Let x1 x2 x100 denote the actual net weights in pounds o

Let x1, x2, . . . , x100 denote the actual net weights (in pounds) of 100 randomly selected bags of fertilizer. Suppose that the weight of a randomly selected bag has a distribution with mean 65 lb and variance 1 lb2. Let x be the sample mean weight (n = 100).

(a) What is the probability that the sample mean is between 64.65 lb and 65.35 lb? (Round your answer to four decimal places.)
P(64.65 x 65.35) =

(b) What is the probability that the sample mean is greater than 65 lb?

Solution

Mean ( u ) =65
Standard Deviation ( sd )=1
Number ( n ) = 100
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 64.65) = (64.65-65)/1/ Sqrt ( 100 )
= -0.35/0.1
= -3.5
= P ( Z <-3.5) From Standard Normal Table
= 0.00023
P(X < 65.35) = (65.35-65)/1/ Sqrt ( 100 )
= 0.35/0.1 = 3.5
= P ( Z <3.5) From Standard Normal Table
= 0.99977
P(64.65 < X < 65.35) = 0.99977-0.00023 = 0.9995                  
b)
P(X > 65) = (65-65)/1/ Sqrt ( 100 )
= 0/0.1= 0
= P ( Z >0) From Standard Normal Table
= 0.5