Solution Found!
In the previous exercise, suppose that the mean of the
Chapter 4, Problem 77E(choose chapter or problem)
Problem 77E
In the previous exercise, suppose that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.1 fluid ounce.
(a) At what value should the mean be set so that 99.9% of all cans exceed 12 fluid ounces?
(b) At what value should the mean be set so that 99.9% of all cans exceed 12 fluid ounces if the standard deviation can be reduced to 0.05 fluid ounce?
Questions & Answers
QUESTION:
Problem 77E
In the previous exercise, suppose that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.1 fluid ounce.
(a) At what value should the mean be set so that 99.9% of all cans exceed 12 fluid ounces?
(b) At what value should the mean be set so that 99.9% of all cans exceed 12 fluid ounces if the standard deviation can be reduced to 0.05 fluid ounce?
ANSWER:
Solution:
Step 1 of 2:
We have the standard deviation of 0.1 fluid ounce.
- The claim is to find the mean that 99.9% of all cans exceed 12 fluid ounces.
Then, P(X > 12) = 0.999
P(X > ) = 0.999
= 1 - 0.999
=
(0.001)
= -3.09 ( from the area under normal curve)
12 - = -0.309
= 12.309
Hence, = 12.309.