SOLUTION: 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 monit

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Question 1142019: 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 be between $7000 and $7500?
A) 0.8186
B) 0.2667
C) 0.7333
D) 0.9147
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
SD of the sample = SD of the population divided by the square root of the number of items in the sample:

SD of the sample = 1200%2Fsqrt%2830%29 = 219.09

Take 7500 and subtract the mean of 7200 to get a result of 300. Divide 300 by the SD of the sample (219.09) to get a result of 1.37. Look up +1.37 on a z-table to get a result of 0.9147. This means there is a 0.9147 probability that the mean daily revenue for the next 30 days will be below $7500.

Take 7000 and subtract the mean of 7200 to get a result of -200. Divide -200 by the SD of the sample (219.09) to get a result of -0.91. Look up -0.91 on a z-table to get a result of 0.1814. This means there is a 0.1814 probability that the mean daily revenue for the next 30 days will be below $7000.

To find out the probability that the mean daily revenue for the next 30 days will be between $7000 and $7500, simply subtract the probability that the mean daily revenue is less than $7000 (0.1814) from the probability that the mean daily revenue is less than $7500 (0.9147) to get a result of 0.7333.

So, the answer is (C).