Question

The mean cost of a five pound bag of shrimp is 50 dollars with a standard deviation of 8 8 dollars. If a sample of 56 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 0.9 dollars? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 50

standard deviation = = 88

= / n = 88 / 56 = 11.7595

= P[(-0.9) /11.7595 < ( - ) / < (0.9) / 11.7595)]

= P(-0.08 < Z < 0.08)

= P(Z < 0.08) - P(Z < -0.08)   

= 0.5319 - 0.4681

= 0.0638

Probability = 0.0638