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The mean cost of a five pound bag of shrimp is 40 dollars with a variance...

The mean cost of a five pound bag of shrimp is 40 dollars with a variance of 36.
If a sample of 43 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by more than 2.5 dollars? Round your answer to four decimal places.
Math   Statistics-And-Probability   
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Homework Answers

Answer #1

Solution :

Given that,

mean = = 40

standard deviation = = 6

n = 43

= 40

= / n = 6 / 43

= 1 - P[(2.5) / 6 / 43 < ( - ) / < (2.5) / 6 / 43)]

= 1 - P(-2.73 < Z < 2.73)

= 1 - P(Z < 2.73) - P(Z < -2.73)

= 1 - 0.9937

= 0.0063

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