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Lecture 14.ppt
- 1. Lecture 14
CHAPTER 6
CONTINUOUS RANDOM
VARIABLES AND THE
NORMAL
DISTRIBUTION
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 2. STANDARDIZING A NORMAL
DISTRIBUTION
Converting an x Value to a z Value
For a normal random variable x, a particular value
of x can be converted to its corresponding z value
by using the formula
where μ and σ are the mean and standard
deviation of the normal distribution of x,
respectively.
x
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 3. Example 6-6
Let x be a continuous random variable that
has a normal distribution with a mean of
50 and a standard deviation of 10. Convert
the following x values to z values and find
the probability to the left of these points.
a) x = 55
b) x = 35
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 4. Example 6-6: Solution
a) x = 55
P(x < 55) = P(z < .50) = .6915
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
50
.
10
50
55
x
z
- 5. Figure 6.31 z value for x = 55.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 6. Example 6-6: Solution
b) x = 35
P(x < 35) = P(z < -1.50) = .0668
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
35 50
1.50
10
x
z
- 7. Figure 6.32 z value for x = 35.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 8. Example 6-7
Let x be a continuous random variable
that is normally distributed with a mean
of 25 and a standard deviation of 4.
Find the area
a) between x = 25 and x = 32
b) between x = 18 and x = 34
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 9. Example 6-7: Solution
a)
The z value for x = 25 is 0
The z value for x = 32 is
P (25 < x < 32) = P(0 < z < 1.75)
= .4599
75
.
1
4
25
32
x
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 10. Figure 6.33 Area between x = 25 and x = 32.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 11. Example 6-7: Solution
b)
For x = 18:
For x = 34:
P (18 < x < 34) = P (-1.75 < z < 2.25 )
= .9878 - .0401 = .9477
75
.
1
4
25
18
z
25
.
2
4
25
34
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 12. Figure 6.34 Area between x = 18 and x = 34.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 13. Example 6-8
Let x be a normal random variable with
its mean equal to 40 and standard
deviation equal to 5. Find the following
probabilities for this normal distribution
a) P (x > 55)
b) P (x < 49)
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 14. Example 6-8: Solution
a)
For x = 55:
P (x > 55) = P (z > 3.00) = 1.0 - .9987
= .0013
00
.
3
5
40
55
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 15. Figure 6.35 Finding P (x > 55).
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 16. Example 6-8: Solution
b)
For x = 49:
P (x < 49) = P (z < 1.80) = .9641
80
.
1
5
40
49
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 17. Figure 6.36 Finding P (x < 49).
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 18. Example 6-9
Let x be a continuous random variable that
has a normal distribution with μ = 50 and σ
= 8. Find the probability P (30 ≤ x ≤ 39).
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 19. Example 6-9: Solution
For x = 30:
For x = 39:
P (30 ≤ x ≤ 39) = P (-2.50 ≤ z ≤ -1.38)
= .0838 - .0062 = .0776
50
.
2
8
50
30
z
38
.
1
8
50
39
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 20. Figure 6.37 Finding P (30 ≤ x ≤ 39).
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 21. Example 6-10
Let x be a continuous random variable
that has a normal distribution with a
mean of 80 and a standard deviation of
12. Find the area under the normal
distribution curve
a) from x = 70 to x = 135
b) to the left of 27
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 22. Example 6-10: Solution
a)
For x = 70:
For x = 135:
P (70 ≤ x ≤ 135) = P (-.83 ≤ z ≤ 4.58)
= 1 - .2033
= .7967 approximately
83
.
12
80
70
z
58
.
4
12
80
135
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 23. Figure 6.38 Area between x = 70 and x = 135.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 24. Example 6-10: Solution
b)
For x = 27:
P (x < 27) = P (z < -4.42)
=.00 approximately
42
.
4
12
80
27
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 25. Figure 6.39 Area to the left of x = 27.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 26. APPLICATIONS OF THE NORMAL
DISTRIBUTION
Section 6.2 through 6.4 discussed the
normal distribution, how to convert a
normal distribution to the standard normal
distribution, and how to find areas under a
normal distribution curve. This section
presents examples that illustrate the
applications of the normal distribution.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 27. Example 6-11
According to a Sallie Mae and credit bureau
data, in 2008, college students carried an
average of $3173 debt on their credit cards
(USA TODAY, April 13, 2009). Suppose
that current credit card debts for all college
students have a normal distribution with a
mean of $3173 and a standard deviation of
$800. Find the probability that credit card
debt for a randomly selected college
student is between $2109 and $3605.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 28. Example 6-11: Solution
For x = $2109:
For x = $3605:
P ($2109 < x < $3605)
= P (-1.33 < z < .54)
= .7054 - .0918
= .6136 = 61.36%
2109 3173
1.33
800
z
3605 3173
.54
800
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 29. Figure 6.40 Area between x = $2109 and x =
$3605.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 30. Example 6-12
A racing car is one of the many toys
manufactured by Mack Corporation. The
assembly times for this toy follow a
normal distribution with a mean of 55
minutes and a standard deviation of 4
minutes. The company closes at 5 p.m.
every day. If one worker starts to
assemble a racing car at 4 p.m., what is
the probability that she will finish this job
before the company closes for the day?
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 31. Example 6-12: Solution
For x = 60:
P(x ≤ 60) = P(z ≤ 1.25) = .8944
Thus, the probability is .8944 that this
worker will finish assembling this racing
car before the company closes for the day.
60 55
1.25
4
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 32. Figure 6.41 Area to the left of x = 60.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 33. Example 6-13
Hupper Corporation produces many types of soft
drinks, including Orange Cola. The filling machines
are adjusted to pour 12 ounces of soda into each
12-ounce can of Orange Cola. However, the actual
amount of soda poured into each can is not exactly
12 ounces; it varies from can to can. It has been
observed that the net amount of soda in such a
can has a normal distribution with a mean of 12
ounces and a standard deviation of .015 ounce.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 34. Example 6-13
a) What is the probability that a randomly
selected can of Orange Cola contains 11.97 to
11.99 ounces of soda?
b) What percentage of the Orange Cola cans
contain 12.02 to 12.07 ounces of soda?
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 35. Example 6-13: Solution
a)
For x = 11.97:
For x = 11.99:
P (11.97 ≤ x ≤ 11.99)
= P (-2.00 ≤ z ≤ -.67) = .2514 - .0228
= .2286
00
.
2
015
.
12
97
.
11
z
67
.
015
.
12
99
.
11
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 36. Figure 6.42 Area between x = 11.97 and x = 11.99.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 37. Example 6-13: Solution
b)
For x = 12.02:
For x = 12.07:
P (12.02 ≤ x ≤ 12.07)
= P (1.33 ≤ z ≤ 4.67) = 1 - .9082
= .0918
33
.
1
015
.
12
02
.
12
z
67
.
4
015
.
12
07
.
12
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 38. Figure 6.43 Area from x = 12.02 to x = 12.07.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 39. Example 6-14
The life span of a calculator manufactured by
Texas Instruments has a normal distribution with a
mean of 54 months and a standard deviation of 8
months. The company guarantees that any
calculator that starts malfunctioning within 36
months of the purchase will be replaced by a new
one. About what percentage of calculators made
by this company are expected to be replaced?
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 40. Example 6-14: Solution
For x = 36:
P(x < 36) = P (z < -2.25)
= .0122
Hence, 1.22% of the calculators are
expected to be replaced.
25
.
2
8
54
36
z
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved
- 41. Figure 6.44 Area to the left of x = 36.
Prem Mann, Introductory Statistics, 7/E
Copyright © 2010 John Wiley & Sons. All right reserved