Answer to Question #262822 in Statistics and Probability for deepak

Question

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

Answer 1

Below is a summary of the data given above in order to find the variance and the mean deviation.

x f fx fx²

2 1 2 4

3 1 3 9

4 2 8 32

5 4 20 100

6 4 24 144

7 3 21 147

8 7 56 448

9 5 45 405

10 3 30 300

The mean is given by,

$\bar{x}=\sum(fx)/\sum(f)=209/30=6.966667$

To find the mean deviation, let us first make the summary below.

$x$ $|x_i-\bar{x}|$ $f$ $f|x-\bar{x}|$

2 4.96667 1 4.96667

3 3.966667 1 3.966667

4 2.96667 2 5.93333

5 1.966667 4 7.86667

6 0.966667 4 3.86667

7 0.03333 3 0.1

8 1.03333 7 7.23333

9 2.03333 5 10.16667

10 3.03333 3 9.1

Now, the mean deviation is given as,

$M.D=\displaystyle\sum^9_{i-1}f|x_i-\bar{x}|/(\sum(f))$

$M.D=53.2/30=1.773333$

The variance is given by the formula,

$variance=(\sum(fx^2)-(\sum(fx))^2/\sum(f))/(\sum(f)-1)$

$variance=(1589-(209^2/30))/(30-1)=132.9667/29=4.585057$

Therefore, the M.D and variance for the distribution are 1.733333 and 4.585057 respectively.