Question

The mean cost of a five pound bag of shrimp is 47 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 greater than 1.4  dollars?

Round your answer to four decimal places.

Answer 1

Solution :

= / n = 6/ 43 = 0.9150

= 1 - P[(-1.4) / 0.9150 < ( - ) / < (1.4) / 0.9150)]

= 1 - P(-1.53 < Z < 1.53)

= 1 - P(Z < 1.53) - P(Z < -1.53)

= 1 - P(0.937 - 0.063)

= 1 - 0.874

= 0.1260

Probability = 0.1260