Normal Distribution Real Life Examples

In this class, We discuss Normal Distribution Real Life Examples.

The reader should have prior knowledge of normal distribution. Click Here.

Example 1:

A sales tax officer has reported that the average sales of 500 businesses he has to deal with during a year are 36000 with a standard deviation of 10000.

Sales data follows a normal distribution.

1) Find the number of businesses as the sales of which are above 40000.

2) Find the percentage of businesses that are likely between 30000 and 40000.

Solution:

1) Random variable X > 40000.

Z = (X – μ)/σ

Z = (40000 – 36000)/10000 = 0.4

P(Z > 0.4) = 0.5 – P(Z = 0.4)

From the normal table P(Z = 0.4) = 0.1554

= 0.5 – 0.1554

= 0.3446

Number of business = 500 * 0.3446 = 172

2) P(30000 < x < 40000)

Z1 = (30000 – 36000)/10000 = -0.6

Z2 = 0.4

From the basics P(Z= -0.6) = P( Z = 0.6)

Total probability = P(0.6) + P(0.4)

= 0.1554 + 0.2257 = 0.3811

= 38.11 percent

Example 2:

The weights of 300 students have a normal distribution with a mean of 68 kg and a standard deviation of 3 kg.

How many students have a weight greater than 72 kg?

Solution:

Random variable X = 72.

Z = (72 – 68)/3 = 1.33

P(Z > 1.33) = 0.5 – P(Z = 1.33)

= 0.5 – 0.4082 = 0.0912