Exercise 13.1 (Revised) - Chapter 15 - Statistics - Ncert Solutions class 11 - Maths

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Chapter 13 - Statistics | NCERT Solutions Class 11 Maths

Find the mean deviation about the mean for the data in Exercises 1 and 2.
Ex 13.1 Question 1.

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

Answer.

Mean of the given data $\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10$

Mean deviation about mean $=\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right|=\frac{1}{8} \times 24=3$
Ex 13.1 Question 2.

$38,70,48,40,42,55,63,46,54,44$

Answer.

Mean of the given data
$
\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}=50
$

$
\begin{aligned}
& \text { Mean deviation about mean }=\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right| \\
& =\frac{1}{10} \times 84=8.4
\end{aligned}
$

Find the mean deviation about the median for the data in Exercises 3 and 4.
$3.13,17,16,14,11,13,10,16,11,18,12,17$
Ans. Arranging the data in ascending order, $10,11,11,12,13,13,14,16,16,17,17,18$
Here, $n=12$ (even number)
$
\begin{aligned}
& \therefore \text { Median }=\frac{1}{2}\left(6^n \text { term }+7^n \text { term }\right) \\
& =\frac{1}{2}(13+14)=\frac{27}{2}=13.5
\end{aligned}
$

Mean deviation about median $=\frac{1}{n} \sum_{i=1}^n\left|x_i-\mathrm{M}\right|$
$
=\frac{1}{12} \times 28=2.33
$
Ex 13.1 Question 4.

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

Answer.

Arranging the data in ascending order, $36,42,45,46,46,49,51,53,60,72$
Here, $n=10$ (even number)
$
\therefore \text { Median }=\frac{1}{2}\left(5^{\text {th }} \text { term }+6^{\text {th }} \text { term }\right)=\frac{1}{2}(46+49)=\frac{95}{2}=47.5
$

Mean deviation about median $=\frac{1}{n} \sum_{i=1}^n\left|x_i-\mathrm{M}\right|$
$
=\frac{1}{10} \times 70=7
$

Find the mean deviation about the mean for the data in Exercises 5 and 6.
Ex 13.1 Question 5.

Answer. 

Mean $(\bar{x})=\frac{1}{\mathrm{~N}} \sum f_i x_i$
$
=\frac{1}{25} \times 350=14
$

Mean deviation about mean $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|$
$
=\frac{158}{25}=6.32
$
Ex 13.1 Question 6.

Answer. 

Mean $(\bar{x})=\frac{1}{\mathrm{~N}} \sum f_i x_i$
$
=\frac{1}{80} \times 4000=50
$

Mean deviation about mean $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|$

$
=\frac{1}{80} \times 1280=16
$

Find the mean deviation about the median for the data in Exercises 7 and 8 .
Ex 13.1 Question 7.

Answer.

Here, $\mathrm{N}$ = 26 (even number)
$\therefore$ Median $=7$
Mean deviation about median $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\mathrm{M}\right|$
$
=\frac{1}{26} \times 84=3.23
$
Ex 13.1 Question 8.

Answer.

Here, $\mathrm{N}=29$ (odd number)
Median $=\left(\frac{29+1}{2}\right)^{\text {th }}$ term $=15^{\text {th }}$ term $=30$
$\therefore$ Median $=30$
Mean deviation about median $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\mathrm{M}\right|$
$
=\frac{1}{29} \times 148=5.1
$

Find the mean deviation about the mean for the data in Exercises 9 and 10 .
Ex 13.1 Question 9 .

Answer.

Mean $(\bar{x})=\frac{1}{\mathrm{~N}} \sum f_i x_i$
$
=\frac{1}{50} \times 17900=358
$

Mean deviation about mean $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|$
$
=\frac{1}{50} \times 7896=157.92
$
Ex 13.1 Question 10.

Answer.

$
\begin{aligned}
& \text { Mean }(\bar{x})=\frac{1}{\mathrm{~N}} \sum f_i x_i \\
& =\frac{1}{100} \times 12530=125.3 \\
& \text { Mean deviation about mean }=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right| \\
& =\frac{1}{100} \times 1128.8=11.288
\end{aligned}
$
Ex 13.1 Question 11.

Find the mean deviation about median for the following data:

Answer.

Here, $\frac{\mathrm{N}}{2}=\frac{50}{2}=25$
$
\begin{aligned}
& \therefore \text { Median class }=20-30 \\
& \therefore \text { Median }(M)=20+\frac{25-14}{14} \times 10 \\
& M=20+7.86=27.86
\end{aligned}
$

Mean deviation about median $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\mathrm{M}\right|$
$
=\frac{1}{50} \times 517.16=10.34
$
Ex 13.1 Question 12.

Calculate the mean deviation about median for the age distribution of 100 persons given below:

Answer.

Firstly the data is to be made continous by making classes 15.5-20.5, 20.5-25.5, 25.5$30.5,30.5-35.5,35.5-40.5,40.5-45.5,45.5-50.5,50.5-55.5$
otal

Here, $\frac{\mathrm{N}}{2}=\frac{100}{2}=50$
$
\begin{aligned}
& \therefore \text { Median class }=35.5-40.5 \\
& \therefore \text { Median }=35.5+\frac{50-37}{26} \times 5 \\
& =35.5+2.5=38
\end{aligned}
$

Mean deviation about median $=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\mathrm{M}\right|$
$
=\frac{1}{100} \times 735=7.35
$