Question

Which is false?
(a) If $\overline{x}$ is the mean of $x_1,x_2,...,x_n$, then $\sum _{i=1}^n(x_i-\overline{x})=0$.
(b) If the mean of $x_1,x_2,...,x_n$ is $\overline{x}$, then the mean of $(x_1+a),(x_2+a),...,(x_n+a)is(\overline{x}+a)$.
(c) If the mean of $x_1,x_2,...,x_n$ is $\overline{x}$ and $a\ne 0$, then the mean of $a{x}_1,a{x}_2,...,a{x}_n$ is $a\overline{x}$.
(d) If M is the median of $x_1,x_2,...,x_n$ and $a\ne 0$, then aM is the median of $a{x}_1,a{x}_2,...a{x}_n$.

Answer 1

(a)

$$\begin{gathered} \mathrm{If}\overline{\overline{x}}\mathrm{is}\mathrm{the}\mathrm{mean}\mathrm{of}{x}_1,x_2,...,x_{n,}\mathrm{then}\mathrm{we}\mathrm{have}: \\ \mathrm{Mean}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{observations}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{observations}} \\ \overline{\overline{x}}=\frac{x_1+x_2+...+x_n}{n} \end{gathered}$$
$\Rightarrow x_1+x_2+.....+x_n=nx........\left(1\right)$

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \sum \left(x_i-x\right)=\left(x_1-x\right)+\left(x_2-x\right)+.......+\left(x_n-x\right) \\ =\left(x_1+x_2+....+x_n\right)-nx \\ =nx-nx\left[\mathrm{From}\left(1\right)\right] \\ =0 \end{gathered}$$
Hence, (a) is correct.

(b)
$$\begin{gathered} \mathrm{Now},\mathrm{the}\mathrm{numbers}\mathrm{are}(x_1+a),(x_2+a),...(x_n+a). \\ \mathrm{Mean}=\frac{x_1+a+x_2+a+....x_n+a}{n} \\ =\frac{(x_1+x_2.....+x_n)+na}{n} \\ =\frac{(x_1+x_2.....+x_n)}{n}+a \\ =\overline{x}+a \end{gathered}$$

Hence, (b) is correct.

(c)
$$\begin{gathered} \mathrm{Now},\mathrm{the}\mathrm{numbers}\mathrm{are}a{x}_1,a{x}_2,.....a{x}_n. \\ \mathrm{Mean}=\frac{a{x}_1+a{x}_2+....a{x}_n}{n} \\ =\frac{a(x_1+x_2+....x_n)}{n} \\ =a\overline{\overline{x}} \\ \mathrm{Hence},\left(\mathrm{c}\right)\mathrm{is}\mathrm{correct}. \end{gathered}$$

(d) M is the median of x₁, x₂,...x_n.
We know that median is the middle term.
Hence, on multiplying the numbers by a, aM would be the median.
Examples:
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{numbers}\mathrm{be}1,2,3\mathrm{and}4. \\ \mathrm{Here},n\mathrm{is}4,\mathrm{which}\mathrm{is}\mathrm{even}. \\ \therefore \mathrm{Median}=\frac{1}{2}(2+3)=2.5 \\ \mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{multiply}\mathrm{the}\mathrm{numbers}\mathrm{by}5. \\ \mathrm{The}\mathrm{new}\mathrm{numbers}\mathrm{are}5,10,15\mathrm{and}20. \\ \mathrm{Median}=\frac{1}{2}(10+15)=12.5=5\times 2.5 \\ \mathrm{Now},\mathrm{considering}\mathrm{the}\mathrm{case}\mathrm{when}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{terms}\mathrm{is}\mathrm{odd}. \\ \mathrm{Let}\mathrm{the}\mathrm{numbers}\mathrm{be}1,2\mathrm{and}3. \\ \mathrm{Here},n\mathrm{is}3,\mathrm{which}\mathrm{is}\mathrm{odd}. \\ \therefore \mathrm{Median}=2 \\ \mathrm{Now},\mathrm{multiply}\mathrm{the}\mathrm{numbers}\mathrm{by}5. \\ \mathrm{The}\mathrm{new}\mathrm{numbers}\mathrm{are}5,10and15. \\ \mathrm{Median}=10=5\times 2 \\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{examples},\mathrm{w}e\mathrm{can}\mathrm{say}\mathrm{that}\mathrm{if}M\mathrm{is}\mathrm{the}\mathrm{median}\mathrm{of}{x}_1,x_{2,...}{x}_n\mathrm{and}a\ne 0,\mathrm{then}aM\mathrm{is}\mathrm{the}\mathrm{median}\mathrm{of}a{x}_1,a{x}_2,...a{x}_n. \end{gathered}$$

Hence, (d) is correct.