The mean cost of a five pound bag of shrimp is 42 dollars wi

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.

Solution

Mean ( u ) =42
Standard Deviation ( sd )=7
Number ( n ) = 54
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)  

Sample mean differ less than 1.4 Dollor = [ 42 - 1.4 < Mean < 42 + 1.4 ] = [ 40.6 < X < 43.4 ]
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 40.6) = (40.6-42)/7/ Sqrt ( 54 )
= -1.4/0.9526
= -1.4697
= P ( Z <-1.4697) From Standard Normal Table
= 0.07082
P(X < 43.4) = (43.4-42)/7/ Sqrt ( 54 )
= 1.4/0.9526 = 1.4697
= P ( Z <1.4697) From Standard Normal Table
= 0.92918
P(40.6 < X < 43.4) = 0.92918-0.07082 = 0.8584