Respuesta :
Answer:
(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}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0[/tex]
Condition 1 is fulfilled.
Check condition 2:
[tex]\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1[/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.
