Normal distribution word problems

Find here some normal distribution word problems or some applications of the normal distribution.

Example #1

Suppose the current annual salary of all teachers in the United States have a normal distribution with a mean of 51000 dollars and a standard deviation of 6000 dollars. Find the probability that the  annual salary of a randomly selected teacher would be between 42000 and 65000.

Solution

The probability that the annual salary of a randomly selected teacher is between 42000 and 65000 is given by the area under the normal curve between x = 42000 and x = 65000.

For x = 42000, z =  
42000 - 51000 / 6000

z =  
-9000 / 6000
= -1.5

For x = 65000,  z =  
65000 - 51000 / 6000

z =  
14000 / 6000
= 2.33

The required probability is given by the area under the normal curve between z = -1.5 and z = 2.33.

This is obtained by adding the area between z = -1.5 and z = 0 and the area between z = 0 and z = 2.33

Using the standard normal distribution table, we see that the area between z = -1.5 and z = 0 is 0.4332 and the area between z = 0 and z = 2.33 is 0.4901

P(42000 < x < 65000)= P(-1.5 < z < 2.33) = 0.4332 + 0.4901 = 0.9233

This means that about 92.33% of all teachers in the USA earn between 42000 and 65000.

Example #2

The time it takes a computer repair company to diagnose a computer follows a normal distribution with a mean of 50 minutes and a standard deviation of 12 minutes. The company closes at 4 pm every day. If the technician starts diagnosis at 3 pm, what is the probability he will finish his diagnosis before he closes for the day?

Solution

You are looking for the probability that the technician will finish his diagnosis in 60 minutes or less.

This area is given by the area under the normal curve to the left of x = 60.

For x = 60, z =  
60 - 50 / 12

z =  
10 / 12
= 0.83

You are looking for P( z < 0.83)

P(z < 0.83) = P(z < 0) + P(0 < z < 0.83) = 0.5 + 0.2967 = 0.7967

There is an 79.67% chance that the technician will finish before 4 pm.

Example #3

The life span of scientific calculator has a normal distribution with a mean of 58 months and a standard deviation of 10 months. The company gives a warranty of 36 months to replace any defective calculator with a new one. Suppose the company makes 1 million calculators per year, how many calculators may be replaced?

Solution

First, find P(x < 36) or the probability that a randomly selected calculator will be defective in less than 36 months.

When x  = 36,  z =  
36 - 58 / 10

z =  
-22 / 10
= -2.2

We are looking for  P(z < -2.2). Since the normal curve is symmetric, P( z < -2.2) = P( z > 2.2)

P(0 < z < 2.2) + P(z > 2.2) = 0.5

P( z > 2.2) = 0.5 - P( 0 < z < 2.2)

P(0 < z < 2.2) is the area under the normal curve between 0 and 2.2 which is equal to 0.4861

P( z > 2.2) = 0.5 - 0.4861 = 0.0139

1.39% of all calculators may malfunction within 36 months. The number of such calculators is 0.0139 times 1 million or

13,900

If you did not quite understand these normal distribution word problems, review area under the normal curve.

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