# A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

###### Question:

(a) What is the value of k?

(b) What is the probability that at most three forms are required?

(c) What is the probabilty that between two and four forms (inclusive) are required?

(d) Could p(y)=y^2/50 for y=1,...,5 be the pmf of Y?

## Answers

a) c = 1/15

b) =2/5

c) = 3/5

d) No, it cannot be the PMF of y.

Step-by-step explanation:

We have that

[tex]p_X(x)\geq0[/tex]

(a) The value of k is [tex]\frac{1}{15}[/tex].

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) [tex]P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5[/tex] is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

[tex]P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right[/tex]

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

[tex]\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}[/tex]

Thus, the value of k is [tex]\frac{1}{15}[/tex].

(b)

Compute the value of P (Y ≤ 3) as follows:

[tex]P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40[/tex]

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

[tex]P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60[/tex]

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for [tex]P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5[/tex] to be the pmf of Y it has to satisfy the conditions:

[tex]P(y)=\frac{y^{2}}{50}0;\ for\ all\ values\ of\ y \\[/tex][tex]\sum P(y)=1[/tex]Check condition 1:

[tex]y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.020\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.080 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.180\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.320 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.500[/tex]

Condition 1 is fulfilled.

Check condition 2:

[tex]\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.11[/tex]

Condition 2 is not satisfied.

Thus, [tex]P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5[/tex] is not the pmf of y.