# Mean/Standard Deviation question

• May 25th 2009, 11:58 AM
chobozor
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)?
• May 25th 2009, 01:06 PM
CaptainBlack
Quote:

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
• May 25th 2009, 01:39 PM
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?
• May 25th 2009, 07:18 PM
CaptainBlack
Quote:

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