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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

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Answer:

(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      

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