Refer to exercise 5.10 on page 238 of the text book where you are told that the
ID: 3206172 • Letter: R
Question
Refer to exercise 5.10 on page 238 of the text book where you are told that the arrival time of requests to a web server within each hour can be modeled by a uniform distribution. The number of seconds from the start of the hour that a (randomly selected) request is made is uniformly distributed between 0 seconds and 3, 600 seconds. Answer the questions below (not the questions in the textbook). What is the probability that this request is made to the web server between 100 and 500 seconds from the start of the hour? Round your final answer to 4 decimals, if necessary. What is the probability that this request is made to the web server within 1,000 seconds from the start of the hour? Round your final answer to 4 decimals, if necessary. What is the probability that this request is made to the web server sometime within the last 15 minutes of the hour? Round your final answer to 4 decimals, if necessary. Let x be the number of seconds from the start of the hour that this randomly selected request is made. What is the 38^th percentile of the distribution of x? Round your final answer to 1 decimal, if necessary. What is the mean (i.e. expected value) of x? Round your final answer to 1 decimal, if necessary. What is the standard deviation of x? Round your final answer to 1 decimal, if necessary. Now suppose that you will select a random sample of size n = 40 requests. Describe the sampling distribution of x, the sample mean number of seconds from the start of the hour of these 40 requests. Include the shape, mean, and standard deviation of this distribution. Round any final values to 1 decimal, if necessary. What is the probability that the sample mean number of seconds from the start of the hour, x, will be between 1, 500 seconds and 1, 700 seconds? Round your answer to 4 decimals, if necessary.Explanation / Answer
from uniform distribution P(x1<X<x2)=(x2-x1)/(3600-0) =(x2-x1)/3600
a)P(100<X<500) =(500-100)/3600 =0.1111
b)P(X<1000) =1000/3600=0.2778
c)as 15 minutes =900 seconds
hence probability =900/3600=0.25
d)as 38th percentile =0.38 =x/3600
x=3600*0.38 =1368 seconds
e) as mean of uniform distribution=(a+b)/2 =(0+3600)/2 =1800
f)also std deviation=(b-a)/(12)1/2 =3600/(12)1/2=1039.23
g) as per central limit theorum shape of this distrbution will be normal
mean =1800
std deviation =population std deviation/(n)1/2 =1039.23/(40)1/2 =164.3
(ii)P(1500<X<1700) =P((1500-1800)/164.3<Z<(1700-1800)/164.3) =P(-1.8257<Z<-0.6086) =0.2714-0.0339=0.2375