1. Elementary Statistics
Larson Farber
5 Normal Probability Distributions

2. Introduction to
Normal
Distributions
Section 5.1

3. Properties of a Normal Distribution
• The mean, median, and mode are equal
• Bell shaped and is symmetric about the mean
• The total area that lies under the curve is one or 100%
x

4. • As the curve extends farther and farther away from the
mean, it gets closer and closer to the x-axis but never
touches it.
• The points at which the curvature changes are called
inflection points. The graph curves downward between the
inflection points and curves upward past the inflection
points to the left and to the right.
x
Inflection pointInflection point
Properties of a Normal Distribution

5. Means and Standard Deviations
2012 15 1810 11 13 14 16 17 19 21 229
12 15 1810 11 13 14 16 17 19 20
Curves with different means, different standard deviations
Curves with different means, same standard deviation

6. Empirical Rule
About 95% of the area
lies within 2 standard
deviations
About 99.7% of the area lies within
3 standard deviations of the mean
About 68% of the area
lies within 1 standard
deviation of the mean
68%

7. 4.2 4.5 4.8 5.13.93.63.3
Determining Intervals
An instruction manual claims that the assembly time for a
product is normally distributed with a mean of 4.2 hours
and standard deviation 0.3 hour. Determine the
interval in which 95% of the assembly times fall.
x
4.2 – 2 (0.3) = 3.6 and 4.2 + 2 (0.3) = 4.8.
95% of the assembly times will be between 3.6 and 4.8 hrs.
95% of the data will fall within 2 standard deviations of the mean.

8. The Standard
Normal
Distribution
Section 5.2

9. The Standard Score
The standard score, or z-score, represents the number of
standard deviations a random variable x falls from the
mean.
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of
7. Find the standard z-score for a person with a score of:
(a) 161 (b) 148 (c) 152
(a) (b) (c)

10. The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
–4 –3 –2 –1 0 1 2 3 4 z

11. Cumulative Areas
• The cumulative area is close to 1 for z-scores close to
3.49.
0 1 2 3–1–2–3 z
The
total
area
under
the curve
is one.
• The cumulative area is close to 0 for z-scores close
to –3.49.
• The cumulative area for z = 0 is 0.5000.

12. Find the cumulative area for a z-score of –1.25.
0 1 2 3–1–2–3 z
Cumulative Areas
0.1056
Read down the z column on the left to z = –1.25 and across to
the column under .05. The value in the cell is 0.1056, the
cumulative area.
The probability that z is at most –1.25 is 0.1056.

13. Finding Probabilities
To find the probability that z is less than a given value,
read the cumulative area in the table corresponding to
that z-score.
0 1 2 3–1–2–3 z
Read down the z-column to –1.4 and across to .05. The
cumulative area is 0.0735.
Find P(z < –1.45).
P (z < –1.45) = 0.0735

14. Finding Probabilities
To find the probability that z is greater than a given
value, subtract the cumulative area in the table
from 1.
0 1 2 3–1–2–3 z
P(z > –1.24) = 0.8925
Find P(z > –1.24).
The cumulative area (area to the left) is 0.1075. So the area
to the right is 1 – 0.1075 = 0.8925.
0.1075
0.8925

15. Finding Probabilities
To find the probability z is between two given values, find the
cumulative areas for each and subtract the smaller area from
the larger.
Find P(–1.25 < z < 1.17).
1. P(z < 1.17) = 0.8790 2. P(z < –1.25) = 0.1056
3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734
0 1 2 3–1–2–3 z

16. 0 1 2 3-1-2-3 z
Summary
0 1 2 3-1-2-3 z
To find the probability is greater
than a given value, subtract the
cumulative area in the table from 1.
0 1 2 3-1-2-3 z
To find the probability z is
between two given values, find the
cumulative areas for each and
subtract the smaller area from the
larger.
To find the probability that z is less
than a given value, read the
corresponding cumulative area.

17. Normal
Distributions
Finding
Probabilities
Section 5.3

18. Probabilities and Normal Distributions
115100
If a random variable, x is normally distributed, the
probability that x will fall within an interval is equal to the
area under the curve in the interval.
IQ scores are normally distributed with a mean of 100
and a standard deviation of 15. Find the probability that a
person selected at random will have an IQ score less
than 115.
To find the area in this interval, first find the standard
score equivalent to x = 115.

