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

Respuesta :

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 : [tex]\mu=\$7200[/tex]

Standard deviation : [tex]\sigma=\​ $1200[/tex]

Sample size : n=30

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

Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x=$7500

[tex]z=\dfrac{7500-7200}{\dfrac{1200}{\sqrt{30}}}\approx1.37[/tex]

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

[tex]P(x>7500)=P(z>1.37)=1-P(z\leq1.37)[/tex]

[tex]=1- 0.9146565= 0.0853435\approx0.0853[/tex]

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

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

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