Question

8.6 . In Self-Test Problem $8.5$, how many components would one need to have on hand to be approximately $90$percent certain that the stock would last at least $35$days?

Short answer

The components is $n=56$.

Step-by-step solution

Given information

The components need to have on hand, approximately $90$percent. The stock would last at least $35$days.

Explanation

Let $X_i$denote the lifetime of $i$th component. Understand that $X_1,X_2,\dots$ is a sequence of independent and identically distributed random variables, each having a mean.
$\mu =E\left[X_i\right]={\int }_0^{1}xf\left(x\right)dx={\int }_0^{1}2x^{2}dx={\left.2\frac{x^3}{3}\right|}_0^1=\frac{2}{3}$
Then the variance is,
${\sigma }^2=Var\left(X_i\right)={\int }_0^1{x}^{2}f\left(x\right)dx-{\left(\frac{2}{3}\right)}^2={\int }_0^{1}2x^{3}dx-\frac{4}{9}={\left.2\frac{x^4}{4}\right|}_0^1-\frac{4}{9}=\frac{1}{18}$
Let$S_n=\sum _{i=1}^n{X}_i$indicates the time of $n$th failure.

Verify that the value of $n$is satisfied.
$P\left\{S_n\ge 35\right\}\approx .90?$

Explanation

Using the central limit theorem:

$$\begin{gathered} .90\approx P\left\{S_n\ge 35\right\}=1-P\left\{S_n\le 35\right\}=1-P\left\{\frac{S_n-n\mu }{\sigma \sqrt{n}}\le \frac{35-n\mu }{\sigma \sqrt{n}}\right\}\approx \\ \approx 1-\Phi \left(\frac{35-n\mu }{\sigma \sqrt{n}}\right)\Rightarrow \Phi \left(\frac{35-n\mu }{\sigma \sqrt{n}}\right)\approx .10\Rightarrow \\ \frac{35-n\mu }{\sigma \sqrt{n}}\approx {\Phi }^{-1}(.10)=-1.28 \end{gathered}$$

Then,

$$\begin{gathered} \begin{array}{c}\frac{\sqrt{18}(105-2n)}{3\sqrt{n}}\approx -1.28\end{array} \\ n=56 \end{gathered}$$

Hence, the components $n=56$