If this is correct the answer would be
y^2 + 0.1y + 0.0025
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
You just have to set the denominator to zero and solve for x
