Chapter 15: Statistics Exercise 15.1

Ans.

We are provided the numbers

$4,7,8,9,10,12,13,17$.

Therefore, mean of the numbers,

 $ \overline{x}=\dfrac{4+7+8+9+10+12+13+17}{8} $

$ =\dfrac{80}{8} $

$ =10 $

The deviations for the given numbers with the mean $\overline{x}=10$, 

that is ${{x}_{i}}-\overline{x}$, for $i=1,2,...,8$ are 

$-6,-3,-2,-1,\,\,0,\,\,2,\,\,3,\,\,7$.

Also, the absolute values of the obtained deviations, that is $\left| {{x}_{i}}-x \right|$ are

$6,3,2,1,0,2,3,7$.

Thus, the mean deviation about the mean $\overline{x}$ for the given numbers is

$ M.D.\left( \overline{x} \right)=\dfrac{\sum\limits_{i=1}^{8}{\left| {{x}_{i}}-overline{x} \right|}}{8}\\ $

 $ =\dfrac{6+3+2+1+0+2+3+7}{8}\\ $

$ =\dfrac{24}{8}\\ $

$ =3\\ $

Ans.

We are provided that numbers

$\text{38,70,48,40,42,55,63,46,54,44}$.

Therefore, the mean of the numbers,

$ \overline{x}=\dfrac{38+70+48+40+42+55+63+46+54+44}{10}\\ $

$ =\dfrac{500}{10}\\ $

$ =50\\ $

Now, the deviations for the given numbers with the mean $\overline{x}=50$, that is

${{x}_{i}}-x$, for $i=1,2,3,...,10$ are 

$-12,20,-2,-10,-8,5,13,-4,4,-6$.

So, the absolute values of the obtained deviations for the given numbers, that is $\left| {{x}_{i}}-x \right|$ are

$12,20,2,10,8,5,13,4,4,6$.

Thus, the mean deviation about the mean $\overline{x}$ for the given numbers is

$M.D.\left( \overline{x} \right)=\dfrac{\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-x \right|}}{10}$

$=\dfrac{12+20+2+10+8+5+13+4+4+6}{10}$

$=\dfrac{84}{10}$

$=8.4$

Ans.

We are provided the numbers

$\text{13,17,16,14,11,13,10,16,11,18,12,17}$.

There is total $12$ numbers, and it is even.

Now, arranging the numbers in the ascending order, gives

$10,11,11,12,13,13,14,16,16,17,17,18$.

It is known that, 

Median, $M=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\,\,number+{{\left( \dfrac{n}{2}+1 \right)}^{th}}number}{2}$, when $n$ is even.

Therefore, the median

$M=\dfrac{{{\left( \dfrac{12}{2} \right)}^{th}}number+{{\left( \dfrac{12}{2}+1 \right)}^{th}}number}{2}$

$=\dfrac{{{6}^{th}}\,number+{{7}^{th}}\,\,number}{2}$

$=\dfrac{13+14}{2}$

$=\dfrac{27}{2}$

$=13.5$

Now, the deviations about the median $M$ for the given numbers, that is ${{x}_{i}}-M$, for $i=1,2,3,...,12$ are

$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,\,0.5,\,2.5,\,3.5,\,3.5,\,3.5,\,4.5$

So, the absolute values of the resulted deviations, that is $\left| {{x}_{i}}-M \right|$, for $i=1,2,3,...,12$ are

$3.5,\,2.5,\,2.5,\,1.5,\,0.5,\,0.5,\,0.5,\,2.5,\,2.5,\,3.5,\,3.5,\,4.5\,$.

Thus, the mean deviation about the median $M$ for the given numbers,

$M.D.\left( M \right)=\dfrac{\sum\limits_{i=1}^{12}{\left| {{x}_{i}}-M \right|}}{12}$

$=\dfrac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+4.5}{12}$

$=\dfrac{28}{12}$

$=2.33$

Ans.

We are provided the numbers

$\text{36,}\,\text{72,}\,\text{46,}\,\text{42,}\,\text{60,}\,\text{45,}\,\text{53,}\,\text{46,}\,\text{51,}\,\text{49}$.

There is total $10$ numbers, and it is even.

Now, arranging the numbers in the ascending order, gives

$36,\,42,\,45,\,46,\,46,\,49,\,51,\,53,\,60,\,72$.

It is known that, 

Median, $M=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\,\,number+{{\left( \dfrac{n}{2}+1 \right)}^{th}}number}{2}$, when $n$ is even.

