If the mean of the following probability distribution of a radam variable X :
is 46/9, then the variance of the distribution is

Solution

Maths sir · 3 steps

Two unknowns need two equations: probabilities add to 1, and the given mean gives the second one.

1. Two conditions. The probabilities must add to 1, so a + 2a + (a+b) + 2b + 3b = 4a + 6b = 1. The mean is Σ X P(X) = 0 + 4a + 4(a+b) + 12b + 24b = 8a + 40b, and this must equal 46/9.

   4a + 6b = 1, 8a + 40b = 46/9

2. Solve for a, b. Double the first equation to get 8a = 2 - 12b. Put this in the second: 2 + 28b = 46/9, so 28b = 28/9 and b = 1/9. Then 4a = 1 - 6/9 = 1/3, so a = 1/12.

   X	P(X)	X P(X)	X² P(X)
   0	1/12	0	0
   2	1/6	1/3	2/3
   4	7/36	7/9	28/9
   6	2/9	4/3	8
   8	1/3	8/3	64/3
   Total	1	46/9	298/9

3. So. Var(X) = E(X²) - (E(X))² = 298/9 - 2116/81 = (2682 - 2116)/81 = 566/81, which is option (b).

The idea to keep. In any distribution with unknown probabilities, first use 'total probability is 1' together with the given data to find them. Then variance is always E(X²) minus the square of the mean.

Answer from the NTA final answer key. Solution by PrepWiser.