- ARemains the same.
- BIs increased by 5.
- CIs decreased by 5.
- DBecomes 5 times the original mean.
Solution:
Is increased by 5
Then old mean x old $=\frac{\displaystyle\sum_{\text{i}=1}^{\text{n}}\text{x}_{\text{i}}}{\text{n}}$
Now, adding 5 in each observation, the new mean becomes
$\overline{\text{x}}_\text{New}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+....+(\text{x}_\text{n}+5)}{\text{n}}$
$\Rightarrow\overline{\text{x}}_\text{New}=\frac{(\text{x}_1 + \text{x}_2+....+\text{x}_\text{n})+5\text{n})}{\text{n}}$
$\Rightarrow\overline{\text{x}}_\text{New}=\frac{\displaystyle\sum_{\text{i}=1}^{\text{n}}\text{x}_{\text{i}}}{\text{n}}+5=\overline{\text{x }}{\text{old}}+5$
$\Rightarrow\overline{\text{x}}_\text{New}=\overline{\text{x}}_\text{old}+5$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

