A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 85 months with a variance of 64. If the claim is true, what is the probability that the mean monitor life would be greater than 86.6 months in a sample of 122 monitors? Round your answer to four decimal places.

Step-by-step explanation:

Hello!

X: the lifespan of a new computer monitor of Glotech.

The average life is μ= 85 months and the variance δ²= 64

And a sample of 122 monitors was taken.

You need to calculate the probability that the sample mean is greater than 86.6 months.

Assuming that the variable has a normal distribution X~N(μ;δ²), then the distribution of the sample mean is X[bar]~N(μ;δ²/n)

To calculate this probability you have to work using the sampling distribution and the following formula Z= (X[bar]-μ)/δ/√n ~N(0;1)

P(X[bar]>86.6)= 1 - P(X[bar]≤86.6)

1 - P(Z≤(86.6-85)/(8/√122))= 1 - P(Z≤2.21)= 1 - 0.98645= 0.013355

The probability of the sample mean is greater than 0.013355

I hope this helps!

the answer is c just got it on ed

step-by-step explanation:

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a