Q1. 4, 7, 8, 9, 10, 12, 13, 17

Solution

The given data consists of the observations \(4,\; 7,\; 8,\; 9,\; 10,\; 12,\; 13,\; 17\). Since the data are ungrouped, each value is treated as occurring once, so \(N=8\).

\(x_i\)	\(|x_i-\overline{x}|\)
4	6
7	3
8	2
9	1
10	0
12	2
13	3
17	7
\(\sum x_i=80\)	\(\sum |x_i-\overline{x}|=24\)

First, we calculate the arithmetic mean of the observations.

\[ \begin{aligned} \overline{x}&=\dfrac{1}{N}\sum_{i=1}^{n} x_i\\ &=\dfrac{80}{8}\\ &=10 \end{aligned} \]

Using this mean, the absolute deviations \(|x_i-\overline{x}|\) are computed for each observation as shown in the table. Their total is \(24\).

The mean deviation about the mean is obtained by dividing the sum of absolute deviations by the total number of observations.

\[ \begin{aligned} \text{Mean Deviation about Mean}&=\dfrac{1}{N}\sum_{i=1}^{n}|x_i-\overline{x}|\\ &=\dfrac{24}{8}\\ &=3 \end{aligned} \]

Hence, the mean deviation about the mean for the given data is \(3\).

Q2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution

The given data consists of the observations \(38,\; 70, 48,\; 40,\; 42,\; 55,\; 63,\; 46,\; 54,\; 44\). Since the data are ungrouped, each value is treated as occurring once, so \(N=8\).

\(x_i\)	\(|x_i-\overline{x}|\)
38	12
70	20
48	2
40	10
42	8
55	5
63	13
46	4
54	4
44	6
\(\sum x_i=500\)	\(\sum |x_i-\overline{x}|=84\)

First, we calculate the arithmetic mean of the observations.

\[ \begin{aligned} \overline{x}&=\dfrac{1}{N}\sum_{i=1}^{n} x_i\\ &=\dfrac{500}{10}\\ &=10 \end{aligned} \]

Using this mean, the absolute deviations \(|x_i-\overline{x}|\) are computed for each observation as shown in the table. Their total is \(84\).

The mean deviation about the mean is obtained by dividing the sum of absolute deviations by the total number of observations.

\[ \begin{aligned} \text{Mean Deviation about Mean}&=\dfrac{1}{N}\sum_{i=1}^{n}|x_i-\overline{x}|\\ &=\dfrac{84}{10}\\ &=3 \end{aligned} \]

Hence, the mean deviation about the mean for the given data is \(8.4\).

Q3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution

The observations are first arranged in ascending order as shown below. Since the total number of observations is even, the median is obtained by taking the average of the sixth and seventh terms.

$$ \begin{aligned} x_i &:10,\;11,\;11,\;12,\;13,\;13,\;14,\;16,\;16,\;17,\;17,\;18\\ |x_i-M| &:3.5,\;2.5,\;2.5,\;1.5,\;0.5,\;0.5,\;0.5,\;2.5,\;2.5,\;3.5,\;3.5,\;4.5 \end{aligned} $$ $$ \begin{aligned} N&=12\\ \dfrac{N}{2}&=\dfrac{12}{2}=6 \end{aligned} $$

Hence, the median is the average of the sixth and seventh terms.

\[ M=\dfrac{\text{sixth term}+\text{seventh term}}{2}=\dfrac{13+14}{2}=13.5 \]

Using this value of the median, the absolute deviations \(|x_i-M|\) are calculated for each observation as listed above, and their sum is

$$ \sum |x_i-M|=28 $$

The mean deviation about the median is obtained by dividing this total by the number of observations.

$$ \begin{aligned} \text{Mean Deviation about Median}&=\dfrac{\sum |x_i-M|}{N}\\ &=\dfrac{28}{12}\\ &=\dfrac{7}{3}=2.33 \end{aligned} $$

Therefore, the mean deviation about the median for the given data is \(2.33\) (approximately).

