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

To find the 80th percentile of the distribution of the average, we apply the Normal approximation to the distribution of the average of the 80 parts. The standard deviation of the sample average is calculated using the standard deviation of the original distribution and the square root of the sample size. The value of c is found by subtracting and adding the standard deviation of the sample average to the value of x that corresponds to the 80th percentile.

Step-by-step explanation:

To find the 80th percentile of the distribution of the average, we need to apply the Normal approximation to the distribution of the average of the 80 parts. Since the distribution of the length of time the part can work is Exponential, the mean (E(X)) of the sample average is equal to the mean of the original distribution, which is 26 days. To find the value of c, we need to calculate the standard deviation of the sample average. The standard deviation of the sample average (standard error) is equal to the standard deviation of the original distribution divided by the square root of the sample size. Since the distribution is Exponential and the variance (σ²) of an Exponential distribution is equal to the square of the mean (μ), the standard deviation (σ) of the original distribution is also 26 days. Therefore, the standard deviation of the sample average is 26 / √80 = 2.9067 days.

To find the 80th percentile of the distribution of the average, we can use the z-score formula: z = (x - μ) / σ, where z is the z-score, x is the value of interest, μ is the mean, and σ is the standard deviation. Since we want to find the value of x that corresponds to the 80th percentile, we need to find the z-score that corresponds to the 80th percentile. We can use a standard normal distribution table or a calculator to find this z-score. Once we have the z-score, we can use it to find the corresponding value of x using the formula x = z * σ + μ. Finally, we can find the 80th percentile of the distribution of the average by subtracting and adding the value of c (2.9067) from the value of x.

Answer 2

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) = (1)/(n) \sum \limits^n_(i=1) E( {x_i})= 26`

`V(\bar X) = (1)/(n^2) \sum \limits ^n_(i=1) U(X_i) = ((26)^L)/(n)`

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

`\bar X \approx N (26, ((26))/(n)^L)`

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

However, let's determine c such that:

`P (-c \leq \bar x \leq c) = 0.8 \\ \\ P ( (-c-26)/((26)/(\sqrt n)) \leq (\bar x -26 )/((26)/(n) ) \leq (c-26)/((26)/(\sqrt n))) =0.8`

`P ( (-c-26)/((26)/(\sqrt n)) \leq Z \leq (c-26)/((26)/(\sqrt n))) =0.8`

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

`P ( Z > (c-26)/((26)/(\sqrt n))) =0.15 = P ( Z < (-c-26)/((26)/(\sqrt n)))`

`(c-26)/((26)/(\sqrt n))= 1.036`

`c= 26+(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)