Ex 13.1 - Question 1

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

Solution:
Step 1: Find the mean ($\bar{x}$) of the given data.
The given observations are $x_i = 4, 7, 8, 9, 10, 12, 13, 17$ and the number of observations $n = 8$. $$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8} = \frac{80}{8} = 10$$
Step 2: Calculate the absolute deviations from the mean, $|x_i - \bar{x}|$.

• $|4 - 10| = |-6| = 6$
• $|7 - 10| = |-3| = 3$
• $|8 - 10| = |-2| = 2$
• $|9 - 10| = |-1| = 1$
• $|10 - 10| = 0$
• $|12 - 10| = 2$
• $|13 - 10| = 3$
• $|17 - 10| = 7$

Step 3: Find the sum of absolute deviations. $$\sum |x_i - \bar{x}| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24$$
Step 4: Calculate the mean deviation about the mean. $$\text{Mean Deviation}(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{24}{8} = 3$$

Ex 13.1 - Question 2

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

Solution:
Step 1: Find the mean ($\bar{x}$) of the given data.
The observations are $x_i = 38, 70, 48, 40, 42, 55, 63, 46, 54, 44$ and $n = 10$. $$\text{Sum} = 38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44 = 500$$ $$\bar{x} = \frac{500}{10} = 50$$
Step 2: Calculate the absolute deviations $|x_i - \bar{x}|$ and their sum.

• $|38 - 50| = 12$
• $|70 - 50| = 20$
• $|48 - 50| = 2$
• $|40 - 50| = 10$
• $|42 - 50| = 8$
• $|55 - 50| = 5$
• $|63 - 50| = 13$
• $|46 - 50| = 4$
• $|54 - 50| = 4$
• $|44 - 50| = 6$

$$\sum |x_i - \bar{x}| = 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84$$
Step 3: Calculate the mean deviation. $$\text{Mean Deviation}(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{84}{10} = 8.4$$

Ex 13.1 - Question 3

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

Solution:
Step 1: Arrange the data in ascending order.
The given data: $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$ ($n = 12$).
Ascending order: $10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$
Step 2: Find the median ($M$).
Since $n = 12$ (even), the median is the mean of the $(\frac{n}{2})^{\text{th}}$ and $(\frac{n}{2} + 1)^{\text{th}}$ observations. $$\text{Median} = \frac{6^{\text{th}}\text{ term} + 7^{\text{th}}\text{ term}}{2} = \frac{13 + 14}{2} = \frac{27}{2} = 13.5$$
Step 3: Calculate absolute deviations $|x_i - M|$ and their sum.

• $|10 - 13.5| = 3.5$
• $|11 - 13.5| = 2.5$
• $|11 - 13.5| = 2.5$
• $|12 - 13.5| = 1.5$
• $|13 - 13.5| = 0.5$
• $|13 - 13.5| = 0.5$
• $|14 - 13.5| = 0.5$
• $|16 - 13.5| = 2.5$
• $|16 - 13.5| = 2.5$
• $|17 - 13.5| = 3.5$
• $|17 - 13.5| = 3.5$
• $|18 - 13.5| = 4.5$

$$\sum |x_i - M| = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28$$
Step 4: Calculate the mean deviation about the median. $$\text{Mean Deviation}(M) = \frac{\sum |x_i - M|}{n} = \frac{28}{12} = \frac{7}{3} \approx 2.33$$

Ex 13.1 - Question 4

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

Solution:
Given data points: $x_i = \{36, 72, 46, 42, 60, 45, 53, 46, 51, 49\}$
Number of observations, $n = 10$.

Step 1: Find the mean ($\bar{x}$) of the data.
$$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$$ $$\bar{x} = \frac{36 + 72 + 46 + 42 + 60 + 45 + 53 + 46 + 51 + 49}{10} = \frac{500}{10} = 50$$ Step 2: Calculate the absolute deviations from the mean, $|x_i - \bar{x}|$.

• $|36 - 50| = 14$
• $|72 - 50| = 22$
• $|46 - 50| = 4$
• $|42 - 50| = 8$
• $|60 - 50| = 10$
• $|45 - 50| = 5$
• $|53 - 50| = 3$
• $|46 - 50| = 4$
• $|51 - 50| = 1$
• $|49 - 50| = 1$

Step 3: Find the sum of absolute deviations and compute the Mean Deviation ($\text{M.D.}(\bar{x})$).
$$\sum |x_i - \bar{x}| = 14 + 22 + 4 + 8 + 10 + 5 + 3 + 4 + 1 + 1 = 72$$ $$\text{M.D.}(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{72}{10} = 7.2$$

