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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?

Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

mean = = 40

standard deviation = = 6

= / n = 6/ 43 = 0.9150

= 1 - P[(-2.5) /0.9150 < ( - ) / < (2.5) / 0.9150)]

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

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

= 1 - P(0.9968 - 0.0032)   

= 1 - 0.9936

= 0.0064

Probability = 0.0064

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