Wutao is working hard at his job in the state of Washington while finishing up t
ID: 3302354 • Letter: W
Question
Wutao is working hard at his job in the state of Washington while finishing up the writing on his PhD dissertation. On days that he spends a full day at his office, he is not able to work on his dissertation. That happens with a probability of 0.1. On the days where he does not go to the office or is only there for part of the day, he will complete up to a maximum of 4 pages. Assume that Wutao will complete 0 to 4 full pages each day. (To simplify we will round partial pages up).
The probability that Wutao writes two pages is twice the probability that he writes one page. The probability he writes three pages is three times the probability that he writes one page. Finally, the probability Wutao writes 4 pages is 4 times the probability that he writes one page.
a) Create a PMF table for the number of pages of his dissertation that Wutao completes in one day.
b) What is the probability that Wutao will complete fewer than 2 pages in a day?
c) On a particular day, Wutao must complete at least 2 pages. Knowing this, what is the probability he writes 4
pages?
d) What is the expected number of pages Wutao will complete in a day?
e) What is the standard deviation of the number of pages Wutao will complete in a day?
Explanation / Answer
from above probability of writing 0 pages =P(X=0) =0.1
probability that Wutao writes two pages is twice the probability that he writes one page or P(X=2) =2*P(X=1)
probability he writes three pages is three times the probability that he writes one page or P(X=3)=3*P(X=1)
also P(X=4)=4*P(X=1)
as sum of probability =1
P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1
0.1+P(X=1)+2*P(X=1)+3*P(X=1)+4*P(X=1)=1
P(X=1) =0.09
P(X=2)=0.18
P(X=3) =0.27
P(X=4) =0.36
a) therefore below is PMF table for the number of pages(X) of his dissertation:
b)probability that Wutao will complete fewer than 2 pages in a day =P(X=0)+P(X=1) =0.1+0.09=0.19
c)
probability he writes 4 pages given he must write at least 2 pages =P(X=4|X>=2) =0.36/(0.18+0.27+0.36)
=0.4444
d)
from above expected number of pages Wutao will complete in a day =E(X) =2.7
e)standard deviation of the number of pages Wutao will complete in a day =(E(X2) -(E(X))2)1/2 =(9-2.72)1/2 =1.3077
x P(x) 0 0.1 1 0.09 2 0.18 3 0.27 4 0.36Related Questions
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