Question

The mean cost of a five pound bag of shrimp is 42 dollars with a variance of 49.

If a sample of 54 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by less than  1.4 dollars? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 42

standard deviation = = 7

= / n = 7 / 54 = 0.9526

= P[(-1.4) /0.9526 < ( - ) / < (1.4) / 0.9526)]

= P(-1.47 < Z < 1.47)

= P(Z < 1.47) - P(Z < -1.47)

= 0.9292 - 0.0708

= 0.8584

Probability = 0.8584