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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?

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3 votes

Answer:

The required probability is 0.0855

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)=(\sigma)/(√(n))


\sigma_(\bar x)=(1200)/(√(30))=219.089

As we know
Z=(\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=(7500-7200)/(219.089)=1.3693

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

Hence, the required probability is 0.0855

User Slabofguinness
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