Question 4 An insurance company offers its policyholders a number of different payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: F(x)= 0, ifr <0 F(x)=0.3, if 0≤x<3 F(x)=0.55, if 3 6 Using just the cdf, compute P(3≤X ≤ 6). O 0.8 O 0.3 O 0.25 O something else O 0.5

A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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**Question 4**

An insurance company offers its policyholders a number of different payment options. For a randomly selected policyholder, let \( X \) = the number of months between successive payments. The cdf of \( X \) is as follows:

- \( F(x) = 0 \), if \( x < 0 \)
- \( F(x) = 0.3 \), if \( 0 \leq x < 3 \)
- \( F(x) = 0.55 \), if \( 3 \leq x < 4 \)
- \( F(x) = 0.8 \), if \( 4 \leq x \leq 6 \)
- \( F(x) = 1 \), if \( x > 6 \)

Using just the cdf, compute \( P(3 \leq X \leq 6) \).

- \( \bigcirc \) 0.8
- \( \bigcirc \) 0.3
- \( \bigcirc \) 0.25
- \( \bigcirc \) something else
- \( \bigcirc \) 0.5
Transcribed Image Text:**Question 4** An insurance company offers its policyholders a number of different payment options. For a randomly selected policyholder, let \( X \) = the number of months between successive payments. The cdf of \( X \) is as follows: - \( F(x) = 0 \), if \( x < 0 \) - \( F(x) = 0.3 \), if \( 0 \leq x < 3 \) - \( F(x) = 0.55 \), if \( 3 \leq x < 4 \) - \( F(x) = 0.8 \), if \( 4 \leq x \leq 6 \) - \( F(x) = 1 \), if \( x > 6 \) Using just the cdf, compute \( P(3 \leq X \leq 6) \). - \( \bigcirc \) 0.8 - \( \bigcirc \) 0.3 - \( \bigcirc \) 0.25 - \( \bigcirc \) something else - \( \bigcirc \) 0.5
Expert Solution
Step 1: Definitions

CDF : Cummulative distribution function is defined as follows :

F left parenthesis x right parenthesis space equals space P left parenthesis X less or equal than x right parenthesis

CDF of a random variable X is defined as has following properties :

1. F left parenthesis a less than X less or equal than b right parenthesis space equals space F left parenthesis b right parenthesis space minus space F left parenthesis a right parenthesis
2. text F is right continious  end text
3. F left parenthesis negative infinity right parenthesis space equals space 0 space a n d space F left parenthesis plus infinity right parenthesis space equals space 1


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