[SOLVED] Normal Distrubtion...

**looi76** · April 19th 2008, 04:28 AM

Question:
The fluorescent light tubes made by the company Well-lit have lifetimes which are normally distrusted with mean 2010 hours and standard deviation 20 hours. The company decides to promote its sales of the tubes by guaranteeing a minimum life. If the company wishes to have to replace free only 3% if the tubes sold, find the guaranteed minimum it must set.

Attempt:

$\mu = 2010$ , $\sigma = 20$

How can I get the value of $\phi^{-1} (0.3\%)$ out of the table?

**flyingsquirrel** · April 19th 2008, 05:17 AM

Hi

Let's call $T$ the random variable associated to the lifetime.
We're looking for $t=\phi^{-1}(3\%)$ which means $P(T\leq t)=\int_{-\infty}^t\frac{1}{\sigma\sqrt{2\pi}}\exp \frac{(x-\mu)^2}{\sigma^2}\,\mathrm{d}x=0.997$

As we know the value taken by the normal distribution for $\sigma=1$ and $\mu=0$ , we substitute $u=\frac{x-\mu}{\sigma} \Rightarrow \mathrm{d}u=\frac{\mathrm{d}x}{\sigma}$ :
$P(T\leq t)=\int_{-\infty}^{\frac{t-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp u^2\,\mathrm{d}u=0.997$

Using the table, you can find the value of $\frac{t-\mu}{\sigma}$ corresponding and then deduce $\phi^{-1}(3\%)=t$ .

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