Mathematics: The quality control manager at a computer manufacturing company believes that the mean life of a computer is 83 months, with a standard deviation of 9 months. if he is correct, what is the probability that the mean of a sample of 78 computers would be less than 82.06 months? round your answer to four decimal places.

The quality control manager at a computer manufacturing

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08.01.2022

0.1788

Step-by-step explanation:

As per given , we have

$\mu=83,\ \ \sigma=9$

Let x be the random variable that represents the life of a computer.

Sample size : n= 78

Then, the z-score corresponds to x= 82.06 will be :-

$z=\dfrac{82.06-83}{\dfrac{9}{\sqrt{78}}}$ [∵ $z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$ ]

$z=-0.92242835715\approx-0.92$

Required probability ( using standard z-value table ) :-

P(x

Hence, the probability that the mean of a sample of 78 computers would be less than 82.06 months = 0.1788

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