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

Step-by-step 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 : $z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x=$7500

$z=\dfrac{7500-7200}{\dfrac{1200}{\sqrt{30}}}\approx1.37$

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=

Answer 2

The probability that the mean daily revenue for the next 30 days will exceed​ $7500 is 0.0853

What is z score?

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / (standard deviation / √ sample)

Given mean​ $7200 and standard deviation​ $1200. Hence:

For > 7500:

z = (7500 - 7200) / (1200/√30) = 1.37

P(z > 1.37) = 1 - P(z < 1.37) = 1 - 0.9147 = 0.0853

The probability that the mean daily revenue for the next 30 days will exceed​ $7500 is 0.0853

Find out more on z score aT: https://brainly.com/question/25638875