Q4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution

The observations are first arranged in ascending order as shown below. Since the total number of observations is even, the median is obtained by taking the average of the sixth and seventh terms.

$$ \begin{aligned} x_i &:36,\; 42,\; 45,\; 46,\;46,\; 49,\; 51,\;53,\; 60,\; 72\\ |x_i-M| &:11.5+5.5+2.5+1.5+1.5+1\cdot 5+3.5+5.5+12.5+24 \end{aligned} $$ $$ \begin{aligned} N&=10\\ \dfrac{N}{2}&=\dfrac{10}{2}=5 \end{aligned} $$

Hence, the median is the average of the fifth and sixth terms.

\[ M=\dfrac{\text{fifth term}+\text{sixth term}}{2}=\dfrac{46+49}{2}=47.5 \]

Using this value of the median, the absolute deviations \(|x_i-M|\) are calculated for each observation as listed above, and their sum is

$$ \sum |x_i-M|=70 $$

The mean deviation about the median is obtained by dividing this total by the number of observations.

$$ \begin{aligned} \text{Mean Deviation about Median}&=\dfrac{\sum |x_i-M|}{N}\\ &=\dfrac{70}{10}\\ &=7 \end{aligned} $$

Therefore, the mean deviation about the median for the given data is \(7\).

Q5.

\(x_i\)	5	10	15	20	25
\(f_i\)	7	4	6	3	5

Solution

The data are given in discrete form with corresponding frequencies. To find the mean deviation about the mean, we first compute \(f_ix_i\), then determine the mean \(\overline{x}\), and finally calculate \(|x_i-\overline{x}|\) and \(f_i|x_i-\overline{x}|\) for each value of \(x_i\), as shown in the table.

\(x_i\)	\(f_i\)	\(f_ix_i\)	\(|x_i-\overline{x}|\)	\(f_i|x_i-\overline{x}|\)
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
	\(N=\sum f_i=25\)	\(\sum f_ix_i=350\)		\(\sum f_i|x_i-\overline{x}|=158\)

First, we calculate the arithmetic mean using \(\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}\).

$$ \begin{aligned} \overline{x}&=\dfrac{350}{25}\\ &=14 \end{aligned} $$

Using this mean, the absolute deviations \(|x_i-\overline{x}|\) are obtained for each observation and multiplied by their respective frequencies. The total of these products is \(158\).

$$ \begin{aligned} \text{Mean Deviation about Mean}&=\dfrac{1}{N}\sum f_i|x_i-\overline{x}|\\ &=\dfrac{1}{25}\times 158\\ &=6.32 \end{aligned} $$

Hence, the mean deviation about the mean for the given data is \(6.32\).

Q6.

\(x_i\)	10	30	50	70	90
\(f_i\)	4	24	28	16	8

Solution

The data are given in discrete form with corresponding frequencies. To find the mean deviation about the mean, we first compute \(f_ix_i\), then determine the mean \(\overline{x}\), and finally calculate \(|x_i-\overline{x}|\) and \(f_i|x_i-\overline{x}|\) for each value of \(x_i\), as shown in the table.

\(x_i\)	\(f_i\)	\(f_ix_i\)	\(|x_i-\overline{x}|\)	\(f_i|x_i-\overline{x}|\)
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
	\(N=\sum f_i=80\)	\(\sum f_ix_i=4000\)		\(\sum f_i|x_i-\overline{x}|=1280\)

First, we calculate the arithmetic mean using \(\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}\).

$$ \begin{aligned} \overline{x}&=\dfrac{4000}{80}\\ &=50 \end{aligned} $$

Using this mean, the absolute deviations \(|x_i-\overline{x}|\) are obtained for each observation and multiplied by their respective frequencies. The total of these products is \(1280\).

$$ \begin{aligned} \text{Mean Deviation about Mean}&=\dfrac{1}{N}\sum f_i|x_i-\overline{x}|\\ &=\dfrac{1280}{80}\\ &=16 \end{aligned} $$

Hence, the mean deviation about the mean for the given data is \(16\).