19. 0 1
Probabilities and Normal Distributions
Find P(z < 1).
115100
Standard Normal
Distribution
Find P(x < 115).
Normal Distribution
P(z < 1) = 0.8413, so P(x <115) = 0.8413
SAME
SAME

20. Monthly utility bills in a certain city are normally
distributed with a mean of $100 and a standard deviation
of $12. A utility bill is randomly selected. Find the
probability it is between $80 and $115.
P(80 < x < 115)
Normal Distribution
P(–1.67 < z < 1.25)
0.8944 – 0.0475 = 0.8469
The probability a utility bill is
between $80 and $115 is 0.8469.
Application

21. Normal
Distributions
Finding Values
Section 5.4

22. z
From Areas to z-Scores
Locate 0.9803 in the area portion of the table. Read the
values at the beginning of the corresponding row and at
the top of the column. The z-score is 2.06.
Find the z-score corresponding to a cumulative area of 0.9803.
z = 2.06 corresponds
roughly to the
98th percentile.
–4 –3 –2 –1 0 1 2 3 4
0.9803

23. Finding z-Scores from Areas
Find the z-score corresponding to the 90th percentile.
z0
.90
The closest table area is .8997. The row heading is
1.2 and column heading is .08. This corresponds to
z = 1.28.
A z-score of 1.28 corresponds to the 90th percentile.

24. Find the z-score with an area of .60 falling to its right.
.60
.40
0 z
z
With .60 to the right, cumulative area is .
40. The closest area is .4013. The row
heading is 0.2 and column heading is .05.
The z-score is 0.25.
A z-score of 0.25 has an area of .60 to its right.
It also corresponds to the 40th percentile
Finding z-Scores from Areas

25. Find the z-score such that 45% of the area under the
curve falls between –z and z.
0 z–z
The area remaining in the tails is .55. Half this area is
in each tail, so since .55/2 = .275 is the cumulative area
for the negative z value and .275 + .45 = .725 is the
cumulative area for the positive z. The closest table
area is .2743 and the z-score is 0.60. The positive z
score is 0.60.
.45
.275.275
Finding z-Scores from Areas

26. From z-Scores to Raw Scores
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of 7.
Find the test score for a person with a standard score of:
(a) 2.33 (b) –1.75 (c) 0
(a) x = 152 + (2.33)(7) = 168.31
(b) x = 152 + (–1.75)(7) = 139.75
(c) x = 152 + (0)(7) = 152
To find the data value, x when given a standard score, z:

27. Finding Percentiles or Cut-off Values
Monthly utility bills in a certain city are normally distributed
with a mean of $100 and a standard deviation of $12. What is
the smallest utility bill that can be in the top 10% of the bills?
10%
90%
Find the cumulative area in the table that is closest to
0.9000 (the 90th percentile.) The area 0.8997 corresponds
to a z-score of 1.28.
x = 100 + 1.28(12) = 115.36.
$115.36 is the smallest
value for the top 10%.
z
To find the corresponding x-value, use

28. The Central Limit
Theorem
Section 5.5

29. Sample
Sampling Distributions
A sampling distribution is the probability distribution of a
sample statistic that is formed when samples of size n are
repeatedly taken from a population. If the sample statistic is
the sample mean, then the distribution is the sampling
distribution of sample means.
Sample
The sampling distribution consists of the values of the sample
means,
Sample
Sample
Sample
Sample

30. x
the sample means will have a normal distribution
The Central Limit Theorem
and standard deviation
If a sample n  30 is taken from a population with
any type distribution that has a mean =
and standard deviation =

31. the distribution of means of sample size n, will be normal
with a mean
standard deviation
The Central Limit Theorem
x
If a sample of any size is taken from a population with a
normal distribution with mean = and standard
deviation =

32. Application
Distribution of means of sample size 60,
will be normal.
The mean height of American men (ages 20-29) is
inches. Random samples of 60 such men are selected. Find the mean and
standard deviation (standard error) of the sampling distribution.
mean
Standard deviation
69.2

