Problem 1 A college professor never finishes his lecture before the end of the h

Question

Problem 1 A college professor never finishes his lecture before the end of the hour and always finishes his lecture within 2 minutes after the hour. Let X be the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of X is kr f(r) 0, otherwise (1) (2 points) Find the value of k. (2) (4 points) Compute the cdf of X (3) (2 points) What is the probability that the lecture ends within 1 minute of the end of the hour? (4) (2 points) Compute the expected value of X (5) (2 points) Compute the standard deviatioon of X

Explanation / Answer

Problem 1 Part (1)

Since X is a continuous random variable, integration of the pdf over the range of values of X must be 1. So, for the present problem,

int(0, 2) kx2 must be 1. => (k/3)x3 from 0 to 2 = 1 or  (k/3)23 = 1 or k = 3/8 ANSWER

Problem 1 Part (2)

cdf(t) represents the cumulative probabilty upto t = integarl of pdf from lowest value to t. In this case, cdf(t) = integral of kx2 over (0, t) = (k/3)t3 = t3/8, [substituing the value of k], So, cdf(t) = t3/8 ANSWER

Problem 1 Part (3)

Probabilty of lecture ending within 1 minute = cdf(1) = 1/8 ANSWER

Problem 1 Part (4)

Expected Value = E(X) = integral of {x.f(x)} over the range = integral over (0, 2) of (x.kx2)

= integral over (0, 2) of (kx3) = (k/4)x4 over (0, 2 )= (3/32)24 = 3/2. ANSWER

Problem 1 Part (5)

Variance of X = V(X) = integral of {x2.f(x)} over the range = integral over (0, 2) of (x2.kx2)

= integral over (0, 2) of (kx4) = (k/5)x5 over (0, 2 )= (3/40)25= 12/5. And standard deviation = square root of V(X) = sq.rt of 12/5 ~ 1.549 ANSWER

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