Given that fx(x) = 0 (x <= 10) = 10/x^2 (x >10) Find the 60-th percentile of X. How do we solve this?
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Originally Posted by dingdong Given that fx(x) = 0 (x <= 10) = 10/x^2 (x >10) Find the 60-th percentile of X. How do we solve this? You have to solve the equation: $\displaystyle \int_{-\infty}^x f(\xi)\ d\xi=0.6$ In this case this reduces to solving: $\displaystyle \int_{10}^x \frac{10}{\xi^2}\ d\xi=1-\frac{10}{x}=0.6$ CB
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