Answer to Question #262822 in Statistics and Probability for deepak

Question #262822

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 


1
Expert's answer
2021-11-15T18:22:51-0500

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.


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