Question

People’s monthly electric bills in Chennai are normally distributed with a mean of Rs. 225 and a standard deviation of Rs. 55. Those people spend a lot of time online. In a group of 500 customers, how many would we expect to have a bill that is Rs. 100 or less?

Answer 1

Let X be the normal variable denoting the monthly bills in rupees.

Given mean µ = 225 and s.d σ = 55 

Now the probability that the bill will be Rs.100 or less is P (X ≤ 100)

= P(Z ≤ (100 - 225)/55) 

= P(Z ≤ -2.27) 

= 0.5 – P(-2.27 < Z < 0) 

= 0.5 – P(0 < Z < 2.27) 

= 0.5 – 0.4884 

= 0.0116 

Thus, in a group of 500 customers, we expect to have 500 × 0.0116 = 5.8 ~ 6 customers whose electric bills will be Rs. 100 or less.