In: Statistics and Probability
The mean cost of a five pound bag of shrimp is 50 dollars with a standard deviation of 8 dollars. If a sample of 56 bags of shrimp is randomly selected, what is the probability that the sample mean would be greater than 49.1 dollars? Round your answer to four decimal places.
Solution :
Given that,
mean =
= 50
standard deviation =
= 8
n=56
=
=50
=
/
n = 8/
56 = 1.0690
P( >49.1 ) = 1 - P(
<49.1 )
= 1 - P[(
-
) /
< (49.1-50 ) /1.0690 ]
= 1 - P(z <-0.84 )
Using z table
= 1 - 0.2005
= 0.7995
probability= 0.7995