The mean cost of a five pound bag of shrimp is 50 dollars with a variance of 64. If a sample of 43 bags of shrimp is randomly selected, what is the probability that the sample mean would be greater than 49 dollars? Round your answer to four decimal places.
Solution :
Given that ,
mean = = 50
Variance = 64
So ,standard deviation = = 8
_{} = / n = 8 / 43 = 1.2200
P( > 49) = 1 - P( < 49)
= 1 - P[( - _{} ) / _{} < (49 - 50) / 1.2200]
= 1 - P(z < -0.82)
= 1 - 0.2061
0.7939
Probability = 0.7939
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