Answer to Question #263729 in Statistics and Probability for Rohit

Question

Find the M. D. & variance of the following distribution. X 2 3 4 5 6 7 8 9 10 Y 1 1 2 4 4 3 7 5 3

Answer 1

$$mean=\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}$$

$$=\dfrac{2(1)+3(1)+4(2)+5(4)+6(4)}{1+1+2+4+4+3+7+5+3}$$

$$+\dfrac{7(3)+8(7)+9(5)+10(3)}{1+1+2+4+4+3+7+5+3}$$

$$=\dfrac{209}{30}$$

i)

$$\sum f_i|x_i-\bar{x}|=|2-\dfrac{209}{30}|(1)+|3-\dfrac{209}{30}|(1)$$

$$+|4-\dfrac{209}{30}|(2)+|5-\dfrac{209}{30}|(4)+|6-\dfrac{209}{30}|(4)$$

$$+|7-\dfrac{209}{30}|(3)+|8-\dfrac{209}{30}|(7)+|9-\dfrac{209}{30}|(5)$$

$$+|10-\dfrac{209}{30}|(3)=\dfrac{1596}{30}=\dfrac{532}{10}$$

$$MD=\dfrac{\sum f_i|x_i-\bar{x}|}{\sum f_i}=\dfrac{\dfrac{532}{10}}{30}=\dfrac{532}{300}$$

$$=\dfrac{133}{75}\approx1.773333$$

ii)

$$Var(x)=\sum f_i(x_i-\bar{x})^2$$

$$=(2-\dfrac{209}{30})^2(1)+(3-\dfrac{209}{30})^2(1)+(4-\dfrac{209}{30})^2(2)$$

$$+(5-\dfrac{209}{30})^2(4)+(6-\dfrac{209}{30})^2(4)+(7-\dfrac{209}{30})^2(3)$$

$$+(8-\dfrac{209}{30})^2(7)+(9-\dfrac{209}{30})^2(5)+(10-\dfrac{209}{30})^2(3)$$

$$=\dfrac{119670}{900}=\dfrac{3989}{30}\approx132.9667$$