Measures of Dispersion — Maths STD 11 Science — Question

$\bar{x}_1=\frac{\sum f _{1 i } x_{ i }}{ N _1}=\frac{120}{60}=2$

$\sigma_{x_1}^2=\frac{1}{ N _1} \sum f _{1 l } x_i^2-\left(\bar{x}_1\right)^2=\frac{394}{60}-2^2=6.5666-4=2.5666$

$\sigma_{x_1}=\sqrt{2.5666}=1.60$

C.V. of team $A=100 \times \frac{\sigma_{x_1}}{\bar{x}}=100 \times \frac{1.60}{2}=80 \%$

$\bar{x}_2=\frac{\sum f _{21} x_1}{ N _3}=\frac{140}{70}=2$

$\sigma_{x_2}^2=\frac{1}{ N _2} \sum f _{2 i} x_i^2-\left(\bar{x}_2\right)^2=\frac{436}{70}-2^2=6.2285-4=2.2285$

$\sigma_{x_2}=\sqrt{2.2285}=1.49$

C.V. of team $B=100 \times \frac{\sigma_{x_2}}{\bar{x}_2}=100 \times \frac{1.49}{2}=74.5 \%$

Since C.V. of team A > C.V. of team B,

Team B is more consistent.