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Find k and then E(X)
written 6.2 years ago by | • modified 3.9 years ago |
Find k and then E(X) if X has the p.d.f
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written 6.2 years ago by | • modified 3.9 years ago |
Find k and then E(X) if X has the p.d.f
written 6.1 years ago by | • modified 6.1 years ago |
Since the total probability is 1
$∫_{-∞}^∞f(x)$ = 1
$∫_0^2kx(2-x) \ dx = k∫_0^2(2x-x^2 )\ dx = k[x^2-\frac{x^3}{3}]_0^2 = k . \frac{4}{3} = 1$
∴ k = $\frac{3}{4}$
∴ f(x) = $ \begin{cases} \frac{3}{4}x(2-x) \,\,\, if \ 0\leq x \leq 2, \ k \gt 2 \\ 0 \,\,\,\, elsewhere \end{cases} $
By definition, $E(X) =∫_0^2xf(x) \ dx = ∫_0^2x.\frac{3}{4} x(2-x) \ dx = ∫_0^2 \frac{3}{4}(2x^2-x^3) \ dx$
= $ \frac{3}{4} [\frac{2x^3}{3}-\frac{x^4}{4}]_0^2$ = 1