# Finding probability 400 components last 5012 or less hours using standard deviation?

The lifetime of certain electronic components is a random variable with the expectation of 5000 hours and a standard deviation of 100 hours. What is the probability that the average lifetime of 400 components is less than 5012 hours?

When I calculated this answer I was not sure how to include 400 items into what I know about standard deviation but assuming the probability works much like calculating consecutive coin flips (Ex: 1 heads = 1/2^1, 2 heads = 1/2^2, n heads = 1/2^n) I just used the standard deviation of .12 and then added 400 as the exponent to get a ridiculously small number which makes me feel very unconfident.

My assumption:

P(400 items <= 5012 hours) = I(.12)^400 = .5478^400 = 2.810678e-105

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Possibly helpful: If the failure times for the components are independent, then $Pr(400 fail <5012)=(Pr(1fail<5012))^{400}$. – Daryl Nov 8 '12 at 21:38
@Daryl: No, because there are many ways to choose which $400$ fail. – Ross Millikan Nov 8 '12 at 21:39
@Ross, Jeremy Out of how many then are you choosing the 400 to fail? In response to your comment, this is likely to be important. – Daryl Nov 8 '12 at 22:17
hint: use chebyshev's inequality – jay-sun Nov 8 '12 at 22:18

Hint: Let the individual lifetimes be $X_1, X_2, X_3,\dots, X_n$ where $n=400$. Then the average lifetime $Y$ is given by $$Y=\frac{1}{n}\left(X_1+X_2+X_3+\cdots+X_n\right).$$ The mean of $Y$ is just $\dfrac{1}{n}$ times the sum of the means of the $X_i$. Each $X_i$ has mean $5000$, so the mean of $Y$ is $5000$.

The variance of an independent sum is the sum of the variances, and the variance of $kW$ is $k^2$ times the variance of $W$. It follows after a while that the variance of $Y$ is $\dfrac{10000}{400}$, and therefore the standard deviation of $Y$ is $\dfrac{100}{20}$.

Finally and most importantly, by the Central Limit Theorem we can probably safely assume that $Y$ has nearly normal distribution. I expect you can handle the rest. What is the probability that a normal with mean $5000$ and standard deviation $5$ is less than $5012$?

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I think mean of 5000 for 400 random variables is a bit simplistic. Isn't it? I would use chebyshev's inequality (and I am pretty sure we can solve this via that) – jay-sun Nov 8 '12 at 22:25
Certainly can use the Chebyshev Inequality. It will often give a result which is of course correct, but much less accurate than the normal approximation. The Chebyshev Inequality will also use the fact that the mean of $Y$ is $5000$, and that the variance is $10000/400$. So it cannot be that you think these simplistic. If you think normal approximation is simplistic, well, as a rule of thumb $400$ is very safe for normal approximation. – André Nicolas Nov 8 '12 at 22:38