I am stuck on this question, I don't exactly know how to approach it. Thanks for any advice.
The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed. Suppose that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.1 ounce.
a) At what value should the mean be set so that 99.9% of all cans exceed 12 ounces?
b) At what value should the mean be set so that 99.9% of all cans exceed 12 ounces if the standard deviation can be reduced to 0.05 fluid ounce?
Thanks. I have another quick question. It isn't related to one posted earlier. I want to know if I am doing this correctly.
Assume X is normally distributed with a mean of 13 and a standard deviation of 2. Determine the value for x that solves each of the following
If the question asks P(X > x) = 0.95
My steps(not sure if they are correct)
P(X > x) = 0.95
1 - P(X < x) = 0.95
1 - P(Z < (x - 13)/2) = 0.95
P(Z < (x - 13)/2) = 0.05
now, looking at the z-chart to find the value which corresponds to the probability 0.05 I find it to be approximately -1.64. So,
(x - 13)/2 = -1.64, x = 9.72
Are these steps correct? Or is something wrong? Thanks.