Ex 13.1 - Question 5

Question: 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:
Let us set up a table to compute the mean and absolute deviations: $$\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & f_i x_i & |x_i - \bar{x}| & f_i |x_i - \bar{x}| \\ \hline 5 & 7 & 35 & 9 & 63 \\ 10 & 4 & 40 & 4 & 16 \\ 15 & 6 & 90 & 1 & 6 \\ 20 & 3 & 60 & 6 & 18 \\ 25 & 5 & 125 & 11 & 55 \\ \hline \Sigma & 25 & 350 & & 158 \\ \hline \end{array}$$ Step 1: Calculate the mean ($\bar{x}$).
$$N = \sum f_i = 25$$ $$\sum f_i x_i = 350$$ $$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$$ Step 2: Calculate absolute deviations $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$.

• $|5 - 14| = 9 \implies 7 \times 9 = 63$
  $|10 - 14| = 4 \implies 4 \times 4 = 16$
  $|15 - 14| = 1 \implies 6 \times 1 = 6$
  $|20 - 14| = 6 \implies 3 \times 6 = 18$
  $|25 - 14| = 11 \implies 5 \times 11 = 55$
  Sum $\sum f_i |x_i - \bar{x}| = 63 + 16 + 6 + 18 + 55 = 158$.

$$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$$

Ex 13.1 - Question 6

Question: 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:
$$\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & f_i x_i & |x_i - \bar{x}| & f_i |x_i - \bar{x}| \\ \hline 10 & 4 & 40 & 40 & 160 \\ 30 & 24 & 720 & 20 & 480 \\ 50 & 28 & 1400 & 0 & 0 \\ 70 & 16 & 1120 & 20 & 320 \\ 90 & 8 & 720 & 40 & 320 \\ \hline \Sigma & 80 & 4000 & & 1280 \\ \hline \end{array}$$ Step 1: Compute the mean ($\bar{x}$).
$$N = \sum f_i = 80$$ $$\sum f_i x_i = 4000$$ $$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{4000}{80} = 50$$ Step 2: Compute Mean Deviation.
$$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{1280}{80} = 16$$

Ex 13.1 - Question 7

Question: 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:
$$\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & \text{C.F.} & |x_i - M| & f_i |x_i - M| \\ \hline 5 & 8 & 8 & 2 & 16 \\ 7 & 6 & 14 & 0 & 0 \\ 9 & 2 & 16 & 2 & 4 \\ 10 & 2 & 18 & 3 & 6 \\ 12 & 2 & 20 & 5 & 10 \\ 15 & 6 & 26 & 8 & 48 \\ \hline \Sigma & 26 & & & 84 \\ \hline \end{array}$$ Step 1: Find the Median ($M$).
Here, total frequency $N = \sum f_i = 26$ (even).
$$\frac{N}{2} = 13 \quad \text{and} \quad \frac{N}{2} + 1 = 14$$ The 13th and 14th observations both lie in the cumulative frequency 14, which corresponds to $x_i = 7$.
Thus, Median $M = 7$. Step 2: Calculate absolute deviations and Mean Deviation.
$$\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} = \frac{84}{26} \approx 3.23$$

Ex 13.1 - Question 8

Question: Find the mean deviation about the mean for the given data: $$\begin{array}{c|ccccc} x_i & 15 & 21 & 27 & 30 & 35 \\ \hline f_i & 3 & 5 & 6 & 7 & 8 \end{array}$$

Solution:

Let us construct the table to find the mean and the mean deviation about the mean:

$x_i$	$f_i$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
15	3	45	12.66	37.98
21	5	105	6.66	33.30
27	6	162	0.66	3.96
30	7	210	2.34	16.38
35	8	280	7.34	58.72
Total	$N = 29$	$\sum f_i x_i = 802$		$\sum f_i |x_i - \bar{x}| = 150.34$

Step 1: Calculate the mean ($\bar{x}$).
$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{802}{29} \approx 27.655 \approx 27.66$$

Step 2: Calculate Mean Deviation.
$$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{150.34}{29} \approx 5.18 \approx 5.1$$

Ex 13 - Question 9

Question: Find the mean deviation about the mean for the data: $$\begin{array}{c|c|c|c|c|c|c|c|c} \text{Income per day in ₹} & 0-100 & 100-200 & 200-300 & 300-400 & 400-500 & 500-600 & 600-700 & 700-800 \\ \hline \text{Number of persons} & 4 & 8 & 9 & 10 & 7 & 5 & 4 & 3 \end{array}$$

Solution:

First, we find the mid-points ($x_i$) for each class interval and set up the frequency distribution table:

