An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X= the number of months between successive payments. The cdf of X is as follows: 0 0.38 15x<3 0.51 35x<4 0.53 45x<6 0.83 65x< 12 12 % x (a) What is the pmf of X? 1 F(x)= p(x) (b) Using just the cdf, compute P(3 s X s 6) and P(4 s X). P(3 ≤ x ≤ 6) = P(4 SX)= 6 12
An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X= the number of months between successive payments. The cdf of X is as follows: 0 0.38 15x<3 0.51 35x<4 0.53 45x<6 0.83 65x< 12 12 % x (a) What is the pmf of X? 1 F(x)= p(x) (b) Using just the cdf, compute P(3 s X s 6) and P(4 s X). P(3 ≤ x ≤ 6) = P(4 SX)= 6 12
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.9: Combining Functions
Problem 27PPS
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![An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows:
0
x < 1
1<x<3
0.38
0.51 3 ≤ x < 4
0.53 4 ≤ x < 6
6 ≤ x < 12
0.83
1
12 ≤ x
F(x) =
(a) What is the pmf of X?
1
X
p(x)
3
(b) Using just the cdf, compute P(3 ≤ x ≤ 6) and P(4 ≤ X).
P(3 ≤ x ≤ 6) =
P(4 ≤ X) =
4
6
12](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1507ef8-c387-4f95-9ab7-d384c0c17d2a%2F70824585-94cc-41a5-b8df-c81f57912cd5%2Fr6i7dwb_processed.png&w=3840&q=75)
Transcribed Image Text:An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows:
0
x < 1
1<x<3
0.38
0.51 3 ≤ x < 4
0.53 4 ≤ x < 6
6 ≤ x < 12
0.83
1
12 ≤ x
F(x) =
(a) What is the pmf of X?
1
X
p(x)
3
(b) Using just the cdf, compute P(3 ≤ x ≤ 6) and P(4 ≤ X).
P(3 ≤ x ≤ 6) =
P(4 ≤ X) =
4
6
12
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