Question

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation$1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenuefor the next 30 days will exceed $7500

Answer 1

Answer:

The probability that the mean daily revenue for the next 30 days will exceed $7500 is 0.0855

Step-by-step explanation:

It has been given that

$\mu=7200\\\\\sigma=1200,n=30,x=7500$

Now, the formula for z-score of a normal distribution is given by

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

Substituting the known values, we get

$z=\frac{7500-7200}{\frac{1200}{\sqrt{30}}}$

Simplifying, we get

$z=\frac{300}{219.1}\\\\z=1.369238$

Now, we have to find that the daily revenue for next 30 days will exceed $7500.

Thus, we have to find

P(z>1.369238)

From the normal distribution table, we have

P(z>1.369238)= 0.0855

Therefore, required probability is 0.0855