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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.



Answer :

Qwdog

The probability that the sample mean would differ from the true mean by more than $2.5 is 0.0062

Mean cost of the five pound shrimp bag = $40

Variance of the five pound shrimp bag = $36

The number of the shrimp bag = 43

The given mean value = $2.5

The z-score = $2.5

The probability of the sample mean differ by $2.5 from the true mean can be calculated by

  P(z > 2.5) = 1 - P(z < 2.5)

From the z table , we can get the value of  z < 2.5 which is 0.9938

    P(z > 2.5) = 1 - 0.9938

   P(z > 2.5) = 0.0062

Therefore , the probability of sample mean from the true mean by 2.5 is 0.0062

         

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