Answer to Question #104470 in Statistics and Probability for Amra Musharraf

Question

Calculate the standard deviation and mean deviation from mean if the frequency function f(x) has the form:
f(x)={3+2x/18,for2≤x≤4; 0 otherwise

Answer 1

$f(x)=\begin{cases} \frac{3+2x}{18}, \ \ 2\leq x\leq 4 \\ 0, \ otherwise \end{cases}$

Mean value is $\mu =\int\limits _{\R}xf(x)dx=\int\limits_{2}^{4}x \times \frac{3+2x}{18}dx=\int\limits_{2}^{4}(\frac{x}{6}+\frac{x^2}{9})dx= (\frac{x^2}{12}+\frac{x^3}{27})\big|^4_2=$

$(\frac{16}{12}+\frac{64}{27})-(\frac{4}{12}+\frac{8}{27})=\frac{83}{27}$

Standard deviation is $\sigma =\sqrt {\int\limits_{\R} (x-\mu )^2f(x)dx}= \sqrt {\int\limits_{2}^{4}(x-\frac{83}{27} )^2\times \frac{3+2x}{18}dx}=$

$\sqrt {\int\limits_{2}^{4}(\frac{x^3}{9} - \frac{251 x^2}{486 }+ \frac{166 x}{6561 }+\frac{ 6889}{4374} )dx}=\sqrt{(\frac{x^4}{36}-\frac{251x^3}{1458}+\frac{83x^2}{6561}+\frac{6889x}{4374})\big|^4_2}=\sqrt{\frac{239}{729}}=\frac{\sqrt{239}}{27}$

Mean deviation from mean is $\int\limits_{\R} |x-\mu|f(x)=\int\limits_2^4|x-\frac{83}{27}| \ \frac{3+2x}{18}dx=\int\limits_2^{\frac{83}{27}} (\frac{83}{27}-x)\frac{3+2x}{18}dx + \int\limits_{\frac{83}{27}}^4 (x-\frac{83}{27})\frac{3+2x}{18}dx =$

$= \int\limits_2^{\frac{83}{27}} (-\frac{x^2}{9}+\frac{85x}{486} +\frac{83}{162})dx + \int\limits_{\frac{83}{27}}^4 (\frac{x^2}{9}-\frac{85x}{486} -\frac{83}{162})dx= (-\frac{x^3}{27}+\frac{85x^2}{972} +\frac{83x}{162})\big|_2^{\frac{83}{27}}+ +(\frac{x^3}{27}-\frac{85x^2}{972} -\frac{83x}{162})\big|_{\frac{83}{27}}^4=\frac{525625}{1062882}$

Answer: standard deviation =$\frac{\sqrt{239}}{27}$

mean deviation from mean =$\frac{525625}{1062882}$