Let Xbar denote the actual weight of 100 randomly selected 5

Let X(bar) denote the actual weight of 100 randomly selected 50-lb bags of fertilizer. a) If the expected weight of each bag is 50 and the variance is 1, calculate P(49.8<=X(bar)<=50.3). b) If a sample of 100 yielded an average weight if 49.6 pounds, what would you begin to think about the population mean? Why? Answer with complete sentences.

Solution

Mean ( u ) =50
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 < 49.8) = (49.8-50)/1/ Sqrt ( 100 )
= -0.2/0.1
= -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 50.3) = (50.3-50)/1/ Sqrt ( 100 )
= 0.3/0.1 = 3
= P ( Z <3) From Standard Normal Table
= 0.99865
P(49.8 < X < 50.3) = 0.99865-0.02275 = 0.9759