Captain Ts tuna is sold in cans that have a net weight of 80

Captain T\'s tuna is sold in cans that have a net weight of 8.02 ounces. The weights are normally distributed with a mean of 8.015 ounces and a standard deviation of 0.12 ounces. You take a sample of 40 cans. Compute the probability that the sample would have a mean:

a. Greater than 8.01 ounces

b. Less than 8.025 ounces

c. Between 7.995 and 8.04

Solution

Mean ( u ) =8.015
Standard Deviation ( sd )=0.12
Number ( n ) = 40
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X > 8.01) = (8.01-8.015)/0.12/ Sqrt ( 40 )
= -0.005/0.019= -0.2635
= P ( Z >-0.2635) From Standard Normal Table
= 0.6039
b)
P(X < 8.025) = (8.025-8.015)/0.12/ Sqrt ( 40 )
= 0.01/0.019= 0.527
= P ( Z <0.527) From Standard NOrmal Table
= 0.7007                  
c)              
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 7.995) = (7.995-8.015)/0.12/ Sqrt ( 40 )
= -0.02/0.019
= -1.0541
= P ( Z <-1.0541) From Standard Normal Table
= 0.14592
P(X < 8.04) = (8.04-8.015)/0.12/ Sqrt ( 40 )
= 0.025/0.019 = 1.3176
= P ( Z <1.3176) From Standard Normal Table
= 0.90618
P(7.995 < X < 8.04) = 0.90618-0.14592 = 0.7603