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The mean life of a Chevy Volt battery (normally distributed) is $1000$ hours and the standard deviation is $100$. How many hours should GM warranty the battery for so that it has to replace the battert $0.5\%$ of the time?


I tried to use the standard units formula $z = x-\mu/ \sigma$

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A. What have you tried ? B. What about the noraml (or any, in general) distribution do you know ? – Belgi Mar 7 '13 at 1:48
Normal distribution????? Yes!!!!!!!! Maybe?????? – Asaf Karagila Mar 7 '13 at 1:50
I tried to use the standard units formula z = x-μ/ σ – user65434 Mar 7 '13 at 2:10

We give first a quite formal version. Then in a remark, we give a quick informal version.

Let random variable $X$ represent the life of a randomly chosen battery. We want to find the number $a$ such that $\Pr(X\le a)=0.005$.

Recall that if $X$ is normal, mean $\mu$ and standard deviation $\sigma$, then $\frac{X-\mu}{\sigma}$ has standard normal distribution $Z$.

In our case, $$\Pr(X\lt a)=\Pr\left(\frac{X-1000}{100}\lt \frac{X-1000}{100}\right)=\Pr\left(Z\lt \frac{1000-a}{100}\right).$$

Look up, in tables of the standard normal, the place $c$ such that $\Pr(Z\lt c)=0.005$. In most tables, you cannot do that, since the $c$ is negative.

We want the place $c$ such that the probability of the left tail is $0.005$. To find this, we find the place $d$ such that $\Pr(Z\gt d)=0.005$, and change sign. So look for the place $d$ such that $\Pr(Z\le d)=0.995$.

The standard normal table says that $d\approx 2.57$. So $c\approx -2.57$, and we set $$\frac{1000-a}{100}=-2.57.$$

Remark: We want the place on the normal that has $99.5\%$ of the area above it. In the standard normal, the place has has $99.5\%$ of the area below it is at $2.57$.

So in our case, we want to go $2.57$ standard deviation units below the mean. Our guarantee point should be at $1000-(2.57)(100)$.

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Thank You I will try this – user65434 Mar 7 '13 at 2:47

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