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Let X denote a random variable that can take values [0,1,2,3,4]. Assume that all of these numbers a equally likely to be attained by X, in an experiment. Compute the following. P(X>0)= Expected value of X= Variance of X=

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User Dtell
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Answer:

To compute the probabilities and expected value (mean) of the random variable X, which can take values [0, 1, 2, 3, 4] with equal likelihood, we can start by finding the probabilities for each value:

1. Probability of X > 0:

P(X > 0) means the probability that X is greater than 0. In this case, X can take values 1, 2, 3, or 4. Since all values are equally likely:

P(X > 0) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Since each value is equally likely, the probability for each is 1/5:

P(X > 0) = (1/5) + (1/5) + (1/5) + (1/5) = 4/5

2. Expected Value (Mean) of X:

The expected value (mean) of a random variable is calculated by taking the sum of each possible value multiplied by its probability:

E(X) = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2)) + (3 * P(X = 3)) + (4 * P(X = 4))

Since each value is equally likely (1/5 probability for each):

E(X) = (0 * 1/5) + (1 * 1/5) + (2 * 1/5) + (3 * 1/5) + (4 * 1/5)

E(X) = (1/5) * (1 + 2 + 3 + 4)

E(X) = (1/5) * 10

E(X) = 2

So, the expected value of X is 2.

3. Variance of X:

The variance of a random variable is calculated as:

Var(X) = E(X^2) - (E(X))^2

We've already calculated E(X) as 2. Now, let's calculate E(X^2):

E(X^2) = (0^2 * P(X = 0)) + (1^2 * P(X = 1)) + (2^2 * P(X = 2)) + (3^2 * P(X = 3)) + (4^2 * P(X = 4))

Since each value is equally likely (1/5 probability for each):

E(X^2) = (0^2 * 1/5) + (1^2 * 1/5) + (2^2 * 1/5) + (3^2 * 1/5) + (4^2 * 1/5)

E(X^2) = (1/5) * (0 + 1 + 4 + 9 + 16)

E(X^2) = (1/5) * 30

E(X^2) = 6

Now, calculate the variance:

Var(X) = E(X^2) - (E(X))^2

= 6 - (2^2)

= 6 - 4

= 2

So, the variance of X is 2.

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User Keshaun
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