Question

A random variable X has a Uniform distribution over (–3, 3). Compute P(|X|<2). its modulus X​

Answer 1

Answer: P(|X|<2) = 1/2

Concept: Random Variable X is uniformly distributed over (-3,3)

we have

f(x) = 1/(3+3)  (-3<X>3)

P(|X|<2) = ∫f(x)dx

Given: Random Variable X is uniformly distributed over (-3,3)

To find: P(|X|<2)

Step-by-step explanation:

Random Variable X is uniformly distributed over (-3,3)

For Uniform Distribution we have;

we have

f(x) = 1/(3+3)  (-3<X>3)

f(x) = 1/6

we have

P(|X|<2) = P[-2<X>2]

P(|X|<2) = ∫f(x)dx                (limit -2 to 2)

Solving the integration

P(|X|<2) = ∫(1/6)dx               (limit -2 to 2)

P(|X|<2) = x/6                       (limit -2 to 2)

Putting limits in above integration

P(|X|<2) = 2/6 - (-2/6)

P(|X|<2) = 2/6 + 2/6

P(|X|<2) = 3/6

P(|X|<2) = 1/2

Answer: P(|X|<2) = 1/2

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Answer 2

Answer:

We can thus conclude that the value of P(|X|<2) is $\frac{1}{2}$ .

Step-by-step explanation:

We are given that a random variable X is uniformly distributed over the range (-3,3).
We can denote this as $f(x) = \frac{1}{3+3} \;\;\; (-3 < X > 3)$

Thus, we can write this as $P(|X| < 2) = \int\limits^3_-_3 {f(x)} \, dx$

We are asked to find the value of P(|X|<2), which we can calculate as follows.

$f(x) = \frac{1}{3 + 3} \;\;\;\; (-3 < X > 3)\\\\f(x) = \frac{1}{6}$

We can also rewrite the value of P(|X|<2) as P[-2 < X > 2], i.e, $P(|X| < 2) = \int\limits^2_2 {f(x)}$

Upon solving this integration, we get the value to be

$P(|X| < 2) = \int\limits^2_2 {f(x)}\\P(|X| < 2) = \int\limits^2_2 {\frac{1}{6}} \;dx\\P(|X| < 2) = \int\limits^2_2 {\frac{x}{6}} \; dx$

Now, by applying limits in the given integration, we get the end values of the functions to be $\frac{-2}{6} \;and\; \frac{2}{6}$

It can then be written finally as $P(|X| < 2) = \frac{3}{6} = \frac{1}{2}$

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