Cant solve this one If a finite Population is normally distr

If a finite Population is normally distributed with a mean of 50 and a standard deviation f 5, what is the probability that the sample mean lies between 90 and 100, if the sample ize is 40 and sampling is performed with replacement? 0.7286 0.7458 0.0000 1.0000 A quality control manager is concerned about the breaking strength of a metal wire manufactured to stringent specifications. A sample of size 25 is randomly obtained and the

Solution

Mean ( u ) =50
Standard Deviation ( sd )=5
Number ( n ) = 40
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
To find P(a <= Z <=b) = F(b) - F(a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 90) = (90-50)/5
= 40/5 = 8
= P ( Z <8) From Standard Normal Table
= 1
P(X < 100) = (100-50)/5
= 50/5 = 10
= P ( Z <10) From Standard Normal Table
= 1
P(90 < X < 100) = 1-1 = 0                  

ANSWER: 0.0000