The probability distribution of a random variable X is given below : array|c|c|c|c|c|c|c|c|c| X & 4k & 307k & 327k & 347k & 367k & 387k & 407k & 6k \ P(X) & 215 & 115 & 215 & 15 & 115 & 215 & 15 & 115 \ array If E(X) = 26315, then P(X < 20) is equal to :
- 35
- 1415
- 815
- 1115
Answer
(4) 1115
Solution
Check: 2 + 1 + 2 + 3 + 1 + 2 + 3 + 115 = 1
E(X) = xi pi = k105(56 + 30 + 64 + 102 + 36 + 76 + 120 + 42)
= 526k105
526k105 = 26315
k = 263 10515 526 = 72
values of X: 14, 15, 16, 17, 18, 19, 20, 21
P(X < 20) = 1 - P(X = 20) - P(X = 21)
= 1 - 315 - 115 = 1115
Answer: (4)