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A turbine shaft is made up of $4$ different sections. The lengths of those sections are independent and have normal distributions with $\mu$ and $\sigma$: (8.10, 0.22), (7.25, 0.20), (9.75, 0.24), and (3.10, 0.20). What is the probability an assembled shaft meets the speciications $28\pm 0.26$?

My solution:

When we assemble the shaft we get a normal distribution with $\mu=28.2$ and $\sigma=0.43$. Call this random variable $Y$. Now we are trying to find$$P(27.74\leq Y\leq 28.26)$$which is $$P(-1.069\leq Z\leq 0.1395)=0.3995$$

but the textbook gives $0.4314$. This seems too significant to be a roundoff error in the normal distribution table.

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up vote 3 down vote accepted

The standard deviation is about $0.4312772$.

We want the probability that our normal is between $-0.46$ and $0.06$. This is the probability that the standard normal is between $-1.0666$ and $0.1391$.

The probability of being less than $0.1391$ is about $0.557$. The probability of being less than $-1.0666$ is about $0.142$. Subtract.

Remark: The official answer is off by quite a lot. It is very close to the actual standard deviation. This may be accidental, or that intermediate result may have been written down by mistake.

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I get something like $0.412...$. – copper.hat Apr 1 '13 at 16:17
@copper.hat: I got my number by using table crudely. Out of curiosity asked my new buddy Alpha. He says about $0.412$. – André Nicolas Apr 1 '13 at 16:24
I used the norm.cdf from the Python binding of the scipy package. The sensitivity of the answer to the (original) upper bound is $\approx 0.9$, and the lower bound is $\approx -0.5$, so the answer is fairly sensitive to the upper bound, in particular. I suspect either your explanation or a small data entry mistake... – copper.hat Apr 1 '13 at 16:45
Hmm. I'm using a printed-out big sheet from our course notes... – user54609 Apr 1 '13 at 21:14
Hmm. I got $0.431277173$ too. How did copper.hat get 0.412? I just did $\sqrt{0.22^2 + 0.20^2 + 0.24^2 + 0.20^2}$ – user54609 Apr 1 '13 at 21:44

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