# NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1

Go back to  'Statistics'

## Chapter 15 Ex.15.1 Question 1

Find the mean deviation about the mean for the data $$4,7,8,9,10,12,13,17$$.

### Solution

The given data is $$4,7,8,9,10,12,13,17$$

Mean of the data

\begin{align}\bar x &= \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8}\\&= \frac{80}{8}\\&= 10\end{align}

The deviations of the respective observations from the mean $$\bar x$$, i.e., $${x_i} - \bar x$$ are

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

The absolute values of the deviations, $$\left| {{x_i} - \bar x} \right|$$ are $$6,3,2,1,0,2,3,7$$

The required mean deviation about the mean is

\begin{align}M.D.\left( {\bar x} \right)& = \frac{{\sum\limits_{i = 1}^8 {\left| {x_i - \bar x} \right|} }}{8}\\&= \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}\\&= \frac{24}{8}\\&= 3\end{align}

## Chapter 15 Ex.15.1 Question 2

Find the mean deviation about the mean for the data $$38,70,48,40,42,55,63,46,54,44$$.

### Solution

The given data is $$38,70,48,40,42,55,63,46,54,44$$

Mean of the data

\begin{align}\bar x &= \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10}\\&= \frac{500}{10}\\&= 50\end{align}

The deviations of the respective observations from the mean $$\bar x$$, i.e., $${x_i} - \bar x$$ are

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

The absolute values of the deviations, $$\left| {x_i - \bar x} \right|$$ are

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

The required mean deviation about the mean is

\begin{align}M.D.\left( {\bar x} \right) &= \frac{{\sum\limits_{i = 1}^{10} {\left| {x_i - \bar x} \right|} }}{10}\\&= \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}\\&= \frac{84}{10}\\&= 8.4\end{align}

## Chapter 15 Ex.15.1 Question 3

Find the mean deviation about the median for the data $$13,17,16,14,11,13,10,16,11,18,12,17$$.

### Solution

The given data is $$13,17,16,14,11,13,10,16,11,18,12,17$$

Here, the numbers of observations are $${\rm{12}}$$, i.e., even.

Arranging the above data in ascending order, we obtain $$10,11,11,12,13,13,14,16,16,17,17,18$$

Median of the data

\begin{align}M &= \frac{{\left( {\frac{12}{2}} \right)^{th}observation + {\left( {\frac{12}{2} + 1} \right)^{th}}observation}}{2}\\&= \frac{{6^{th}observation + 7^{th}observation}}{2}\\&= \frac{13 + 14}{2}\\&= \frac{27}{2}\\&= 13.5\end{align}

The deviations of the respective observations from the median, i.e., $${x_i} - M$$ are

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

The absolute values of the deviations, $$\left| {{x_i} - M} \right|$$ are

$3.5,2.5,2.5,1.5,0.5,0.5,0.5,2.5,2.5,3.5,3.5,4.5$

The required mean deviation about the median is

\begin{align}M.D.\left( M \right) &= \frac{{\sum\limits_{i = 1}^{12} {\left| {{x_i} - M} \right|} }}{{12}}\\&= \frac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5}{12}\\&= \frac{28}{12}\\&= 2.33\end{align}

## Chapter 15 Ex.15.1 Question 4

Find the mean deviation about the median for the data $$36,72,46,42,60,45,53,46,51,49$$

### Solution

The given data is $$36,72,46,42,60,45,53,46,51,49$$

Here, the numbers of observations are $${\rm{1}}0$$, hence even.

Arranging the above data in ascending order, we obtain $$36,42,45,46,46,49,51,53,60,72$$

Median of the data

\begin{align}M &= \frac{{{{\left( {\frac{10}{2}} \right)}^{th}}observation + {\left( {\frac{{10}}{2} + 1} \right)^{th}}observation}}{2}\\&= \frac{{5^{th}observation \;+\; 6^{th}observation}}{2}\\&= \frac{46 + 49}{2}\\&= \frac{95}{2}\\&= 47.5\end{align}

The deviations of the respective observations from the median, i.e., $${x_i} - M$$ are

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

The absolute values of the deviations, $$\left| {{x_i} - M} \right|$$are

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

The required mean deviation about the median is

\begin{align}M.D.\left( M \right) &= \frac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - M} \right|} }}{{10}}\\&= \frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10}\\&= \frac{70}{10}\\&= 7\end{align}

## Chapter 15 Ex.15.1 Question 5

Find the mean deviation about the mean for the data

 $${x_i}$$ $$5$$ $$10$$ $$15$$ $$20$$ $$25$$ $${f_i}$$ $$7$$ $$4$$ $$6$$ $$3$$ $$5$$

