Question

Observe a one-day cricket match in the year 2019. Prepare a project report with respect to the following points. (i) Prepare team-wise grouped frequency distribution tables showing the number of overs as class intervals and corresponding scores as frequencies. (ii) Represent frequency distribution tables obtained in part (i) with the help of Histogram. (iii) Find range of the scores in the match. (iv) Find team-wise Mean Deviation (M.D) score. (v) Find Standard Deviation (S.D) of the winning team

Answer 1

Answer:

is it any match

Explanation:

Answer 2

Given :

To prepare a project report on one - day cricket match.

To Find :

Mean

Mean Deviation

Standard Deviation

Solution :

Lets take p1, p2, p3 , p4 as players .

The scores of the players are :

p1   -  20

p2   -  40

p3   -  25

p4   -   15

Now, lets find the mean for the above data .

Mean = $\frac{20 + 40 + 25 + 15}{4}$  

         = $= 25$

Deviation from the mean ,

p1   | 25 - 20 |  = 5

p2  | 40 - 25 | = 15

p3   | 25 - 25 | = 0

p4   | 15 - 25 | = 10

Now, lets find mean deviation

Mean Deviation  =    $\frac{5 + 15 + 10 + 0}{4}$

                            $= 7.5$

Now, lets find the Standard deviation

   Formula for Standard deviation = $\sqrt{\frac{1}{N}\sum_{I=1}^{n} (x_{i} - x_{m} )^{2} }$

Standard deviation = $\sqrt{\frac{1}{4}((20 - 25)^{2} + (40 - 25 )^{2} + (25 - 25 )^{2} + (15 - 25 )^{2} ) }$

                               = $\sqrt{\frac{1}{4} (25 + 225 + 0 + 100}$

                               = $\sqrt{\frac{1 }{4} ( 350) }$

                               = $\sqrt{875}$

                              = 9.354

Therefore ,

Mean = 25

Mean Deviation = 7.5

Standard deviation = 9.354