Question

In the following distribution, P(X<2) = 0.35, and expected value is 1.9

Answer 1

Solution :

We have :

X            0     1          2    3   4

P(X)     0.10   A    0.35   B    C

a). P(X < 2) = 0.35

P(X < 2) = P(X = 0) + P(X = 1) = 0.35

⇒ 0.10 + A = 0.35

⇒ A = 0.25

So the value of A is 0.25

b). The total probability = 1

So ,

0.10 + A + 0.35 + B + C = 1

0.10 + 0.25 + 0.35 + B + C = 1

B + C  = 1 - 0.70

B + C  = 0.30  ......(i)

We have the expected value = 1.9

So, $\sum X P(X) = 19$     for x  = 0, 1, 2, 3, 4

⇒ (0 x 0.10) + (1 x 0.25) + (2 x 0.35) + (3 x B) + (4 x C) = 1.9

⇒ 0 + 0.25 + 0.70 + 3B + 4C = 1.9

⇒ 3B + 4C = 1.9 - 0.95

⇒ 3B + 4C = 0.95   ...................(ii)

From (i), we take the value of B = 0.30 - C and substitute it in the equation (i), we get,

⇒ 3( 0.30 - C) + 4C = 0.95

⇒ 0.90 - 3C + 4C = 0.95

⇒ C = 0.95 - 0.90

       = 0.05

Now substituting the value of C = 0.05 in (ii), we get,

⇒ B = 0.05 = 0.30

⇒ B = 0.25

c). The value of C is 0.05