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X has the following cdf

F(x) = 0 if x<0

   =x/2                               if 0≤x<1

   =(x^2−1)/6  +  1/2                 if 1≤x<2

   = 1                                 if  2 ≤ x

Find the mean of X.

I tried the integration using 1 - F(x) but got the wrong answer of 1.80555

  • 2
    Can't you just find the PDF and compute $\int x f(x) dx$? – 2017-02-21
  • 1
    You should show us your work, because integrating over the tail works just fine. I believe the answer is something like 1.028. Without seeing your work, we can't really do much to help you. – 2017-02-21
  • 0
    integral (1-(x/2))dx from 0 to 1 + the integral (1-((x^2-1)/6)-(1/2)) from 1 to 2 , for the first integral , I got 3/4 and for the second I get 1.05555. I add them together and get approx 1.806 – 2017-02-21

1 Answers 1

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You may have made an arithmetic error:

$$\int_1^2 (1-((x^2-1)/6)-(1/2)) \, dx $$ $$=\int_1^2 \left(\frac23-\frac{x^2}{6} \right)\, dx$$ $$=\left[\frac{2x}3-\frac{x^3}{18} \right]_1^2$$ $$=\left(\frac{4}3-\frac{8}{18} \right)-\left(\frac{2}3-\frac{1}{18} \right)$$ $$=\frac{5}{18} \approx 0.27778$$ rather than the $1.05555$ you calculated.

$\dfrac34+\dfrac5{18}=\dfrac{37}{36}\approx 1.027778$ for the final answer