Distribution function of a random variable X is given by F(x) = 1 - 1/(x ^ 2); 1 <= x < ∞ Then P(X <= 2) and respectively P(X > 3/2) are

Question

Distribution function of a random variable X is given by

F(x) = 1 - \frac{1}{x ^ 2}; 1 <= x < ∞
Then P(X <= 2) and respectively P(X > \frac{3}{2}) are

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Solution

The cumulative distribution function (CDF) F(x) of a random variable X is given by F(x) = 1 - 1/(x^2) for 1 <= x < ∞.

To find P(X <= 2), we simply substitute x = 2 into the CDF:

F(2) = 1 - 1/(2^2) = 1 - 1/4 = 3/4.

So, P(X <= 2) = 3/4.
To find P(X > 3/2), we use the property that P(X > x) = 1 - P(X <= x).

Substituting x = 3/2 into the CDF, we get:

F(3/2) = 1 - 1/(3/2)^2 = 1 - 4/9 = 5/9.

So, P(X > 3/2) = 1 - P(X <= 3/2) = 1 - 5/9 = 4/9.

This problem has been solved

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