NIMCET 2020 — Mathematics PYQ

Concept:

For a set $x_1,x_2,\dots ,x_n$ of $n$ observations:
- Mean: $\bar{x}=\frac{x_1+x_2+\dots +x_n}{n}$
- Variance: ${\sigma }^2=\frac{(x_1-\bar{x}{)}^2+(x_2-\bar{x}{)}^2+\dots +(x_n-\bar{x}{)}^2}{n}=\frac{\sum (x_i-\bar{x}{)}^2}{n}=\frac{\sum x_i^2}{n}-{\bar{x}}^2$
- Standard Deviation: $\sigma =\sqrt{{\sigma }^2}=\sqrt{\text{Variance}}$.

Calculation:

Total number of items in the distribution $=\sum f_i=3+6+9+13+8+5+4=48$.

The Mean $(\bar{x})$ of the given set $=\frac{\sum f_i{x}_i}{\sum f_i}$.

$\Rightarrow \bar{x}=\frac{6\times 3+7\times 6+8\times 9+9\times 13+10\times 8+11\times 5+12\times 4}{48}=\frac{432}{48}=9$.

Let's calculate the variance using the formula: ${\sigma }^2=\frac{\sum x_i^2}{n}-{\bar{x}}^2$.

$\frac{\sum x_i^2}{n}=\frac{6^2\times 3+7^2\times 6+8^2\times 9+9^2\times 13+10^2\times 8+11^2\times 5+12^2\times 4}{48}=\frac{4012}{48}=83.58$.

$\therefore {\sigma }^2=83.58-9^2=83.58-81=2.58$.

And, Standard Deviation $(\sigma )=\sqrt{{\sigma }^2}=\sqrt{\text{Variance}}=\sqrt{2.58}\approx 1.607$.