Question

The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be normally distributed with mean equal to $87.00 a month and standard deviation of $36.00. If a statistical sample of n = 100 customers is selected at random, what is the probability that the average bill for those sampled will exceed $75.00? g

Answer 1

Answer:  0.9996

Step-by-step explanation:

Given : The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be normally distributed with  a mean $\mu= \ 87.00$

Standard deviation : $\sigma= \ 36.00$

Sample size : n=100

Let X be the random variable that represents the electricity utility bill for a randomly selected month .

z-score : $z=\dfrac{X-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For X = $75.00

$z=\dfrac{75.00-87.00}{\dfrac{36}{\sqrt{100}}}\approx-3.33$

Now, the probability that the average bill for those sampled will exceed $75.00 will be :-

$P(X>75)=P(z>-3.33)=1-P(z\leq-3.33)\\\\=1-  0.0004342=0.9995658\approx0.9996$

Hence, the probability that the average bill for those sampled will exceed $75.00 =0.9996