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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 |
Evaluate p(x < 6), p(x ≥ 6) and p(0 < x < 5)
Solution
P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)
= 0 + k + 2k +2k + 3k + k2
= 8k + k2
= `8(1/10) + (1/10)^2`
= `8/10 + 1/100`
= `(80 + 1)/100`
P(x < 6) = `81/100`
P(x ≥ 6) = P(x = 6) + p(x = 7)
= 2k2 + 7k2 + k
= 9k2 + k
= `9(1/10)^2 + 1/10`
= `9/100 + 1/10`
= `(9 + 10)/100`
= `19/100`
∴ P(x ≥ 6) = 19/100`
P(0 < x < 5) = P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)
= k + 2k + 2k + 3k
= 8k
= `8(1/10)`
∴ P(0 < x < 5) = `8/10`
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