Question

If the probability density function of a random variable X is given as

x1	–2	–1	0	1	2
P(X = x1)	0.2	0.3	0.15	0.25	0.1

then F(0) is equal to

पर्याय

• P(X < 0)

• P(X > 0)

• 1 – P(X > 0)

• 1 – (X < 0)

Answer 1

1 – P(X > 0)

Explanation:

Since, cumulative distribution function

F(x) = P(X ≤ x)

So, F(0) = P(X ≤ 0)

= P(X = 0) + P(X = –1) + P(X = –2)

= 0.15 + 0.3 + 0.2

= 0.65

(a) P(X < 0) = P(X = –1) + P(X = –2)

= 0.3 + 0.2

= 0.5

Therefore, P(X < 0) ≠ F (0)

(b) P(X > 0) = P(X = 1) + P(X = 2)

= 0.25 + 0.1

= 0.35

Thus, P(X > 0) ≠ F (0)

(c) 1 – P(x > 0)

= 1 – 0.35

= 0.65

Hence, 1 – P(X > 0) = F(0)