Assume that the weights of all packages of a certain brand o

Assume that the weights of all packages of a certain brand of cookies are no distributed with a mean of 32 ounces and a standard deviation of 0.3 ounces Determine the probability that the mean weight of a random sample of 20 packages of this brand of cookies will be: Between 31.8 and 31.9 ounces At least 22 ounces

Solution

Sample Mean ( u ) =32
Sample Standard Deviation ( sd )= 0.3/ Sqrt ( 20 ) = 0.0671
Number ( n ) = 20
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
              
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 31.8) = (31.8-32)/0.3/ Sqrt ( 20 )
= -0.2/0.0671
= -2.9814
= P ( Z <-2.9814) From Standard Normal Table
= 0.00143
P(X < 31.9) = (31.9-32)/0.3/ Sqrt ( 20 )
= -0.1/0.0671 = -1.4907
= P ( Z <-1.4907) From Standard Normal Table
= 0.06802
P(31.8 < X < 31.9) = 0.06802-0.00143 = 0.0666                  

b)
P(X < 22) = (22-32)/0.3/ Sqrt ( 20 )
= -10/0.0671= -149.0712
= P ( Z <-149.0712) From Standard NOrmal Table
= 0                  
P(X > = 22) = 1 - P(X < 22)
= 1 - 0 = 1