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Find the expected value of the continuous random variable X

associated with the probability density function over the indicated
interval.
f(x)= 1/3 ; [5,8]

1 Answer

4 votes

Explanation:

first, the simple practical approach :

the PDF is simply a constant, a flat line. so, all values in the interval have the same probability (1/3).

therefore the expected value is simply the middle of the interval :

(5 + 8)/2 = 13/2 = 6.5 or 6 ½

now for the formal approach :

as the expected values of a set of discreet numbers is the sum of the products of each of these numbers with their probabilities, the expected value of a continuous interval of numbers ( [a, b] ) is simply the integral over the products of the numbers and their probabilities.

EX = integral[a, b] of x×f(x)×dx

f(x) = 1/3

a = 5

b = 8

so, we have

EX = integral[5, 8] of x/3 × dx =

= [5, 8] of x²/6

as the integral of x is x²/2.

x²/6 between 5 and 8 is

64/6 - 25/6 = 39/6 = 13/2 = 6.5

EX = 6.5 or 6 ½

we got the same result practically and theoretically.

these are the best results ever !

User Javier Cobos
by
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