Question 1. Show using integration that k = 3/2 2. Using this value of k, determine the expected value or mean of X. 3. Determine the following probabilities, show all your calculations. 3. Consider the following probability density function of a continuous random variable X: f(x) = =) = {1x06 {kova tore or fete si elsewhere 1 p(x = 2 i. P|X< 2 1 <x< 1 2 Determine the cumulative distribution function of X, F(x). (Hint: F(x) will also be a piecewise-defined function.]
1. Show using integration that
k = 3/2
2. Using this value of k, determine the expected value or mean of X.
3. Determine the following probabilities, show all your calculations.
3. 
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I) P\left(x=\frac{1}{2}\right)P\left(x=\frac{1}{2}\right)=0, because, The given probability density function of a continuals random variable x, if satisfies P(x=x)=0 for each x. this means that X assumes, continuors values.IT)\text { II) } \begin{aligned}& P\left(x \leqslant \frac{1}{2}\right) \\= & \int_{0}^{1 / 2} \frac{3}{2} \sqrt{x} d x \\= & \frac{3}{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1 / 2} \\= & \frac{3}{2}[0.2357] \\= & 0.3535 \\\because & P\left(x \leqslant \frac{1}{2}\right)=0.3535\end{aligned}III) C \cdot D \cdot F of x\begin{aligned}F(x) & =\int_{0}^{x} f(x) d x \\& =\int_{0}^{x} \frac{3}{2} \sqrt{x} d x \\& =\frac{3}{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{x} \\& =\frac{3}{2} \times \frac{2}{3}\left[x^{3 / 2}-0\right] \\F(x) & =x^{3 / 2} ; x<0 \\\therefore & F(x)=\left\{\begin{array}{ll}0 & ; x>1 \\x^{3 / 2} & 0<1 \\1 & x<1\end{array}\right.\end{aligned}Solution:\begin{aligned}f(x) & =k \sqrt{x} & & ; 0 \leq x \leq 1 \\& =0 & & ; 0 . \omega .\end{aligned}\begin{array}{l}\int_{0}^{1} f(x) d x=1 \\\int_{0}^{1} k \sqrt{x}=1 \\k \int_{0}^{1} \sqrt{x}=1 \\k\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}=1 \\k\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}=1 \\k\left[\frac{2}{3}\right]=1 \text { or } k=1.5 \\\therefore k=\frac{3}{2}\end{array}\begin{array}{rlrl}\therefore f(x) & =\frac{3}{2} \sqrt{x} \quad ; 0 \leq x \leq 1 \\& =0 & ; 0 . \omega .\end{array}E(x)\begin{array}{l}E(x)=\int_{0}^{1} x \cdot f(x) d x \\=\int_{0}^{1} x \cdot \frac{3}{2} x^{1 / 2} d x \\=\frac{3}{2} \int_{0}^{1} x^{3 / 2} d x \\=\frac{3}{2}\left[\frac{x^{5 / 2}}{5 / 2}\right]_{0}^{1} \\=\frac{3}{2}\left[\frac{1}{\frac{5}{2}}-0\right] \\=\frac{3}{2} \times \frac{2}{5} \\=\frac{3}{5} \\\therefore E(x)=\frac{3}{5} \text { or } E(x)=0.6 \\\end{array}  ...