The mean cost of a five pound bag of shrimp is 46 dollars with a variance of 64. If a sample of 53 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by greater than 2.1 dollars? Round your answer to four decimal places.

The probability that sample mean differ the true greater than 2.1 will be 2.8070 %

Step-by-step explanation:

Given:

Sample mean =46 dollars

standard deviation=8

n=53

To Find :

Probability that sample mean would  differ from true mean by greater than 2.1

Solution;

This sample distribution mean problem,

so for that

calculate Z- value

Z=(sample mean - true mean)/(standard deviation/Sqrt(n))

Z=-2.1/(8/Sqrt(53))

Z=-2.1*Sqrt(53)/8

Z=-1.91102

Now for P(X≥2.1)=P(Z≥-1.91102)

Using Z-table,

For Z=-1.91

P(X>2.1)=0.02807

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step-by-step explanation:

sorry no step by step