### Solution

 $${x_i}$$ $${f_i}$$ $${f_i}{x_i}$$ $$\left| {{x_i} - \bar x} \right|$$ $${f_i}\left| {{x_i} - \bar x} \right|$$ $$5$$ $$7$$ 35 9 63 $$10$$ $$4$$ $$40$$ $$4$$ $$16$$ $$15$$ $$6$$ $$90$$ $$1$$ $$6$$ $$20$$ $$3$$ $$60$$ $$6$$ $$18$$ $$25$$ $$5$$ $$125$$ $$11$$ $$55$$  $$25$$ $$350$$  $$158$$

$$N = \sum\limits_{i = 1}^5 {{f_i}} = 25$$ and $$\sum\limits_{i = 1}^5 {{f_i}{x_i} = 350}$$

Therefore,

\begin{align}\overline x &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} \\&= \frac{1}{{25}} \times 350\\&= 14\end{align}

\begin{align}M.D.\left( {\bar x} \right) &= \frac{1}{N}\left| {{x_i} - \bar x} \right|\\&= \frac{1}{{25}} \times 158\\&= 6.32\end{align}

## Chapter 15 Ex.15.1 Question 6

Find the mean deviation about the mean for the data

 $${x_i}$$ $$10$$ $$30$$ $$50$$ $$70$$ $$90$$ $${f_i}$$ $$4$$ $$24$$ $$28$$ $$16$$ $$8$$

### Solution

 $${x_i}$$ $${f_i}$$ $${f_i}{x_i}$$ $$\left| {{x_i} - \overline x } \right|$$ $${f_i}\left| {{x_i} - \overline x } \right|$$ $$10$$ $$4$$ $$40$$ $$40$$ $$160$$ $$30$$ $$24$$ $$720$$ $$20$$ $$480$$ $$50$$ $$28$$ $$1400$$ $$0$$ $$0$$ $$70$$ $$16$$ $$1120$$ $$20$$ $$320$$ $$90$$ $$8$$ $$720$$ $$40$$ $$320$$ $$80$$ $$4000$$ $$1280$$

$$N = \sum\limits_{i = 1}^5 {{f_i}} = 80$$ and $$\sum\limits_{i = 1}^5 {{f_i}{x_i} = 4000}$$

Therefore,

\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} \\&= \frac{1}{{80}} \times 4000\\&= 50\end{align}

\begin{align}M.D.\left( {\overline x } \right) &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} \\&= \frac{1}{{80}} \times 1280\\&= 16\end{align}

## Chapter 15 Ex.15.1 Question 7

Find the mean deviation about the median for the data

 $${x_i}$$ $$5$$ $$7$$ $$9$$ $$10$$ $$12$$ $$15$$ $${f_i}$$ $$8$$ $$6$$ $$2$$ $$2$$ $$2$$ $$6$$

### Solution

The given observations are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

 $${x_i}$$ $${f_i}$$ $$c.f$$ $$5$$ $$8$$ $$8$$ $$7$$ $$6$$ $$14$$ $$9$$ $$2$$ $$16$$ $$10$$ $$2$$ $$18$$ $$12$$ $$2$$ $$20$$ $$15$$ $$6$$ 26

Here, $$N = 26$$, which is even.

Hence, Median is the mean of 13th and 14th observations. Both the observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

\begin{align} \text{Median}&= \frac{{{13^{th}}\,observation + {14^{th}}observation}}{2}\\& = \frac{7 + 7}{2}\\&= 7 \end{align}

The absolute values of the deviations, $$\left| {{x_i} - M} \right|$$are

 $$\left| {{x_i} - M} \right|$$ 2 0 2 3 5 8 $${f_i}$$ 8 6 2 2 2 6 $${f_i}\left| {{x_i} - M} \right|$$ 16 0 4 6 10 48

$$\sum\limits_{i = 1}^6 {{f_i}} = 26$$ and $$\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} = 84$$

Hence, the mean deviation about the median

\begin{align}M.D.\left( M \right) &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} \\&= \frac{1}{26} \times 84\\&= 3.23\end{align}

## Chapter 15 Ex.15.1 Question 8

Find the mean deviation about the median for the data

 $${x_i}$$ $$15$$ $$21$$ $$27$$ $$30$$ $$35$$ $${f_i}$$ $$3$$ $$5$$ $$6$$ $$7$$ $$8$$

### Solution

The given observations are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

 $${x_i}$$ $${f_i}$$ $$c.f$$ $$15$$ $$3$$ $$3$$ $$21$$ $$5$$ $$8$$ $$27$$ $$6$$ $$14$$ $$30$$ $$7$$ $$21$$ $$35$$ $$8$$ $$29$$

Here, $$N = 29$$, which is odd.

$$\text{Median} = {\left( {\frac{{29 + 1}}{2}} \right)^{th}} = {15^{th}}$$observation

This observation lies in the cumulative frequency $${\rm{21}}$$, for which the corresponding observation is$$30$$.

