Answer to Question #103785 in Statistics and Probability for Biraj Chhetri

Question

Calculate the standard deviation and mean deviation from mean if the frequency function f(X) has the form:
f(X) =(3+2x)/18. For 2≤x≤4
=0. Otherwise

Answer 1

$EX=\int\limits_{\mathbb R} xf(x)dx=\int\limits_2^4 x\frac{3+2x}{18}dx=\int\limits_2^4 \left(\frac{x}{6}+\frac{x^2}{9}\right)dx=$

$=\left(\frac{x^2}{12}+\frac{x^3}{27}\right)\bigl|_2^4=\frac{83}{27}$

$EX^2=\int\limits_{\mathbb R} x^2f(x)dx=\int\limits_2^4 x^2\frac{3+2x}{18}dx=\int\limits_2^4 \left(\frac{x^2}{6}+\frac{x^3}{9}\right)dx=$

$=\left(\frac{x^3}{18}+\frac{x^4}{36}\right)\bigl|_2^4=\frac{88}{9}$

a)Mean deviation from mean is $\mu=E|X-EX|=E\left|X-\frac{83}{27}\right|=Eg(X)$, where $g(x)=\left|x-\frac{83}{27}\right|$. Then

$\mu=Eg(X)=\int\limits_{\mathbb R}g(x)f(x)dx=\int\limits_2^4\left|x-\frac{83}{27}\right|\frac{3+2x}{18}dx=\\ \int\limits_2^\frac{83}{27}\left(\frac{83}{27}-x\right)\frac{3+2x}{18}dx+\int\limits_\frac{83}{27}^4\left(x-\frac{83}{27}\right)\frac{3+2x}{18}dx=\\ =\frac{1}{486}\int\limits_2^\frac{83}{27}(249+85x-54x^2)dx+\\ +\frac{1}{486}\int\limits_\frac{83}{27}^4(-249-85x+54x^2)dx=\\ =\frac{1}{486}\left(249x+\frac{85}{2}x^2-18x^3\right)\bigl|_2^\frac{83}{27}+\\ +\frac{1}{486}\left(-249x-\frac{85}{2}x^2+18x^3\right)\bigl|_\frac{83}{27}^4=\frac{525625}{1062882}$

b)Standard deviation is $\sigma=\sqrt{EX^2-(EX)^2}=\sqrt{\frac{88}{9}-\left(\frac{83}{27}\right)^2}=\frac{\sqrt{239}}{27}$

Answer: mean deviation is $\mu=\frac{525625}{1062882}$, standard deviation is $\sigma=\frac{\sqrt{239}}{27}$