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- The formula required to solve this problem are : Mean $ A+ \large\frac{\sum f_i x_i}{\sum f_i} $
- Standard deviation $\sigma= \sqrt {\large\frac{\sum f_i x_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_ix_i}{\sum f_i} \bigg)^2 }$$ \times h$
- Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$

Step 2:

For Team A: : Mean $ A+ \large\frac{\sum f_i x_i}{\sum f_i} $

$\qquad= \large\frac{50}{25}$

$\qquad= 2$

Step 3:

Standard deviation $\sigma= \sqrt {\large\frac{\sum f_i x_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_ix_i}{\sum f_i} \bigg)^2 }$$ \times h$

$\qquad= \sqrt { \large\frac{130}{25} - \bigg( \large\frac{50}{25}\bigg)^2}$

$\qquad= \sqrt{\large\frac {30}{25}}$

$\qquad= \large\frac{5.48}{5} $$=1.096$

Step 4:

Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$

$\qquad= \large\frac{1.096 }{2}$$ \times 100$

$\qquad= \large\frac{109.6}{2}$

$\qquad=54.8$

For team B:

Mean $\bar {X} =2$(given)

$\sigma= 1.25$ (given)

Step 1:

Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$

$\qquad= \large\frac{1.25}{2} $$\times 100$

$\qquad= \large\frac{125}{2}$

$\qquad= 62.5$

COMPARISON:

Consistency of the team depends upon the coefficient of variation.

Lesser the coefficient of variation, more consistent the team is

$\therefore $ coefficient of variation team A is lesser as compared to coefficient of variation of team B.

Team A is more consistent.

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