Section 5.2 Normal Distributions: Finding Probabilities 1
Larson/Farber 4th ed Slide 2 Section 5.2 Objectives Find
probabilities for normally distributed variables 2 Larson/Farber
4th ed Slide 3 Probability and Normal Distributions If a random
variable x is normally distributed, you can find the probability
that x will fall in a given interval by calculating the area under
the normal curve for that interval. P(x < 600) = Area = 500 =
100 600 =500 x 3 Larson/Farber 4th ed Slide 4 Probability and
Normal Distributions P(x < 500) = P(z < 1) Normal
Distribution 600 =500 P(x < 600) = 500 = 100 x Standard Normal
Distribution 1 = 0 = 0 = 1 z P(z < 1) 4 Larson/Farber 4th ed
Same Area Slide 5 Example: Finding Probabilities for Normal
Distributions A survey indicates that people use their computers an
average of 2.4 years before upgrading to a new machine. The
standard deviation is 0.5 year. A computer owner is selected at
random. Find the probability that he or she will use it for fewer
than 2 years before upgrading. Assume that the variable x is
normally distributed. 5 Larson/Farber 4th ed Slide 6 Solution:
Finding Probabilities for Normal Distributions P(x < 2) = P(z
< -0.80) = 0.2119 Normal Distribution 22.4 P(x < 2) = 2.4 =
0.5 x Standard Normal Distribution -0.80 0 = 0 = 1 z P(z <
-0.80) 0.2119 6 Larson/Farber 4th ed Slide 7 Example: Finding
Probabilities for Normal Distributions A survey indicates that for
each trip to the supermarket, a shopper spends an average of 45
minutes with a standard deviation of 12 minutes in the store. The
length of time spent in the store is normally distributed and is
represented by the variable x. A shopper enters the store. Find the
probability that the shopper will be in the store for between 24
and 54 minutes. 7 Larson/Farber 4th ed Slide 8 Solution: Finding
Probabilities for Normal Distributions P(24 < x < 54) =
P(-1.75 < z < 0.75) = 0.7734 0.0401 = 0.7333 2445 P(24 < x
< 54) x Normal Distribution = 45 = 12 0.0401 54 -1.75 z Standard
Normal Distribution = 0 = 1 0 P(-1.75 < z < 0.75) 0.75 0.7734
8 Larson/Farber 4th ed Slide 9 Example: Finding Probabilities for
Normal Distributions Find the probability that the shopper will be
in the store more than 39 minutes. (Recall = 45 minutes and = 12
minutes) 9 Larson/Farber 4th ed Slide 10 Solution: Finding
Probabilities for Normal Distributions P(x > 39) = P(z >
-0.50) = 1 0.3085 = 0.6915 3945 P(x > 39) x Normal Distribution
= 45 = 12 Standard Normal Distribution = 0 = 1 0.3085 0 P(z >
-0.50) z -0.50 10 Larson/Farber 4th ed Slide 11 Example: Finding
Probabilities for Normal Distributions If 200 shoppers enter the
store, how many shoppers would you expect to be in the store more
than 39 minutes? Solution: Recall P(x > 39) = 0.6915 200(0.6915)
=138.3 (or about 138) shoppers 11 Larson/Farber 4th ed Slide 12 Are
you here? Are you listening? The lengths of Altantic croaker fish
are normally distributed with a mean of 10 inches and a standard
deviation of 2 inches. A fish is randomly selected. If 500 croaker
fish are caught, how many would you expect to be over 15 inches in
length? Slide 13 Example: Assume that cholesterol levels of men in
the United States are normally distributed, with a mean of 215
milligrams per deciliter and a standard deviation of 25 milligrams
per deciliter. You randomly select a man from the United States.
What is the probability that his cholesterol level is less than
175? Use a technology tool to find the probability. 13
Larson/Farber 4th ed Slide 14 Example: 175 215 = -1.6 25 P (x <
175) = P (z < -1.60) =.0548 Slide 15 Section 5.2 Summary Found
probabilities for normally distributed variables 15 Larson/Farber
4th ed
