Question

An individual who has automobile insurance from a certain company is randomly selected. Let X be the number of moving violations for which the individual was cited during the last 3 years. The probability mass function of X is given below as; Number of moving violations (X) 0 1 2 3 4 P(X) 0.17 0.23 0.27 0.24 0.09 What is the cumulative distribution function

Answer 1

(X) 0 1 2 3 4

P(X) 0.17 0.23 0.27 0.24 0.09

F(x) 0.17 0.04 0.65 0.91 1

Step-by-step explanation:

Given that;

(X) 0 1 2 3 4

P(X) 0.17 0.23 0.27 0.24 0.09

cumulative distribution function can be calculated by; be cumulatively up the value of p(x) with the values before it;

so

x F(x)

0 P(X = 0) = 0.17

1 P(X = 0) + P(X = 1) = 0.17 + 0.23 = 0.4

2 P(X = 0) + P(X = 1) + P(X = 2) = 0.17 + 0.23 + 0.27 = 0.65

3 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.17 + 0.23 + 0.27 + 0.24 = 0.91

4 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.17 + 0.23 + 0.27 + 0.24 + 0.09 = 1

Therefore, cumulative distribution function f(x) is;

(X) 0 1 2 3 4

P(X) 0.17 0.23 0.27 0.24 0.09

F(x) 0.17 0.04 0.65 0.91 1