The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,637 miles, with a variance of

11,282,880.

What is the probability that the sample mean would differ from the population mean by greater than 778 miles in a sample of 143 tires if the

manager is correct? Round your answer to four decimal places.

0.0028

Step-by-step explanation:

Given that:

Population Mean mileage (m) = 48,637

Variance = 11282880 ;

Standard deviation (σ) = √variance

σ = 3358.9998 = 3359

Sample size (n) = 143

The sample standard deviation = standard error = σ / √n

= 3359 / √143

S. E= 280.89

Zscore = (x - m) / S.E

Difference between sample and population mean = (x - m) > 778 miles

Zscore = 778 / 280.89

Zscore = 2.769

P(Z > 2.769) = 0.0028 (using the z probability calculator)

easy math just plug in a and b

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the eight graders whould be the answer

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