Question

People’s monthly electric bills in Chennai are normally distributed with a mean of ₹ 225 and a standard deviation of ₹ 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 ₹ 100 or less?

Answer 1

Let X be a normally distributed variable with a mean of ₹ 225 and a standard deviation of ₹ 55

Here µ = 225 and σ = 55

The standard normal variate z = `(x - mu)/sigma = (x - 225)/55`

P(a bill have ₹ 100 or less) = P(X ≤ 100)

When x = 100

z = `(100 - 25)/55`

= `(-125)/55`

= – 2.27

P(X ≤ 100) = P(Z < – 2.27)

P(z < – 2.27) = P(`-oo` < z < 0) – P(– 2.27 < z < 0)

= 0.5 – P(0 < z < 2.27)

= 0.5 – 04884

= 0.0116