Question

According to a report by Scarborough Research, the average monthly household cellular phone bill is $73. Suppose local monthly household cell phone bills are normally distributed with a standard deviation of $11.
a. What is the probability that a randomly selected monthly cell phone bill is less than $95?

Answer 1

Answer:

The probability that a randomly selected monthly cell phone bill is less than $95 is 0.9772

Step-by-step explanation:

The average monthly household cellular phone bill is $73.

$\mu = 73$

Local monthly household cell phone bills are normally distributed with a standard deviation of $11.

$\sigma = 11$

We are supposed to find the probability that a randomly selected monthly cell phone bill is less than $95 i.e.P(x<95)

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{95-73}{11}$

Z=2

Refer the z table for p value

So,p value = 0.9772

Hence the probability that a randomly selected monthly cell phone bill is less than $95 is 0.9772