Alpes regarde un canard de basse-cour.
Pauline et lui se dirigeaient vers les coteaux dont l’Yonne
réfléchit, à l’ouest, les murs crayeux. Un sentier qui s’élève entre des
buissons les attira ; ils allaient sans causer, Pauline étant faite aux
habitudes silencieuses du professeur toujours absorbé dans ses
élucubrations. Elle marchait plus vite que lui, et s’animait à gravir
cette colline malingre pour la seule joie de monter, d’atteindre de
l’étendue.
Mais, devant elle, sur la croupe du tertre, une église rustique
terminait l’horizon ; la jeune fille avait l’air de s’y rendre en
pèlerinage. Si sa vue s’abaissait, elle découvrait, dans les champs,
près de la rivière, la chapelle et l’enclos d’une abbaye. Si elle se
tournait du côté de Sens, la cathédrale commandait la plaine ; sa
tour des cloches, accrue du campanile, semblait une formidable tour
de guetteur ; les toits bruns et les arbres nus se brouillaient au-
dessous d’elle dans des vapeurs pareilles à de la suie délayée ; la

ville n’existait qu’autour du donjon massif et par lui. Pauline distingua
cependant un autre clocher, Saint-Pierre-le-Rond, voisin de la rue
qu’elle habitait.
« Que d’églises dans ce pays ! songea-t-elle. On le croirait peuplé
de prêtres et de nonnes. »
Elle s’arrêta au bord du talus, attendant son père ; et elle suivait,
sur l’eau plate de l’Yonne, une péniche qui remontait doucement,
halée par des haridelles. Puis sa pensée se dissipa vers le nord,
parmi la confusion des brumes ; et, sans savoir pourquoi, elle se mit
à fredonner la chanson de Miarka : Nuages, nuages, que dites-
vous ? La voix de M. Ardel l’interrompit :
— Le site est médiocre, opinait-il avec un hochement de tête
dédaigneux. Ça ne vaut pas les environs de Roanne. Ces coteaux
parallèles, ces terres pâles, ces peupliers rangés comme des
soldats à la parade, c’est sans relief et sans énergie, du classique
régulier et morne. Je te montrerai, au Louvre, des paysages de Van
der Meulen, disposés pour y peindre les batailles de Louis XIV ; ils
furent peut-être inspirés de celui-ci.
— Décidément, répondit Pauline d’un ton câlin qui atténuait ses
paroles, tu ne seras jamais content, nulle part ; tu souhaitais la
proximité de Paris, tu l’as, et, déjà, tu voudrais être ailleurs. Si je
voyais les choses comme toi, il ne nous resterait qu’à nous pendre.
— Oh ! ma foi…
Le reproche inattendu de sa fille atteignait M. Ardel au vif de ses
tristesses latentes. C’était trop vrai : l’inquiétude, avec le non-espoir,
faisait son âme stérile ; et, devant tout spectacle, l’esprit critique en
lui tuait la jouissance.
— Ce que tu dis là, pauvre père, répliqua Pauline, tu ne le
penses pas ; autrement, ce serait à croire que je ne suis rien pour
toi. Et je me demande, continua-t-elle plus sérieuse, si je suis
beaucoup pour toi ; ton talent, tes travaux, voilà ce qui compte dans
ta vie ; mais ta fille…
— Tais-toi donc, fit-il en haussant les épaules, tu sais bien que tu
es mon tout !

Deux larmes mouillèrent ses prunelles qu’exalta soudain une
tendresse désespérée, et, attirant Pauline à lui, il l’embrassa.
Ils arrivaient au sommet du tertre et passaient contre l’église de
Saint-Martin dont les fenêtres closes par des planches avouaient le
délabrement, lorsqu’au tournant d’un chemin, à droite du cimetière,
parut une famille de promeneurs. En tête montaient un jeune homme
et une adolescente ; plus bas, s’avançait le père, homme d’un
aspect majestueux qu’il devait non seulement à sa grande barbe
presque blanche, aux larges bords de son feutre, à l’ampleur de son
ulster, mais à son allure de bonhomie patriarcale. Une petite fille lui
donnait la main. Dès qu’il aperçut M. Ardel, il hâta le pas dans
l’intention manifeste de l’aborder.
M. Rude, professeur de dessin, était l’un des rares collègues de
M. Ardel avec qui une liaison lui parût possible. Il en recevait
l’impression d’une nature d’artiste, gaillarde et forte, que son métier
de pédagogue ne parvenait pas à déprimer. De son côté, M. Rude
estimait chez M. Ardel une séduisante intelligence des anciens
peintres ; il l’avait entendu définir, mieux que lui-même ne l’aurait su
faire, le Maître de Moulins. S’il le connaissait libre-penseur, il sentait
pourtant qu’une formation croyante avait dû empreindre dans sa
personne des touches indélébiles ; à des mots brusques de
Victorien, il entrevoyait quelqu’un de fier, d’âpre, de douloureux,
ayant comme lui l’horreur des bassesses. Aussi, en se rencontrant à
l’improviste sur la colline, eurent-ils l’un et l’autre un mouvement de
plaisir.
Les enfants de M. Rude conquirent tout de suite Pauline par leur
simplicité d’accueil.
Julien Rude entrait dans sa vingtième année : haut et flexible, il
laissait sa tête se pencher en avant ; une démarche un peu
traînante, la négligence d’un col de veste dépassant celui du
pardessus, son chapeau rabattu sur son nez aquilin lui donnaient un
air indolent, bizarre. Mais quand il se trouva en présence de Pauline,
elle fut saisie de ses manières et de son visage. Il ne lui rappelait
pas un seul des étudiants qu’elle avait pu voir chez son père. Il mit
dans son salut une aisance grave, réservée, et échangea avec la

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