Refer to exercise 5.10 on page 238 of the text book where you are told that the
ID: 3205882 • 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 >s uniformly distributed between 0 seconds and 3, 600 seconds. Answer the questions below (not the questions in the textbook). a) What is the probability that this request is made to the web server between seconds from the start of the hour? Round your final answer to 4 decimals, n necessary. b) 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. c) 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. d) 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. e) What is the mean (i.e. expected value) of x? Round your final answer to 1 decimal, if necessary. f) What is the standard deviation of x? Round your final answer to 1 decimal, if necessaryExplanation / Answer
PDF of Uniform Distribution f(x) = 1 / ( b - a ) for a < x < b
b = Maximum Value
a = Minimum Value
Mean = a + b / 2
Standard Deviation = Sqrt ( ( b - a ) ^ 2 / 12 )
f(x) = 1/(b-a) = 1 / (3600-0) = 1 / 3600 = 0.0003
a.
To find P(a < X < b) =( b - a ) * f(x)
P(100 < X < 500) = (500-100) * f(x)
= 400*0.0003
= 0.12
b.
P(X < 1000) = (1000-0) * f(x)
= 1000*0.0003
= 0.3
c.
P( last 15 minuetes) = P(X > 2100) = (3600-2100) * f(x)
= 1500*0.0003
= 0.45
e & f.
Mean = a + b / 2 = 1800
Standard Deviation = Sqrt ( ( b - a ) ^ 2 / 12 ) = 1039.23