2. A college professor never finishes his lecture before the end of the hour and always finishes his lecture within 2 min 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 f(x) = kx² for 0

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I need help with parts D and E please. 

2. A college professor never finishes his lecture before the end of the hour and always
finishes his lecture within 2 min 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 f(x) = kx²
for 0 <x<2 and f(x) = 0 elsewhere.
(a) Find the value of k.
(b) What is the probability that the lecture ends within 1 min of the end of the hour?
(c) What is the probability that the lecture continues beyond the hour for between 60
and 90 sec?
(d) What is the probability that the lecture continues for at least 90 sec beyond the end
of the hour?
(e) Find the cdf of X, F(x), and repeat parts (b), (c), and (d) using F(x) instead of f (x).
Transcribed Image Text:2. A college professor never finishes his lecture before the end of the hour and always finishes his lecture within 2 min 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 f(x) = kx² for 0 <x<2 and f(x) = 0 elsewhere. (a) Find the value of k. (b) What is the probability that the lecture ends within 1 min of the end of the hour? (c) What is the probability that the lecture continues beyond the hour for between 60 and 90 sec? (d) What is the probability that the lecture continues for at least 90 sec beyond the end of the hour? (e) Find the cdf of X, F(x), and repeat parts (b), (c), and (d) using F(x) instead of f (x).
Expert Solution
Step 1

 Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts

Part A. 

f(x) always integrate to 1.

f(x)dx=1

Plug the value 

02kx2dx=1

Take constant value outside 

k·02x2dx=1

Using Power rule we have :

kx2+12+102=1

kx3302=1

k233-033=1

k83-03=1

k83=1

k=38

 

 

 

 

 

 

 

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