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Question
The discrete random variable X has the probability function.
| Value of X = x |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
If P(X ≤ x) > `1/2`, then find the minimum value of x.
Solution
We want the minimum value of x for which P(X ≥ x) > `1/2`
Now P(X ≤ 0) = `0 < 1/2`
P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k
= `1/10 < 1/2`
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k = 3k
= `3/10 = 1/2`
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)
= 0 + k + 2k + 2k = 5k
= `5/10 = 1/2`
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)
= 0 + k + 2k + 2k + 3k = 8k
= `8/10 > 1/2`
This Shows that the minimum value of X for which P(X ≤ x) > `1/2` is 4
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