A data consists of 20 observations x_1,x_2, ,x_20.

Question

A data consists of 20 observations ${x}_{1},{x}_{2},\ldots ,{x}_{20}$. If $\sum _{i=1}^{20}{({x}_{i}+5)}^{2}=2500$ and $\sum _{i=1}^{20}{({x}_{i}-5)}^{2}=100$, then the ratio of mean to standard deviation of this data is :

Accepted Answer

Let the mean of the 20 observations be $\vec{x}$ and the standard deviation be $\sigma $.

We are given:

$\sum _{20}^{i=1}({x}_{i}+5{)}^{2}=2500$

and

$\sum _{20}^{i=1}({x}_{i}-5{)}^{2}=100$

We will expand both expressions.

First,

$\sum _{20}^{i=1}({x}_{i}+5{)}^{2}=\sum _{20}^{i=1}({x}_{2}^{i}+10{x}_{i}+25)$

$=\sum {x}_{2}^{i}+10\sum {x}_{i}+20\cdot 25$

$=\sum {x}_{2}^{i}+10\sum {x}_{i}+500$

So,

$\sum {x}_{2}^{i}+10\sum {x}_{i}+500=2500$

$\sum {x}_{2}^{i}+10\sum {x}_{i}=2000\, ...(1)$

Now expand the second one:

$\sum _{20}^{i=1}({x}_{i}-5{)}^{2}=\sum _{20}^{i=1}({x}_{2}^{i}-10{x}_{i}+25)$

$=\sum {x}_{2}^{i}-10\sum {x}_{i}+500$

So,

$\sum {x}_{2}^{i}-10\sum {x}_{i}+500=100$

$\sum {x}_{2}^{i}-10\sum {x}_{i}=-400\, ...(2)$

Now add (1) and (2):

$2\sum {x}_{2}^{i}=1600$

$\sum {x}_{2}^{i}=800$

Now subtract (2) from (1):

$20\sum {x}_{i}=2400$

$\sum {x}_{i}=120$

Hence mean:

$\vec{x}=\frac{\sum {x}_{i}}{20}=\frac{120}{20}=6$

Now use the formula for standard deviation:

$\sigma =\sqrt{\frac{\sum {x}_{2}^{i}}{20}-{\vec{x}}^{2}}$

$=\sqrt{\frac{800}{20}-{6}^{2}}$

$=\sqrt{40-36}$

$=\sqrt{4}=2$

Therefore, the ratio of mean to standard deviation is

$\vec{x}:\sigma =6:2=3:1$

So, the correct option is:

$3:1$

Option B

Solutions · 1