Answer:
a. The probability that a sample of 50 bottles will have a mean volume less than 32 ounces is 0.0000
.
b. The number of times  we expect the quality department to shut down the machine and hold that shift’s production = 0.
Step-by-step explanation:
a. Letting X be the volume of orange juice in a bottle with u = 32.082,0 = 0.01 
 
The distribution of X was not given, 
 
Let \bar{X} be the sample mean of 50 bottles 
 
Using the Central Limit Theorem, the sampling distribution of the sample mean \bar{X} follow Normal with 
 
mean = 32.080 ( population mean ) 
 
Standard error = o/Vn=0.01/V50 = 0.0014 
 
as the sample size is large 
 
that is, X~ N(32.082, 0.00142 
 
then _X - 32.082 ~ N(0.1) 0.0014 
 
Finding the value of P(X <32) 
 
P(X <32) 
 
= P:<- 32 - 32.082 0.0014 
 
= P(z < -58.57) 
 
P(X <32) = 0.0000 (from z table) 
 
The probability that a sample of 50 bottles will have a mean volume of fewer than 32 ounces is 0.0000.
 
b. The number of times we expect the quality department to shut down the machine 
= 1008*P(X <32) 
= 1008*(0) 
= 0
Thus, the number of times we expect the quality department to shut down the machine and hold that shift’s production = 0.