If the mean and the variance of the data gathered 4-8 8-12 12-16 16-20 3 4 7 gathered are …

Question

If the mean and the variance of the data

$\begin{gathered} 4-8 \quad  8-12 \quad  12-16 \quad  16-20 \ 3 \quad  \lambda  \quad  4 \quad  7 \end{gathered}$

are $\mu $ and 19 respectively, then the value of $\lambda +\mu $ is :

Accepted Answer

Class

fi

xi

fi\cdot xi

fi\cdot xi2

4--8

108

8--12

\lambda

10\lambda

100\lambda

12--16

784

16--20

2268

Total

14 + $\lambda $

200 + 10$\lambda $

3160 + 100$\lambda $

$\begin{gathered} \sum {f}_{i}=14+\lambda \ \sum {f}_{i}{x}_{i}=18+10\lambda +56+126=200+10\lambda \ \mu =\frac{\sum {f}_{i}{x}_{i}}{\sum {f}_{i}}=\frac{200+10\lambda }{14+\lambda } \ \sum {f}_{i}{x}_{2}^{i}=3({6}^{2})+\lambda ({10}^{2})+4({14}^{2})+7({18}^{2})=108+100\lambda +784+2268=3160+100\lambda \end{gathered}$

$\begin{gathered} =\frac{\sum {f}_{i}{x}_{2}^{i}}{\sum {f}_{i}}-{\mu }^{2}=19 \ 19=\frac{3160+100\lambda }{14+\lambda }-{(\frac{200+10\lambda }{14+\lambda })}^{2} \ 19(14+\lambda {)}^{2}=(3160+100\lambda )(14+\lambda )-(200+10\lambda {)}^{2} \ 19(196+28\lambda +{\lambda }^{2})=44240+4560\lambda +100{\lambda }^{2}-40000-4000\lambda -100{\lambda }^{2} \ 3724+532\lambda +19{\lambda }^{2}=4240+560\lambda \ 19{\lambda }^{2}-28\lambda -516=0 \end{gathered}$

Solving for $\lambda $ using the quadratic formula:

$\lambda =\frac{28\pm \sqrt{{28}^{2}-4(19)(-516)}}{38}=\frac{28\pm 200}{38}$

Taking the positive value, $\lambda =\frac{228}{38}=6$

Finding the mean $\mu $ :

$\mu =\frac{200+10(6)}{14+6}=\frac{260}{20}=13$

The value of $\lambda +\mu $ is:

$6+13=19$

Solutions · 1