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

(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      

step-by-step explanation:

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!