AP EAMCET202120 Aug 2021Morning ShiftMathematicsProbabilityActual
For the random variable X with probability distribution is given by the table X = x 0 1 2 3 P ( X = x ) K K + 1 7 2 K 2 5 The mean of X is
Options
- A31 35
- B57 35
- C63 35
- D67 35
Correct answer
D. 67 35
Step-by-step solution
Given the distribution, X = x 0 1 2 3 P ( X = x ) K K + 1 7 2   K 2 5 As we know sum of all probabilities of an event is unity, ⇒ K + K + 1 7 + 2 K + 2 5 = 1 4 K = 1 - 1 7 + 2 5 = 1 - 19 35 4 K = 16 35 ⇒ K = 4 35 X = x 0 1 2 3 P ( X = x ) 4 35 9 35 8 35 2 5 Mean = ∑ i = 0 3 x i p x i = 0 × 4 35 + 1 × 9 35 + 2 × 8 35 + 3 × 2 5 = 0 + 9 35 + 16 35 + 6 5 = 67 35