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.