Q7.

\(x_i\)	5	7	9	10	12	15
\(f_i\)	8	6	2	2	2	6

Solution

The data are arranged in ascending order and the cumulative frequencies are calculated to locate the median. Since the total frequency is \(N=26\), the median lies between the \(13^{\text{th}}\) and \(14^{\text{th}}\) observations. From the cumulative frequency column, both these observations correspond to \(x=7\). Hence, the median \(M=7\). Using this value, the absolute deviations \(|x_i-M|\) and the products \(f_i|x_i-M|\) are obtained as shown in the table.

\(x_i\)	\(f_i\)	\(cf\)	\(|x_i-M|\)	\(f_i|x_i-M|\)
5	8	8	2	16
7	6	14	0	0
9	2	16	2	4
10	2	18	3	6
12	2	20	5	10
15	6	26	8	48
	\(N=\sum f_i=26\)			\(\sum f_i|x_i-M|=84\)

Thus, the median is obtained from the \(13^{\text{th}}\) and \(14^{\text{th}}\) observations, both equal to \(7\), so \(M=7\).

The mean deviation about the median is calculated using the formula \(\dfrac{1}{N}\sum f_i|x_i-M|\).

\[ \begin{aligned} \text{M D about Median}&=\dfrac{1}{N}\sum_{i=1}^{n} f_i|x_i-M|\\ &=\dfrac{1}{26}\times 84\\ &=3.23 \end{aligned} \]

Hence, the mean deviation about the median for the given data is \(3.23\) (approximately).

Q8.

\(x_i\)	15	21	27	30	35
\(f_i\)	3	5	6	7	8

Solution

The data are arranged in ascending order and the cumulative frequencies are calculated to locate the median. Since the total frequency is \(N=29\), the median lies between the \(15^{\text{th}}\) observations. From the cumulative frequency column, both these observations correspond to \(x=30\). Hence, the median \(M=30\). Using this value, the absolute deviations \(|x_i-M|\) and the products \(f_i|x_i-M|\) are obtained as shown in the table.

\(x_i\)	\(f_i\)	\(cf\)	\(|x_i-M|\)	\(f_i|x_i-M|\)
15	3	3	15	45
21	5	8	9	45
27	6	14	3	18
30	7	21	0	0
35	8	29	5	40
	\(N=\sum f_i=29\)			\(\sum f_i|x_i-M|=148\)

Thus, the median is obtained from the \(15^{\text{th}}\) observations, both equal to \(30\), so \(M=30\).

The mean deviation about the median is calculated using the formula \(\dfrac{1}{N}\sum f_i|x_i-M|\).

\[ \begin{aligned} \text{M D about Median}&=\dfrac{1}{N}\sum_{i=1}^{n} f_i|x_i-M|\\ &=\dfrac{1}{29}\times 148\\ &=5.1 \end{aligned} \]

Hence, the mean deviation about the median for the given data is \(5.1\).

Q9.

Income per day in ₹	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
No.of persons	4	8	9	10	7	5	4	3

Solution

The given data are grouped into equal class intervals of width \(h=100\). We first find the class marks \(x_i\) for each class and choose \(a=350\) as the assumed mean corresponding to the class \(300\text{–}400\). Using step-deviation, we compute \(d_i=\dfrac{x_i-a}{h}\), then evaluate \(f_id_i\). After obtaining the mean \(\overline{x}\), the absolute deviations \(|x_i-\overline{x}|\) and the products \(f_i|x_i-\overline{x}|\) are calculated as shown in the table.

