Question

It is claimed that the expected length of time some computer part may work before requiring a reboot is 26 days. In order to examine this claim, 80 identical parts are set to work. Assume that the distribution of the length of time the part can work (in days) is Exponential. The 80% percentile of the distribution of the average is of the form E(X) ± c, where E(X) is the expectation of the sample average. The value of c is ________. (You are asked to apply the Normal approximation to the distribution of the average of the 80 parts that are examined. The answer may be rounded up to 3 decimal places of the actual value.)

Answer 1

Answer:

3.012  

Step-by-step explanation:

Let use x to represent the length of time  some computer part may work

where ; we have a sample of  size n = 80

Say $X_1, X_2, ... X_{80$

$X_i$  $\approx$ $exp (\lambda)$   where ( λ =26 days)

$E( \bar X) = \frac{1}{n} \sum \limits^n_{i=1} E( {x_i})= 26$

$V(\bar X) = \frac{1}{n^2} \sum \limits ^n_{i=1} U(X_i) = \frac{(26)^L}{n}$

So find $\bar X 's$ distribution using the Normal approximation.

$\bar X \approx N (26, \frac{(26)}{n}^L)$

$Z = \dfrac{\bar {X} - 26}{\frac{26}{\sqrt n}} \approx N (0,1)$

However, let's determine c such that:

$P (-c \leq \bar x \leq c) = 0.8 \\ \\ P ( \dfrac{-c-26}{\frac{26}{\sqrt n}} \leq \dfrac{\bar x -26 }{\frac{26}{n} } \leq \dfrac{c-26}{\frac{26}{\sqrt n}}) =0.8$

$P ( \dfrac{-c-26}{\frac{26}{\sqrt n}} \leq Z \leq \dfrac{c-26}{\frac{26}{\sqrt n}}) =0.8$

where; $Z \approx N (0,1)$

$P ( Z > \dfrac{c-26}{\frac{26}{\sqrt n}}) =0.15 = P ( Z < \dfrac{-c-26}{\frac{26}{\sqrt n}})$

$\dfrac{c-26}{\frac{26}{\sqrt n}}= 1.036$

$c= 26+\frac{26}{\sqrt 80}}* 1.036$

c = 29.0115

Given that :

The 80% percentile of the distribution of the average is of the form

E(X) + c

Then;

29.0115-26 = 3.0115

≅ 3.012    (to 3 decimal place)