Income per day	$f_i$	Mid-point ($x_i$)	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$0-100$	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	$N=50$		$\sum f_i x_i = 17900$		$\sum f_i |x_i - \bar{x}| = 7896$

1. Mean ($\bar{x}$):

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{17900}{50} = 358$$

2. Mean Deviation about Mean:

$$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{7896}{50} = 157.92$$

Ex 13 - Question 10

Question: Find the mean deviation about the mean for the data: $$\begin{array}{c|c|c|c|c|c|c} \text{Height in cms} & 95-105 & 105-115 & 115-125 & 125-135 & 135-145 & 145-155 \\ \hline \text{Number of boys} & 9 & 13 & 26 & 30 & 12 & 10 \end{array}$$

Solution:

Let us construct the table to find the mean deviation about mean:

Height	$f_i$	$x_i$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$95-105$	9	100	900	27.3	245.7
$105-115$	13	110	1430	17.3	224.9
$115-125$	26	120	3120	7.3	189.8
$125-135$	30	130	3900	2.7	81.0
$135-145$	12	140	1680	12.7	152.4
$145-155$	10	150	1500	22.7	227.0
Total	$N=100$		$\sum f_i x_i = 12730$		$\sum f_i |x_i - \bar{x}| = 1128.0$

1. Mean ($\bar{x}$):

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{12730}{100} = 127.3$$

2. Mean Deviation about Mean:

$$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{1128}{100} = 11.28$$

Ex 13 - Question 11

Question: Find the mean deviation about median for the following data: $$\begin{array}{c|c|c|c|c|c|c} \text{Marks} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text{Number of Girls} & 6 & 8 & 14 & 16 & 4 & 2 \end{array}$$

Solution:

Let us construct the cumulative frequency table and find the median:

Marks	$f_i$	Cumulative Frequency ($c.f.$)	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
Total	$N=50$				$\sum f_i |x_i - M| = 517.16$

1. Find the median class:

$$\frac{N}{2} = \frac{50}{2} = 25$$

The cumulative frequency just greater than $25$ is $28$, which corresponds to the class interval $20-30$.

Median formula:

$$M = l + \frac{\frac{N}{2} - c}{f} \times h$$

Here, $l = 20$, $c = 14$, $f = 14$, $h = 10$, and $\frac{N}{2} = 25$:

$$M = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} = 20 + 7.86 = 27.86$$

2. Calculate the Mean Deviation about Median ($\text{M.D.}(M)$):

$$\text{M.D.}(M) = \frac{1}{N} \sum f_i |x_i - M| = \frac{517.16}{50} = 10.34$$

Miscellaneous Exercise - Question 12

Question: 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 | $5$ | $6$ | $12$ | $14$ | $26$ | $12$ | $16$ | $9$ |

Solution:
First, we convert the given discontinuous classes into continuous classes by subtracting $0.5$ from the lower limits and adding $0.5$ to the upper limits. We then compute the cumulative frequencies ($C.f.$), mid-points ($x_i$), and absolute deviations from the median. Let's prepare the frequency distribution table: - Total number of observations $N = 100$. - $\frac{N}{2} = \frac{100}{2} = 50$. The cumulative frequency just greater than or equal to $50$ is $63$, which corresponds to the class interval $35.5 - 40.5$. Therefore, the median class is $35.5 - 40.5$. The formula for median is: $$\text{Median} = L + \frac{\frac{N}{2} - C}{f} \times h$$ Where: - $L = 35.5$ (lower limit of median class) - $C = 37$ (cumulative frequency of class preceding the median class) - $f = 26$ (frequency of the median class) - $h = 5$ (class width) $$\text{Median} = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + \frac{13 \times 5}{26} = 35.5 + 2.5 = 38$$ Now, we compute $|x_i - \text{Median}| = |x_i - 38|$ and $f_i |x_i - 38|$:

Continuous Class	$f_i$	Mid-point ($x_i$)	$C.f.$	$|x_i - 38|$	$f_i |x_i - 38|$
$15.5 - 20.5$	5	18	5	20	100
$20.5 - 25.5$	6	23	11	15	90
$25.5 - 30.5$	12	28	23	10	120
$30.5 - 35.5$	14	33	37	5	70
$35.5 - 40.5$	26	38	63	0	0
$40.5 - 45.5$	12	43	75	5	60
$45.5 - 50.5$	16	48	91	10	160
$50.5 - 55.5$	9	53	100	15	135
Total	$N=100$				$\sum f_i |x_i - 38| = 735$

Mean Deviation about Median:

$$\text{M.D.}(\text{Median}) = \frac{\sum f_i |x_i - \text{Median}|}{N} = \frac{735}{100} = 7.35$$