Standard Normal Distribution
Examples

Example 1

Suppose the reaction times of teenage drivers are normally distributed with a mean of 0.53 seconds and a standard deviation of 0.11 seconds.

A. What is the probability that a teenage driver chosen at random will have a reaction time less than 0.65 seconds?

The goal is to find P(x < 0.65).

1. The first step is to convert 0.65 to a standard score.

z = (x - mean) / standard deviation = (0.65 - 0.53) / 0.11 = 1.09

2. The problem now is to find P(z < 1.09). This is a left tail problem as shown in the illustration to the right.

P(z < 1.09) = 0.8621 (see table below)

Therefore, P(x < 0.65) = 0.8621

B. Find the probability that a teenage driver chosen at random will have a reaction time between 0.4 and 0.6 seconds.

The goal is to find P(0.4 < x < 0.6).

1. The first step is to convert 0.4 and 0.6 to the corresponding standard scores.

z1 = (x - mean) / standard deviation = (0.4 - 0.53) / 0.11 = -1.18

z2 = (x - mean) / standard deviation = (0.6 - 0.53) / 0.11 = 0.64

2. The problem now is to find P(-1.18 < z < 0.64). This is a "between" problem as shown in the illustration to the right.

P(-1.18 < z < 0.64)
= P(z < 0.64) - P(z < -1.18)
= 0.7389 - 0.1190 (see table below)
= 0.6199

Therefore, P(0.4 < x < 0.6) = 0.6199

C. What is the probability that a teenage driver chosen at random will have a reaction time greater than 0.8 seconds?

The problem is to find P(x > 0.8).

1. The first step is to find the corresponding standard score.

z = (x - mean) / standard deviation = (0.8 - 0.53) / 0.11 = 2.45

2. The problem now is to find P(z > 2.45). This is a right tail problem as shown in the illustration to the right.

P(z > 2.45)
= 1 - P(z < 2.45)
= 1 - 0.9929 (see table below)
= 0.0071

Therefore, P(x > 0.8) = 0.0071

Example 2

Suppose the chest sizes of 19th century Scottish Militiamen are normally distributed with a mean of 39.83 inches and a standard deviation of 2.05 inches.

A. What percentage of the militiamen had a chest size less than 36 inches? In this context, the percentage is the corresponding probability expressed as a percent.

The problem is to find P(x < 36).

1. First, convert 36 to a standard score.

z = (x - mean) / standard deviation = (36 - 39.83) / 2.05 = -1.87

2. Find P(z < -1.87). This is a left-tail problem as shown in the illustration to the right.

P(z < -1.87) = 0.0307 (see table below)

Therefore, P(x < 36) = 0.0307 = 3.07%

B. What percentage of the militiamen had a chest size between 40 and 44 inches?

Find P(40 <= x <= 44).

1. First, convert 40 and 44 to the corresponding standard scores.

z1 = (x - mean) / standard deviation = (40 - 39.83) / 2.05 = 0.08

z2 = (x - mean) / standard deviation = (44 - 39.83) / 2.05 = 2.03

2. Find P(0.08 <= z <= 2.03). This is a "between" problem as shown in the illustration to the right.

P(0.08 <= z <= 2.03)
= P(z <= 2.03) - P(z <= 0.08)
= 0.9788 - 0.5319 (see table below)
= 0.4469

Therefore, P(40 <= x <= 44) = 0.4469 = 44.69%

C. What percentage of militiamen had a chest size of 37 inches or greater?

Find P(x >= 38)?

1. First, convert 38 to a standard score.

z = (x - mean) / standard deviation = (38 - 39.83) / 2.05 = -0.89

2. Find P(z >= -0.89). This is a right-tail problem as shown in the illustration to the right.

P(z >= -0.89)
= 1 - P(z <= -0.89)
= 1 - 0.1867 (see table below)
= 0.8133

Therefore, P(x >= 37) = 0.8133 = 81.33%

Standard Normal Distribution Examples