A contractor is required by a county planning department to submit anywhere from
ID: 3220269 • Letter: A
Question
A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let y be the number of forms required of the next applicant. The probability that y forms are required is known to be proportional to y; that is, p(y) = ky for y = 1, ..., 6. (a) What is the value of k? (b) What is the probability that at most three forms are required? (Answer as an exact fraction or) P(at most three forms are required) = 6/15 (c) What is the probability that between two and four forms (inclusive) are required? (Answer as an exact fraction or) P(between two and four forms (inclusive) are required) = 9/15 (d) Could p(y) = y^2/90 for y = 1, 2, 3, 4, 5, 6 be the probability distribution of y? Yes NoExplanation / Answer
(a) Given:
p(y) = k y,
y = 1,2,3,4,5
NOTE: In the question, it is written: y = 1,2,..,6. But, it shound be y =1,2,..5 since one to five forms are there.
p(1) = y
p(2) = 2 y
p(3) = 3 y
p(4) = 4 y
p(5) = 5 y
Total probability = 1
So, adding,
15 k = 1
So,
k = 1/15 = 0.067
(b) The pdf of y is given by:
y p(y)
1 1/15
2 2/15
3 3/15
4 4/15
5 5/15
-----------------
Total 1
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Probability that atmost 3 forms are required = p(1) + p(2) + p(3) = 1/15 + 2/15 + 3/15 = 6/15 = 2/5 = 0.4
(c) Probability between 2 and 4 forms are required = p(2) + p(3)+p(4) = 2/15 + 3/15 + 4/15 = 9/15 = 0.6
(d) Given:
p(y) = y2/90,
y = 1,2,3,4,5
So, pdf of y is giiven by:
y p(y)
1 1/90
2 4/90
3 9/90
4 16/90
5 25/90
6 36/90
----------------------
Total 91/90
-------------------
Since the total probability is not 1, this cannot be pdf.
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