The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 44,111 miles, with a variance of 5,943,844. What is the probability that the sample mean would be less than 44,257 miles in a sample of 80 tires if the manager is correct? Round your answer to four decimal places.

And if we use the normal standard table we got this:

Step-by-step explanation:

For this case we have the following info :

represent the true mean

represent the deviation

n= 80 represent the sample size

And we want to find the follwing probability:

For this case since the sample size is larger than 30 we can apply the central limit theorem and we can use the z score formula given by:

The distribution for the sample mean would be:

And if we find the z score for this case we got:

And if we use the normal standard table we got this:

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i