Question

If the p.d.f. of a continuous r.v. X is

`f(x) = {{:((x  +  2)/18",", "for" -2 < x < 4),(= 0",", "otherwise"):}`

then P(|X| < 1) = ______.

पर्याय

• `1/9`

• `2/9`

• `1/27`

• `2/27`

Answer 1

If the p.d.f. of a continuous r.v. X is

`f(x) = {{:((x  +  2)/18",", "for" -2 < x < 4),(= 0",", "otherwise"):}`

then P(|X| < 1) = `underlinebb(2/9)`.

Explanation:

Given p.d.f.:

`f(x) = {{:((x  +  2)/18",", -2 < x < 4),(0",", "otherwise"):}`

We need:

P(|X| < 1)

Step 1: Interpret the condition

|X| < 1

Means –1 < X < 1

Step 2: Set up the probability

`P(-1 < X < 1) = int_(-1)^1 (x + 2)/(18) dx`

Step 3: Integrate

= `1/18 int_(-1)^1 (x + 2) dx`

= `1/18 [x^2/2 + 2x]_(-1)^1`

Evaluate at limits

At x = 1:

`1/2 + 2 = 2.5`

At x = –1:

`1/2 - 2 = -1.5`

Difference:

2.5 – (–1.5) = 4

Step 4: Multiply

`P = 1/18 xx 4`

= `4/18`

= `2/9`