Class	\(f\)	\(x_i\)	\(d=\dfrac{x_i-350}{100}\)	\(f_id_i\)	\(|x_i-\overline{x}|\)	\(f_i|x_i-\overline{x}|\)
0–100	4	50	-3	-12	308	1232
100–200	8	150	-2	-16	208	1664
200–300	9	250	-1	-9	108	972
300–400	10	\(\boxed{\bbox[white,2pt]{\color{blue}350}}\)	0	0	8	80
400–500	7	450	1	7	92	644
500–600	5	550	2	10	192	960
600–700	4	650	3	12	292	1168
700–800	3	750	4	12	392	1176
	\(N=\sum f_i=50\)			\(\sum f_id_i=4\)		\(\sum f_i|x_i-\overline{x}|=7896\)

Using the step-deviation formula, the arithmetic mean is obtained as follows.

$$ \begin{aligned} \overline{x}&=a+\dfrac{\sum f_id_i}{\sum f_i}\times h\\ &=350+\dfrac{4}{50}\times 100\\ &=350+8\\ &=358 \end{aligned} $$

With this mean, the absolute deviations are calculated and multiplied by their respective frequencies. Their total is \(7896\). The mean deviation about the mean is therefore

$$ \begin{aligned} \text{Mean Deviation about Mean}&=\dfrac{1}{N}\sum f_i|x_i-\overline{x}|\\ &=\dfrac{1}{50}\times 7896\\ &=157.92 \end{aligned} $$

Hence, the mean deviation about the mean for the given data is \(157.92\).

Q10.

Height in cms	95-105	105-115	115-125	125-135	135-145	145-155
No.of Boys	9	13	26	30	12	10

Solution

The given data are grouped into equal class intervals of width \(h=10\). We first find the class marks \(x_i\) for each class and choose \(a=350\) as the assumed mean corresponding to the class \(115\text{–}125\). Using step-deviation, we compute \(d_i=\dfrac{x_i-a}{h}\), then evaluate \(f_id_i\). After obtaining the mean \(\overline{x}\), the absolute deviations \(|x_i-\overline{x}|\) and the products \(f_i|x_i-\overline{x}|\) are calculated as shown in the table.

Class	\(f\)	\(x_i\)	\(d=\dfrac{x_i-120}{10}\)	\(f_id_i\)	\(|x_i-\overline{x}|\)	\(f_i|x_i-\overline{x}|\)
95-105	9	100	-2	-18	25.3	227.7
105-115	13	110	-1	-13	15.3	a98.9
115-125	26	\(\boxed{\bbox[white, 2pt]{\color{blue}120}}\)	0	0	5.3	137.8
125-135	30	130	1	30	4.7	141.0
135-145	12	140	2	24	14.7	176.4
145-155	10	150	3	30	24.7	247
	\(N=\sum f_i=100\)			\(\sum f_id_i=53\)		\(\sum f_i|x_i-\overline{x}|=1128.8\)

Using the step-deviation formula, the arithmetic mean is obtained as follows.

$$ \begin{aligned} \overline{x}&=a+\dfrac{\sum f_id_i}{\sum f_i}\times h\\ &=120+\dfrac{53}{100}\times 10\\ &=120+5.3\\ &=125.3 \end{aligned} $$

With this mean, the absolute deviations are calculated and multiplied by their respective frequencies. Their total is \(1128.8\). The mean deviation about the mean is therefore

$$ \begin{aligned} \text{Mean Deviation about Mean}&=\dfrac{1}{N}\sum f_i|x_i-\overline{x}|\\ &=\dfrac{1}{100}\times 1128.8\\ &=11.28 \end{aligned} $$

Hence, the mean deviation about the mean for the given data is \(11.28\).

