In a previous question you determined the value of k that ma

In a previous question, you determined the value of k that makes this a valid probability density function: f(x) = kx^2 for 1 < x < 3. Using that answer, what is the mean of X? Answer to THREE decimal places.

Solution

Since, I don\'t have the vlue of K, I start b y calculating it.

Integral of f(x) from 1 to 3 should give the probability = 1

Thus,

Integral of f(x) = (k/3) (x³)

Putting the limits, I get,

26k/3 = 1

k = 3/26

Now, mean value is obtained as integral of x f(x) from x = 1 to x= 3

f(x) = 3x² / 26

Integral of x f(x) = integral of 3x³ /26

= 3x⁴ / (4*26)

Putting the limits we get:

Mean = 30 / 13 =2.307

Hope this helps. Ask if you have any doubts.