Answer to Question #95786 in Statistics and Probability for Sam

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 r.v. X = the number of forms required of the next applicant. The probability that x forms are required is known to be proportional to x; that is, pX(x) = cx for x = 1, . . . , 5.
(a) (1 mark). What is the value of c?
(b) (1 mark). What is the probability that at most three forms are required?
(c) (1 mark). What is the probability that between two and four forms (inclusive) are required?
(d) (2 marks). Could pX(x) = x^2/50 for x = 1, . . . , 5 be a probability distribution of X? Explain.

Answer 1

a) We have that

$$p_X(x)\geq0, x=1,2,3,4,5$$ $$\displaystyle\sum_{i=1}^5p_X(x_i)=1$$

Given that $p_X(x)=cx, x=1,2,3,4,5.$ Then

$$c\cdot1+c\cdot2+c\cdot3+c\cdot4+c\cdot5=1$$ $$15\cdot c=1$$ $$c={1 \over 15}$$

b) The probability that at most three forms are required

$$P(X\leq3)=P(X=1)+P(X=2)+P(X=3)=$$ $$={1 \over 15}\cdot1+{1 \over 15}\cdot2+{1 \over 15}\cdot3={2 \over 5}$$

c) The probability that between two and four forms (inclusive) are required

$$P(2\leq X\leq4)=P(X=2)+P(X=3)+P(X=4)=$$ $$={1 \over 15}\cdot2+{1 \over 15}\cdot3+{1 \over 15}\cdot4={3 \over 5}$$

d) Check that $\displaystyle\sum_{i=1}^5p_X(x_i)=1$

$${1^2 \over 50} +{2^2 \over 50}+{3^2 \over 50}+{4^2 \over 50}+{5^2 \over 50}={11 \over 10}\not=1$$

Therefore, no. A probability distribution of $x$

$$p_X(x)={x^2 \over 50}, x=1,2,3,4,5$$

could not be.

No.