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 [tex]X_1, X_2, ... X_{80[/tex]
[tex]X_i[/tex] [tex]\approx[/tex] [tex]exp (\lambda)[/tex] where ( λ =26 days)
[tex]E( \bar X) = \frac{1}{n} \sum \limits^n_{i=1} E( {x_i})= 26[/tex]
[tex]V(\bar X) = \frac{1}{n^2} \sum \limits ^n_{i=1} U(X_i) = \frac{(26)^L}{n}[/tex]
So find [tex]\bar X 's[/tex] distribution using the Normal approximation.
[tex]\bar X \approx N (26, \frac{(26)}{n}^L)[/tex]
[tex]Z = \dfrac{\bar {X} - 26}{\frac{26}{\sqrt n}} \approx N (0,1)[/tex]
However, let's determine c such that:
[tex]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[/tex]
[tex]P ( \dfrac{-c-26}{\frac{26}{\sqrt n}} \leq Z \leq \dfrac{c-26}{\frac{26}{\sqrt n}}) =0.8[/tex]
where; [tex]Z \approx N (0,1)[/tex]
[tex]P ( Z > \dfrac{c-26}{\frac{26}{\sqrt n}}) =0.15 = P ( Z < \dfrac{-c-26}{\frac{26}{\sqrt n}})[/tex]
[tex]\dfrac{c-26}{\frac{26}{\sqrt n}}= 1.036[/tex]
[tex]c= 26+\frac{26}{\sqrt 80}}* 1.036[/tex]
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)