Therefore, Median = $$30$$

The absolute values of the deviations, $$\left| {{x_i} - M} \right|$$are

 $$\left| {{x_i} - M} \right|$$ $$15$$ $$9$$ $$3$$ $$0$$ $$5$$ $${f_i}$$ $$3$$ $$5$$ $$6$$ $$7$$ $$8$$ $${f_i}\left| {{x_i} - M} \right|$$ $$45$$ $$45$$ $$18$$ $$0$$ $$40$$

Here, $$\sum\limits_{i = 1}^5 {{f_i}} = 29$$ and $$\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} = 148$$

$$M.D.\left( M \right) = \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} = \frac{1}{{29}} \times 148 = 5.1$$

## Chapter 15 Ex.15.1 Question 9

Find the mean deviation about the mean for the data

 Income per day Number of persons $$0 - 100$$ $$4$$ $$100 - 200$$ $$8$$ $$200 - 300$$ $$9$$ $$300 - 400$$ $$10$$ $$400 - 500$$ $$7$$ $$500 - 600$$ $$5$$ $$600 - 700$$ $$4$$ $$700 - 800$$ $$3$$

### Solution

The following table is formed.

 Income per day Number of persons $${f_i}$$ Mid-point $${x_i}$$ $${f_i}{x_i}$$ $$\left| {{x_i} - \overline x } \right|$$ $${f_i}\left| {{x_i} - \overline x } \right|$$ $$0 - 100$$ $$4$$ $$50$$ $$200$$ $$308$$ $$1232$$ $$100 - 200$$ $$8$$ $$150$$ $$1200$$ $$208$$ $$1664$$ $$200 - 300$$ $$9$$ $$250$$ $$2250$$ $$108$$ $$972$$ $$300 - 400$$ $$10$$ $$350$$ $$3500$$ $$8$$ $$80$$ $$400 - 500$$ $$7$$ $$450$$ $$3150$$ $$92$$ $$644$$ $$500 - 600$$ $$5$$ $$550$$ $$2750$$ $$192$$ $$960$$ $$600 - 700$$ $$4$$ $$650$$ $$2600$$ $$292$$ $$1168$$ $$700 - 800$$ $$3$$ $$750$$ $$2250$$ $$392$$ $$1176$$ Total $$50$$ $$17900$$ $$7896$$

Here, $$N = \sum\limits_{i = 1}^8 {{f_i}} = 50$$ and $$\sum\limits_{i = 1}^8 {{f_i}{x_i} = 17900}$$

Therefore,

\begin{align}\overline x &= \frac{1}{N}\sum\limits_{i = 1}^8 {{f_i}{x_i}} \\ &= \frac{1}{{50}} \times 17900\\ &= 358\end{align}

Hence, the mean deviation about the mean

\begin{align}M.D.\left( {\bar x} \right) &= \frac{1}{N}\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - \bar x} \right|} \\&= \frac{1}{{50}} \times 7896\\&= 157.92\end{align}

## Chapter 15 Ex.15.1 Question 10

Find the mean deviation about the mean for the data

 Height in cms $$95-105$$ $$105-115$$ $$115-125$$ $$125-135$$ $$135-145$$ $$135-145$$ Number of boys $$9$$ 13 26 30 12 10

### Solution

The following table is formed.

 Height in cms Number of boys$${f_i}$$ Mid-point $${x_i}$$ $${f_i}{x_i}$$ $$\left| {{x_i} - \overline x } \right|$$ $${f_i}\left| {{x_i} - \overline x } \right|$$ $$95 - 105$$ $$9$$ $$100$$ $$900$$ $$25.3$$ $$227.7$$ $$105 - 115$$ $$13$$ $$110$$ $$1430$$ $$15.3$$ $$198.9$$ $$115 - 125$$ $$26$$ $$120$$ $$3120$$ $$5.3$$ $$137.8$$ $$125 - 135$$ $$30$$ $$130$$ $$3900$$ $$4.7$$ $$141$$ $$135 - 145$$ $$12$$ $$140$$ $$1680$$ $$14.7$$ $$176.4$$ $$145 - 155$$ $$10$$ $$150$$ $$1500$$ $$24.7$$ $$247$$ Total $$100$$ $$12530$$ $$1128.8$$

Here, $$N = \sum\limits_{i - 1}^6 {f_i = 100}$$ and $$\sum\limits_{i = 1}^6 {{f_i}{x_i} = 12530}$$

Therefore,

\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}} \\&= \frac{1}{{100}} \times 12530\\&= 125.3\end{align}

