a)∫−∞∞f(x)dx=1∫−1223xdx=[23x22]−12=1. So f(x) is pdf.a)\int_{-\infty}^{\infty}f(x)dx=1\\ \int_{-1}^{2}\frac{2}{3}xdx=[\frac{2}{3}\frac{x^2}{2}]_{-1}^2=1.\text{ So } f(x)\text{ is pdf}.a)∫−∞∞f(x)dx=1∫−1232xdx=[322x2]−12=1. So f(x) is pdf.
b)P{0<X≤1}=∫0123xdx=[23x22]01=13.b) P\{0<X\leq 1\}=\int_0^1\frac{2}{3}xdx=[\frac{2}{3}\frac{x^2}{2}]_{0}^1=\frac{1}{3}.b)P{0<X≤1}=∫0132xdx=[322x2]01=31.
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Dear Kat, thank you for leaving a feedback. We found that cases 2x/3 and 1/3*x^2 meet conditions of the question. In other words, expert's solution is not wrong.
I think for part b of this problem, they meant to say (x²)/3, in which case the answer to b is 1/9. Because the integral from 0 to 1 of (x²)/3 is (x³)/9 from 0 to 1 which equals 1/9. At least that's how my stats professor solved this problem. Just in case someone came here for that!
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Dear Kat, thank you for leaving a feedback. We found that cases 2x/3 and 1/3*x^2 meet conditions of the question. In other words, expert's solution is not wrong.
I think for part b of this problem, they meant to say (x²)/3, in which case the answer to b is 1/9. Because the integral from 0 to 1 of (x²)/3 is (x³)/9 from 0 to 1 which equals 1/9. At least that's how my stats professor solved this problem. Just in case someone came here for that!