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The cumulative distribution function of X is given by

{F(x) =

0 if x < 1,

0.1 if 1 ≤ x < 2,

0.4 if 2 ≤ x < 3,

0.6 if 3 ≤ x < 4,

q if 4 ≤ x < 5,

0.9 if 5 ≤ x < 6,

1 if x ≥ 6}

You are also told that P(4 < X^2 ≤ 16) = 0.4

What is the value of q?

I have the answer as P(4 < X^2 ≤ 16) = P(2 < X ≤ 4) = P(X = 3) + P(X = 4) = 0.2 + q − 0.6 = q − 0.4

Hence q = 0.8

Can someone explain why this is the answer to me? I keep thinking that P(X=3) = 0.6 and P(X=4) = q so 0.6 + q = 0.4 but this answer doesn't mathematically make sense, I know.

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    Instead, think $P(X \le 3) = 0.6.$ – 2017-02-26
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    are you given the support of $X$? Because it seems strange to define the CDF in that way if it's a discrete variable taking values in $1, 2, \dots 6$. If it doesn't have that support, however, then the solution is wrong. – 2017-02-26

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