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 revenue for the next 30 days will exceed​ $7500? Round to four decimal places. A. 0.9131 B. 0.0869 C. 0.9147 D. 0.0853

Answer 1

Answer: D.0.0853

Explanation:

We assume that the daily revenue totals for the next 30 days will be follow a normal distribution.

Given : Population mean :
`\mu=\$7200`

Standard deviation :
`\sigma=\​ $1200`

Sample size : n=30

Let X be the random variable that represents the daily revenue .

Z-score :
`z=(x-\mu)/((\sigma)/(√(n)))`

For x=$7500

`z=(7500-7200)/((1200)/(√(30)))\approx1.37`

By using the standard normal distribution table for z , we have

`P(x>7500)=P(z>1.37)=1-P(z\leq1.37)`

`=1- 0.9146565= 0.0853435\approx0.0853`

Hence, the probability that the mean daily revenue for the next 30 days will exceed​ $7500= 0.0853=