Question

a quality control expert at glotech computers wants to test their new monitors. the production manager claims they have a mean life of 93 months with a standard deviation of 9 months. if the claim is true, what is the probability that the mean monitor life would be greater than 91.4 months in a sample of 66 monitors? round your answer to four decimal places. answer

Answer 1

The probability that the mean monitor life  would be greater than 91.4 months is 0.8666.

We have mean life as 93 months, and standard deviation as 9 months.

We need to check mean monitor life  will be greater than 91.4 months in the sample.

It means that on left side mean range is=93-91.4=1.8,and

right side mean range =93+91.4=184.4

So, we need to find the probability if mean range varies in between from 1.8 to 184.4.

So, we use z-score table to find the z-score for 1.8 and 184.4,we get

We use Z-formula which is given by

Z = (X-μ)/SE

where X is the standard mean.

Using z-score table, we get

Z = 0.16917 and -0.161917

Using z-score table, we get that P(0.16917)=0.4333.

Since, both of the probability occupies same space area, due to that total probability =0.4333×2=0.8666

Hence, the probability will be 0.8666.

To know more about probability, visit here:

brainly.com/question/11234923

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