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