Question

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

Answer 1

Answer:

The required probability is 0.0855

Step-by-step explanation:

Consider the provided information.

The daily revenue has mean $7200 and standard deviation $1200.

$\mu_{\bar x}=7200$

$\sigma=1200$

The daily revenue totals for the next 30 days will be monitored.

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x}=\frac{1200}{\sqrt{30}}=219.089$

As we know $Z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}$

Substitute $\bar x=7500, \mu_{\bar x}=7200\ and\ \sigma_{\bar x}=219.089$ in above formula.

$Z=\frac{7500-7200}{219.089}=1.3693$

From the standard normal table P( Z >1.3693) = 0.0855

Hence, the required probability is 0.0855