33. Interpreting the Central Limit Theorem
The mean height of American men (ages 20-29) is =
69.2”. If a random sample of 60 men in this age group
is selected, what is the probability the mean height for
the sample is greater than 70”? Assume the standard
deviation is 2.9”.
Find the z-score for a sample mean of 70:
standard deviation
mean
Since n > 30 the sampling distribution of will be normal

34. 2.14z
There is a 0.0162 probability that a sample of 60
men will have a mean height greater than 70”.
Interpreting the Central Limit Theorem

35. Application Central Limit Theorem
During a certain week the mean price of gasoline in California was
$1.164 per gallon. What is the probability that the mean price for
the sample of 38 gas stations in California is between $1.169 and
$1.179? Assume the standard deviation = $0.049.
standard deviation
mean
Calculate the standard z-score for sample values of $1.169 and
$1.179.
Since n > 30 the sampling distribution of will be normal

36. .63 1.90
z
Application Central Limit Theorem
P( 0.63 < z < 1.90)
= 0.9713 – 0.7357
= 0.2356
The probability is 0.2356 that the mean for the
sample is between $1.169 and $1.179.

37. Normal
Approximation to
the Binomial
Section 5.6

38. Binomial Distribution Characteristics
• There are a fixed number of independent trials. (n)
• Each trial has 2 outcomes, Success or Failure.
• The probability of success on a single trial is p and
the probability of failure is q. p + q = 1
• We can find the probability of exactly x successes out
of n trials. Where x = 0 or 1 or 2 … n.
• x is a discrete random variable representing a count
of the number of successes in n trials.

39. Application
34% of Americans have type A+
blood. If 500 Americans are
sampled at random, what is the probability at least 300 have
type A+
blood?
Using techniques of Chapter 4 you could calculate the
probability that exactly 300, exactly 301… exactly 500
Americans have A+
blood type and add the probabilities.
Or…you could use the normal curve probabilities to
approximate the binomial probabilities.
If np  5 and nq  5, the binomial random variable x is
approximately normally distributed with mean

40. Why Do We Require np  5 and nq  5?
0 1 2 3 4 5
n = 5
p = 0.25, q = .75
np =1.25 nq = 3.75
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
n = 20
p = 0.25
np = 5 nq = 15
n = 50
p = 0.25
np = 12.5
nq = 37.5
0 10 20 30 40 50

41. Binomial Probabilities
The binomial distribution is discrete with a probability
histogram graph. The probability that a specific value of
x will occur is equal to the area of the rectangle with
midpoint at x.
If n = 50 and p = 0.25 find
Add the areas of the rectangles with midpoints at
x = 14, x = 15, x = 16.
14 15 16
0.111 0.089
0.065
0.111 + 0.089 + 0.065 = 0.265

42. 14 15 16
Correction for Continuity
Use the normal approximation to the binomial to
find .
Values for the binomial random variable x
are 14, 15 and 16.

43. 14 15 16
Correction for Continuity
Use the normal approximation to the binomial to
find .
The interval of values under the normal curve is
To ensure the boundaries of each rectangle are
included in the interval, subtract 0.5 from a left-hand
boundary and add 0.5 to a right-hand boundary.

44. Normal Approximation to the Binomial
Use the normal approximation to the binomial to find
Adjust the endpoints to correct for continuity P .
Convert each endpoint to a standard score.
Find the mean and standard deviation using binomial
distribution formulas.
.

45. Application
A survey of Internet users found that 75% favored
government regulations of “junk” e-mail. If 200 Internet
users are randomly selected, find the probability that fewer
than 140 are in favor of government regulation.
Since np = 150  5 and nq = 50  5 use the normal
approximation to the binomial.
The binomial phrase of “fewer than 140” means
0, 1, 2, 3…139.
Use the correction for continuity to translate to the
continuous variable in the interval . Find P( x <
139.5).

46. Application
A survey of Internet users found that 75% favored
government regulations of “junk” e-mail. If 200 Internet
users are randomly selected, find the probability that fewer
than 140 are in favor of government regulation.
Use the correction for continuity P(x < 139.5).
P( z < -1.71) = 0.0436
The probability that fewer than 140 are in favor of
government regulation is approximately 0.0436.