Therefore, the median

$M=\dfrac{{{\left( \dfrac{10}{2} \right)}^{th}}number+{{\left( \dfrac{10}{2}+1 \right)}^{th}}number}{2}$

$=\dfrac{{{5}^{th}}\,number+{{6}^{th}}\,\,number}{2}$

$=\dfrac{46+49}{2}$

$=\dfrac{95}{2}$

$=47.5$

Now, the deviations about the median $M$ for the given numbers, that is ${{x}_{i}}-M$, for $i=1,2,3,...,10$ are

$-11.5,\,-5.5,\,-2.5,\,-1.5,\,-1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$

So, the absolute values of the resulted deviations, that is $\left| {{x}_{i}}-M \right|$, for $i=1,2,3,...,10$ are

$11.5,\,5.5,\,2.5,\,1.5,\,1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$.

Thus, the mean deviation about the median $M$ for the given numbers,

$M.D.\left( M \right)=\dfrac{\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-M \right|}}{10}$

$=\dfrac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

$=\dfrac{70}{10}$

$=7$.

Ans.

Consider the following table of data.

Thus, from the above table we have,

$N=\sum\limits_{i=1}^{5}{{{f}_{i}}}=25$,

$\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}=350}$

Therefore, the mean, 

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}\\ $

$  =\dfrac{1}{25}\times 350\\ $

$ =14\\ $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{25}\times 158\\ $

$  =6.32\\ $

Ans.

Consider the following table of data.

Thus, from the above table we have,

$N=\sum\limits_{i=1}^{5}{{{f}_{i}}}=80$,

$\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}=}4000$

Therefore, the mean, 

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}\\ $

$ =\dfrac{1}{80}\times 4000\\ $

 $ =50\\ $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{80}\times 1280\\ $

$ =16\\ $.

Ans.

Notice that the given data is already in ascending order.

Consider the following table that shows the cumulative frequencies of the provided data.

Thus, from the above table we have,

$N=26$, that is an even number.

Therefore, the median can be obtained by calculating the mean of the ${{13}^{th}}$ and ${{14}^{th}}$ observations and both the observations belongs to the cumulative frequency $14$.

Therefore,

Median,

 $ M=\dfrac{{{13}^{th}}\,\,observation+{{14}^{th}}\,observation}{2}\\ $

$ =\dfrac{7+7}{2}\\ $

$ =7\\ $

Now, the following table shows the absolute values of the deviations about the median, that is $\left| {{x}_{i}}-M \right|$.

Thus, from the above table, we have

$\sum\limits_{i=1}^{6}{{{f}_{i}}}=26$ and $\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-M \right|}=84$.

Hence, the required mean deviation about the median,

$ M.D\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-M \right|}\\ $

$ =\dfrac{1}{26}\times 84\\ $

$ =3.23\\ $

Ans.

Notice that the given data is already in ascending order.

Consider the following table that shows the cumulative frequencies of the provided data.

Thus, from the above table we have,

$N=29$, that is an odd number.

Therefore, the median will be the $\dfrac{29+1}{2}=\dfrac{30}{2}={{15}^{th}}$ observation.

Notice that, the ${{15}^{th}}$ observation belongs to the cumulative frequency $21$. So, the median is the corresponding observation, that is $30$.

Now, the following table shows the absolute values of the deviations about the median, that is $\left| {{x}_{i}}-M \right|$.

Thus, from the above table, we have

$\sum\limits_{i=1}^{5}{{{f}_{i}}}=29$ and $\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-M \right|}=148$.

Hence, the required mean deviation about the median,

  $ M.D\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-M \right|} $

 $ =\dfrac{1}{29}\times 148 $

 $ =5.1 $

Ans.

Consider the following table of data.

Thus, from the above table we have,

\[N=\sum\limits_{i=1}^{8}{{{f}_{i}}}=50\],

$\sum\limits_{i=1}^{8}{{{f}_{i}}{{x}_{i}}=}17900$

Therefore, the mean,  

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}{{x}_{i}}} $

 $ =\dfrac{1}{50}\times 17900 $

 $ =358 $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{50}\times 7896\\ $

$=157.92\\ $

Ans.

Consider the following table of data.

Thus, from the above table we have,

\[N=\sum\limits_{i=1}^{6}{{{f}_{i}}}=100\],

$\sum\limits_{i=1}^{6}{{{f}_{i}}{{x}_{i}}=}12530$

Therefore, the mean,

 $\overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}{{x}_{i}}} \\ $ 

 $=\dfrac{1}{100}\times 12530 \\ $ 

 $=125.3 \\ $ 

Hence, the mean deviation about the mean is given by
  $M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|} \\ $ 

 $=\dfrac{1}{100}\times 1128.8 \\ $ 

 $=11.28 \\ $ 

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