Q11. Find the mean deviation about median for the following data :

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

Solution

The data are grouped into equal class intervals. We first compute the cumulative frequencies to locate the median class. Since the total frequency is \(N=50\), we have \(\dfrac{N}{2}=25\). The cumulative frequency just exceeding 25 is 28, corresponding to the class \(20\text{–}30\), which is therefore the median class. For this class, the lower limit is \(l=20\), the cumulative frequency of the preceding class is \(c=14\), the class frequency is \(f=14\), and the class width is \(h=10\). Using these values, the median \(M\) is obtained, after which the mid-points \(x_i\), absolute deviations \(|x_i-M|\), and products \(f_i|x_i-M|\) are calculated as shown in the table.

Class	\(f\)	\(cf\)	\(\text{Mid-Point }(x_i)\)	\(|x_i-M|\)	\(f_i|x_i-M|\)
0–10	6	6	5	22.86	137.16
10–20	8	14	15	12.86	102.88
20–30	14	28	25	2.86	40.04
30–40	16	44	35	7.14	114.24
40–50	4	48	45	17.14	68.56
50–60	2	50	55	27.14	54.28
	\(N=\sum f_i=50\)				\(\sum f_i|x_i-M|=517.16\)

$$ \begin{aligned} \sum f_i&=50 \end{aligned} $$ $$ \begin{aligned} \dfrac{N}{2}&=25\\ l&=20\\ c&=14\\ h&=10\\ f&=14\end{aligned} $$ $$ \begin{aligned} M&=l+\dfrac{\dfrac{N}{2}-c}{f}\times h\\ &=20+\dfrac{25-14}{14}\times 10\\ &=20+\dfrac{110}{14}\\ &=27.86 \end{aligned} $$

Using this value of the median, the absolute deviations are obtained from the mid-points and multiplied by the corresponding frequencies. Their total is \(517.16\). The mean deviation about the median is therefore

$$ \begin{aligned} \text{Mean Deviation about Median}&=\dfrac{1}{N}\sum f_i|x_i-M|\\ &=\dfrac{1}{50}\times 517.16\\ &=10.34 \end{aligned} $$

Hence, the mean deviation about the median for the given data is \(10.34\).

Q12. 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
Numbers	5	6	12	14	26	12	16	9

Solution

The data are grouped into equal class intervals. We first compute the cumulative frequencies to locate the median class. Since the total frequency is \(N=100\), we have \(\dfrac{N}{2}=50\). The cumulative frequency just exceeding 50 is 63, corresponding to the class \(35.5\text{–}40.5\), which is therefore the median class. For this class, the lower limit is \(l=35.5\), the cumulative frequency of the preceding class is \(c=37\), the class frequency is \(f=26\), and the class width is \(h=5\). Using these values, the median \(M\) is obtained, after which the mid-points \(x_i\), absolute deviations \(|x_i-M|\), and products \(f_i|x_i-M|\) are calculated as shown in the table.

Class	\(f\)	\(cf\)	\(\text{Mid-Point }(x_i)\)	\(|x_i-M|\)	\(f_i|x_i-M|\)
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
	\(N=\sum f_i=100\)				\(\sum f_i|x_i-M|=735\)

$$ \begin{aligned} \sum f_i&=100 \end{aligned} $$ $$ \begin{aligned} \dfrac{N}{2}&=50\\ l&=35.5\\ c&=37\\ h&=5\\ f&=26\end{aligned} $$ $$ \begin{aligned} M&=l+\dfrac{\dfrac{N}{2}-c}{f}\times h\\ &=35.5+\dfrac{50-37}{26}\times 5\\ &=35.5+\dfrac{13}{26}\times 5\\ &=35.5+2.5\\ &=38 \end{aligned} $$

Using this value of the median, the absolute deviations are obtained from the mid-points and multiplied by the corresponding frequencies. Their total is \(735\). The mean deviation about the median is therefore

$$ \begin{aligned} \text{Mean Deviation about Median}&=\dfrac{1}{N}\sum f_i|x_i-M|\\ &=\dfrac{1}{100}\times 735\\ &=7.35 \end{aligned} $$

Hence, the mean deviation about the median for the given data is \(7.35\).