Thread: Mean/Standard Deviation question

I need help on this question, I don't know what equation to use to find the solution.

Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with a mean 2.2 ounces and standard deviation 0.04 ounces.

Question 1) If the manufacturer wants to adjust the production process so that only 1 candy in 1000 weights less than the advertised weight, what should the mean of the actual weights be(assuming that the standard deviation of the weights remain 0.04 ounces)?

Originally Posted by chobozor

I need help on this question, I don't know what equation to use to find the solution.

Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with a mean 2.2 ounces and standard deviation 0.04 ounces.

Question 1) If the manufacturer wants to adjust the production process so that only 1 candy in 1000 weights less than the advertised weight, what should the mean of the actual weights be(assuming that the standard deviation of the weights remain 0.04 ounces)?

The z-score such that P(Z<z)=0.999 is z=3.091, so the required mean is:

m=2.13+3.091*0.04

CB

why is it 0.999 instead of using 0.001? i though we were looking for 1 out of 1000 isnt 0.999 out of 1 means 999 out of 1000?

Originally Posted by chobozor

why is it 0.999 instead of using 0.001? i though we were looking for 1 out of 1000 isnt 0.999 out of 1 means 999 out of 1000?

P(Z>z)=1-P(Z<z)

so if P(Z<z)=0.999, then P(Z>z)=0.001

CB