Thread: Statistics Problem

A manufacturer of a certain component requires 3 different machining operations. Machining time for each operation has a normal distribution and the three times are independent of one another. The mean values are 15, 30, and 20 minutes respectively and the standard deviations are 1, 2, and 1.5 minutes respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

I do not know where to start on this problem, and would greatly appreciate help in solving it. Thank-You.

Originally Posted by jpmckell

A manufacturer of a certain component requires 3 different machining operations. Machining time for each operation has a normal distribution and the three times are independent of one another. The mean values are 15, 30, and 20 minutes respectively and the standard deviations are 1, 2, and 1.5 minutes respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

I do not know where to start on this problem, and would greatly appreciate help in solving it. Thank-You.

The machining time of a random component is normally distributed with mean:

mu=15+30+20 minutes

with standard deviation:

sigma=sqrt{1^2+2^2+1.5^2} minutes

RonL

How do you start with that information, and end up finding the probability that the machining time takes 1 hour or less?

Originally Posted by jpmckell

How do you start with that information, and end up finding the probability that the machining time takes 1 hour or less?

jpmckell,

Calculate zscore, do you know how to do that? Then look up the value in a table. If you don't have that table for the normal distribution, I can provide you my calculator for that number.

I do not know how to calculate zscore and do not have the table available.

Originally Posted by jpmckell

I do not know how to calculate zscore and do not have the table available.

Z-Score

For our problem,

From what CaptainBlack gave you, we have:





Do you have Access to Excel? If so, do this in a cell:

=NORMDIST(60,65,7.25,TRUE)

That is your probability.

If not, I can code up my website quickly to do this negative Z score.

I finished the problem and got a probability of 24.5206%. Thank you for all your help.

Originally Posted by jpmckell

I finished the problem and got a probability of 24.5206%. Thank you for all your help.

I would not quote more than three significant digits for an answer to such a question.

RonL