A contractor is required by a county planning department to submit one, two, three, four, five, six, or seven forms (depending on the nature of the project) in applying for a building permit. Let Y = 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, , 7.
A) What is the value of c?
B) What is the probability that at most three forms are required?
C) What is the probability that between two and four forms (inclusive) are required?
D) Could pX(x) = x^2/50 for x = 1, . . . , 5 be a probability distribution of X? Explain.

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2021-02-16T03:51:20+01:00

Answer:

Step-by-step explanation:

Given that:

P(Y) = ky

where;

y =1,2,...7

To find the value of c or k (constant)

[tex]\sum P(Y) = 1[/tex]

[tex]\sum \limits_{y \to 1}^7 k*y = 1[/tex]

= k(1+2+3+4+5+6+7) = 1

28k = 1

[tex]k = \dfrac{1}{28}[/tex]

b) The required probability is P ( X ≤ 3)

[tex]P(X \le 3) = \sum \limits^3_{y=1 } P(y)[/tex]

[tex]P(X \le 3) = \sum \limits^3_{y=1 } \dfrac{1}{28}(y)[/tex]

[tex]P(X \le 3) = \dfrac{1}{28} (1 +2+3)[/tex]

[tex]P(X \le 3) = \dfrac{6}{28}[/tex]

P ( X ≤ 3) = 0.2143

c) The required probability P(2 ≤ Y ≤ 4)

[tex]P(2 \le Y \le 4) = \sum \limits ^4_{y=2} P(Y)[/tex]

[tex]P(2 \le Y \le 4) = \sum \limits ^4_{y=2} \dfrac{1}{28}(Y)[/tex]

[tex]P(2 \le Y \le 4) = \dfrac{1}{28}(2+3+4)[/tex]

[tex]P(2 \le Y \le 4) = 0.3214[/tex]

d) The required probability:

[tex]P(X) = \dfrac{x^2}{50} ; \ \ \ \ where; \ x= 1,2,...5[/tex]

[tex]\sum \limits ^5_{y =1} P(Y)= \sum \limits ^5_{y =1} \dfrac{1}{50}(x)^2[/tex]

[tex]\sum \limits ^5_{y =1} P(Y)= \sum \limits ^5_{y =1} \dfrac{1}{50}(1+4+9+16+25)[/tex]

[tex]\sum \limits ^5_{y =1} P(Y)=1.1[/tex]