Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Statistics
0 votes

The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

$\begin{array}{1 1}(A)\;8.94\\(B)\;5\\(C)\;7\\(D)\;15\end{array} $

Can you answer this question?

1 Answer

0 votes
  • The formula used is Combined $SD= \sqrt {\large\frac{n_1 \sigma_1^2+n_2 \sigma^2}{n_1+n_2} +\frac{n_1n_2 (\bar {x_1} -\bar {x_2})^2}{(n_1+n_2)^2}}$
Step 1:
Given $n_1=60 \qquad n_2=80$
$\bar{x_1}=650 \qquad \bar{x_2}=660$
$\sigma_1=8 \qquad \sigma_2=7$
The formula used is Combined $SD= \sqrt {\large\frac{n_1 \sigma_1^2+n_2 \sigma^2}{n_1+n_2} +\frac{n_1n_2 (\bar {x_1} -\bar {x_2})^2}{(n_1+n_2)^2}}$
$\qquad= \sqrt { \large\frac{60 \times 8^2 + 80 \times 7^2}{60+80}+ \frac{ 60 \times 80 (650-660)^2}{(60+80)^2}}$
$\qquad= \sqrt {\large\frac{ 3840+3920}{140}+ \frac{4800 \times 100}{(140)^2}}$
$\qquad= \sqrt {\large\frac{7760 \times 140+4800 \times 100}{(140^2)}}$
$\qquad= \large\frac{1}{140}$$ \sqrt {1086400+480000}$
$\qquad= \large\frac{1251.559}{140}$$=8.939$
$\qquad= 8.94$
Hence A is the correct answer.
answered Jul 4, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App