Hence, the mean deviation about the mean

\begin{align}M.D.\left( {\bar x} \right) &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \bar x} \right|} \\&= \frac{1}{{100}} \times 1128.8\\&= 11.28\end{align}

## Chapter 15 Ex.15.1 Question 11

Find the mean deviation about median for the following data:

 Marks $$0 - 10$$ $$10 - 20$$ $$20 - 30$$ $$30 - 40$$ $$40 - 50$$ $$50 - 60$$ No. of girls 6 8 14 16 4 2

### Solution

 Marks No. of girls c.f. Mid-point $$\left( {{x_i}} \right)$$ $$\left| {{x_i} - M} \right|$$ $${f_i}\left| {{x_i} - M} \right|$$ $$0 - 10$$ $$6$$ $$6$$ $$5$$ $$22.85$$ $$137.1$$ $$10 - 20$$ $$8$$ $$14$$ $$15$$ $$12.85$$ $$120.8$$ $$20 - 30$$ $$14$$ $$28$$ $$25$$ $$2.85$$ $$39.9$$ $$30 - 40$$ $$16$$ $$44$$ $$35$$ $$7.15$$ $$114.4$$ $$40 - 50$$ $$4$$ $$48$$ $$45$$ $$17.15$$ $$68.6$$ $$50 - 60$$ $$2$$ $$50$$ $$55$$ $$27.15$$ $$54.3$$ $$50$$ $$517.1$$

Here, $$\sum\limits_{i = 1}^6 {{f_i}} = 50$$ and $$\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} = 517.1$$

$$l = 20,C = 14,f = 14,h = 10,N = 50$$

Median,

\begin{align}&= l + \frac{{\frac{N}{2} - C}}{f} \times h\\&= 20 + \frac{25 - 14}{14} \times 10\\&= 20 + 7.85\\&= 27.85\end{align}

Hence, the mean deviation about median

\begin{align}M.D.\left( M \right) &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} \\&= \frac{1}{50} \times 517.1\\&= 10.34\end{align}

## Chapter 15 Ex.15.1 Question 12

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

 Age (in years) $$16 - 20$$ $$21 - 25$$ $$26 - 30$$ $$31 - 35$$ $$36 - 40$$ $$41 - 45$$ $$46 - 50$$ $$51 - 55$$ Number 5 6 12 14 26 12 16 9

[Hint: Convert the given data into continuous frequency distribution by subtracting $$0.5$$ from the lower limit and adding $$0.5$$ to the upper limit of each class interval]

### Solution

The given data is not continuous.

Therefore, it has to be converted into continuous frequency distribution by subtracting $$0.5$$ from the lower limit and adding $$0.5$$ to the upper limit of each class interval.

The table is formed as follows

 Age (in years) Number of persons $${f_i}$$ $$c.f$$ Mid-point $${x_i}$$ $$\left| {{x_i} - M} \right|$$ $${f_i}\left| {{x_i} - M} \right|$$ $$15.5 - 20.5$$ $$5$$ $$5$$ $$18$$ $$20$$ $$100$$ $$20.5 - 25.5$$ $$6$$ $$11$$ $$23$$ $$15$$ $$90$$ $$25.5 - 30.5$$ $$12$$ $$23$$ $$28$$ $$10$$ $$120$$ $$30.5 - 35.5$$ $$14$$ $$37$$ $$33$$ $$5$$ $$70$$ $$35.5 - 40.5$$ $$26$$ $$63$$ $$38$$ $$0$$ $$0$$ $$40.5 - 45.5$$ $$12$$ $$75$$ $$43$$ $$5$$ $$60$$ $$45.5 - 50.5$$ $$16$$ $$91$$ $$48$$ $$10$$ $$160$$ $$50.5 - 55.5$$ $$9$$ $$100$$ $$53$$ $$15$$ $$135$$ $$100$$ $$735$$

The class interval containing the $${\left( {\frac{N}{2}} \right)^{th}}$$or $${50^{th}}$$ item is $$35.5 - 40.5$$

Therefore, $$35.5 - 40.5$$ is the median class.

It is known that,

Median $$= l + \left( {\frac{{\frac{N}{2} - C}}{f}} \right) \times h$$

Here, $$l = 35.5,C = 37,f = 26,h = 5,N = 100$$

Median,

\begin{align}M &= 35.5 + \frac{{50 - 37}}{{26}} \times 5\\&= 35.5 + \frac{{13 \times 5}}{{26}}\\&= 35.5 + 2.5\\&= 38\end{align}

Thus, mean deviation about the median is given by,

\begin{align}M.D.\left( M \right) = \frac{1}{N}\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - M} \right|} \\= \frac{1}{{100}} \times 735\\= 7.35\end{align}

Related Sections
Related Sections
Instant doubt clearing with Cuemath Advanced Math Program