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Question
A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Determine:
- k
- P(X < 3)
- P( X > 4)
Solution
i. The table gives a probability distribution and therefore `sum_(i = 1)^8 P_i` = 1
∴ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
∴ 10k2 + 9k – 1 = 0
∴ 10k2 + 10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1)(k + 1) = 0
∴ k = `1/10` or k = –1
But k cannot be negative
∴ k = `1/10`
ii. P(X < 3)
= P(X = 0 or X = 1 or X = 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k
= 3k
= `3/10`
iii. P(X > 4)
= P(X = 5 or X = 6 or X = 7)
= P(X = 5) + P(X = 6) + P(X = 7)
= k2 + 2k2 + 7k2 + k
= 10k2 + k
= `10(1/10)^2 + 1/10`
= `1/10 + 1/10`
= `1/5`
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