The probability density function, $$\(\mathrm{f}\)$$, of a random variable $$\(X\)$$ is given by $$\[ f(x)= \begin{cases}k\left(6 x-x^{2}\right) & 0 \leqslant x \leqslant 6, \\ 0 & \text { otherwise }\end{cases} \]$$ where $$\(k\)$$ is a constant. State the value of $$\(\mathrm{E}(X)\)$$ and show that $$\(\operatorname{Var}(X)=\frac{9}{5}\)$$. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ 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Exam No：9709_s21_qp_61 Year：2021 Question No：6
Answer:
\(\mathrm{E}(X)=3\)
\(k \int_{0}^{6}\left(6 x-x^{2}\right) \mathrm{d} x=1\)
\(k\left[3 x^{2}-\frac{x^{3}}{3}\right] \begin{array}{l}6
0\end{array}[=1]\)
\(k\left(108-\frac{216}{3}\right)=1\)
\(k=\frac{3}{108}\) or \(\frac{1}{36}\)
' \(\frac{3}{108} \cdot \int_{0}^{6}\left(6 x^{3}-x^{4}\right) \mathrm{d} x\)
\(=\frac{3}{108}\left[\frac{3 x^{4}}{2}-\frac{x^{5}}{5}\right] \begin{array}{l}6
0\end{array}=10.8\)
\(' 10.8^{\prime}-3^{\prime 2}\)
\(\frac{9}{5}\) or \(1.8\)
\(k \int_{0}^{6}\left(6 x-x^{2}\right) \mathrm{d} x=1\)
\(k\left[3 x^{2}-\frac{x^{3}}{3}\right] \begin{array}{l}6
0\end{array}[=1]\)
\(k\left(108-\frac{216}{3}\right)=1\)
\(k=\frac{3}{108}\) or \(\frac{1}{36}\)
' \(\frac{3}{108} \cdot \int_{0}^{6}\left(6 x^{3}-x^{4}\right) \mathrm{d} x\)
\(=\frac{3}{108}\left[\frac{3 x^{4}}{2}-\frac{x^{5}}{5}\right] \begin{array}{l}6
0\end{array}=10.8\)
\(' 10.8^{\prime}-3^{\prime 2}\)
\(\frac{9}{5}\) or \(1.8\)
Knowledge points:
6.3.1 understand the concept of a continuous random variable, and recall and use properties of a probability density function (For density functions defined over a single interval only; the domain may be infinite,.)
6.3.2 use a probability density function to solve problems involving probabilities, and to calculate the mean and variance of a distribution. (Including location of the median or other percentiles of a distribution by direct consideration of an area using the density function.) (Explicit knowledge of the cumulative distribution function is not included.)
Solution:
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