NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15

NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics

Edited By Sumit Saini | Updated on Jul 18, 2022 12:03 PM IST

NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 mainly deals with the mean deviation topics and other exercises include important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range, etc. In Class 9 and 10 Maths NCERT syllabus, students have already read the basic Statistics like mean, mode, and median. In this NCERT book Class 11 chapter applications of Statistics with advanced concepts are discussed.

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Statistics C lass 11 Chapter 15-Exercise: 15.1
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Exercise 15.1 Class 11 Maths have easy questions. Solving NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 is a must to solve upcoming exercises. Direct questions are asked from this exercise as well in Boards. Also students can refer to the following exercise of NCERT for further information about upcoming exercises.

Statistics C lass 11 Chapter 15-Exercise: 15.1

Question:1 . Find the mean deviation about the mean for the data.

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

Answer:

Mean ( $\overline{x}$ ) of the given data:

$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$

The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are

6, 3, 2, 1, 0, 2, 3, 7

$\therefore$ $\sum_{i=1}^{8}|x_i - 10| = 24$

$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$

$= \frac{24}{8} = 3$

Hence, the mean deviation about the mean is 3.

Question:2. Find the mean deviation about the mean for the data.

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

Answer:

Mean ( $\overline{x}$ ) of the given data:

$\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50$

The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are

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

$\therefore$ $\sum_{i=1}^{8}|x_i - 50| = 84$

$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$

$= \frac{84}{10} = 8.4$

Hence, the mean deviation about the mean is 8.4.

Question:3. Find the mean deviation about the median.

$\small 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$

Answer:

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

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

Now, Median (M)

$\\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5$

The respective absolute values of the deviations from median, $|x_i - M|$ 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

$\therefore$ $\sum_{i=1}^{8}|x_i - 13.5| = 28$

$\therefore$ $M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|$

$= \frac{28}{12} = 2.33$

Hence, the mean deviation about the median is 2.33.

Question:4. Find the mean deviation about the median.

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

Answer:

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

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

Now, Median (M)

$\\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5$

The respective absolute values of the deviations from median, $|x_i - M|$ are

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

$\therefore$ $\sum_{i=1}^{8}|x_i - 47.5| = 70$

$\therefore$ $M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|$

$= \frac{70}{10} = 7$

Hence, the mean deviation about the median is 7.

Question:5 Find the mean deviation about the mean.

$\small \\x_i\hspace{1cm}5\hspace{1cm}10\hspace{1cm}15\hspace{1cm}20\hspace{1cm}25\\\small f_i\hspace{1cm}7\hspace{1cm}4\hspace{1.15cm}6\hspace{1.22cm}3\hspace{1.3cm}5$

Answer:

$x_i$	$f_i$	$f_ix_i$	$\|x_i - \overline{x}\|$	$f_i\|x_i - \overline{x}\|$
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
	$\sum{f_i}$ = 25	$\sum f_ix_i$ = 350		$\sum f_i\|x_i - \overline{x}\|$ =158

$N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and

$\sum f_i|x_i - \overline{x}|$ = 158

$\therefore$ $M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|$

$= \frac{158}{25} = 6.32$

Hence, the mean deviation about the mean is 6.32

Question:6. Find the mean deviation about the mean.

$\small \\x_i\hspace{1cm}10\hspace{1cm}30\hspace{1cm}50\hspace{1cm}70\hspace{1cm}90\\\small f_i\hspace{1cm}4\hspace{1.1cm}24\hspace{1.1cm}28\hspace{1.1cm}16\hspace{1.2cm}8$

Answer:

$x_i$	$f_i$	$f_ix_i$	$\|x_i - \overline{x}\|$	$f_i\|x_i - \overline{x}\|$
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
	$\sum{f_i}$ = 80	$\sum f_ix_i$ = 4000		$\sum f_i\|x_i - \overline{x}\|$ =1280

$N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and

$\sum f_i|x_i - \overline{x}|$ = 1280

$\therefore$ $M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|$

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

Hence, the mean deviation about the mean is 16

Question:7. Find the mean deviation about the median.

$\small x_i$ $\small 5$ $\small 7$ $\small 9$ $\small 10$ $\small 12$ $\small 15$

$\small f_i$ $\small 8$ $\small 6$ $\small 2$ $\small 2$ $\small 2$ $\small 6$

Answer:

$x_i$	$f_i$	$c.f.$	$\|x_i - M\|$	$f_i\|x_i - M\|$
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

Now, N = 26 which is even.

Median is the mean of $13^{th}$ and $14^{th}$ observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M $= \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and

$\sum f_i|x_i - M|$ = 84

$\therefore$ $M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|$

$= \frac{84}{26} = 3.23$

Hence, the mean deviation about the median is 3.23

Question:8 Find the mean deviation about the median.

$\small x_i$ $\small 15$ $\small 21$ $\small 27$ $\small 30$ $\small 35$

$\small f_i$ $\small 3$ $\small 5$ $\small 6$ $\small 7$ $\small 8$

Answer:

$x_i$	$f_i$	$c.f.$	$\|x_i - M\|$	$f_i\|x_i - M\|$
15	3	3	13.5	40.5
21	5	8	7.5	37.5
27	6	14	1.5	9
30	7	21	1.5	10.5
35	8	29	6.5	52

Now, N = 30, which is even.

Median is the mean of $15^{th}$ and $16^{th}$ observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M $= \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and

$\sum f_i|x_i - M|$ = 149.5

$\therefore$ $M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|$

$= \frac{149.5}{29} = 5.1$

Hence, the mean deviation about the median is 5.1

Question:9. Find the mean deviation about the mean.

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

Answer:

Income per day	Number of Persons $f_i$	Mid Points $x_i$	$f_ix_i$	$\|x_i - \overline{x}\|$	$f_i\|x_i - \overline{x}\|$
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
	$\sum{f_i}$ =50		$\sum f_ix_i$ =17900		$\sum f_i\|x_i - \overline{x}\|$ =7896

$N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and

$\sum f_i|x_i - \overline{x}|$ = 7896

$\therefore$ $M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|$

$= \frac{7896}{50} = 157.92$

Hence, the mean deviation about the mean is 157.92

Question:10. Find the mean deviation about the mean.

Height in cms	$\small 95-105$	$\small 105-115$	$\small 115-125$	$\small 125-135$	$\small 135-145$	$\small 145-155$
Number of person	$\small 9$	$\small 13$	$\small 26$	$\small 30$	$\small 12$	$\small 10$

Answer:

Height in cms	Number of Persons $f_i$	Mid Points $x_i$	$f_ix_i$	$\|x_i - \overline{x}\|$	$f_i\|x_i - \overline{x}\|$
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
	$\sum{f_i}$ =100		$\sum f_ix_i$ =12530		$\sum f_i\|x_i - \overline{x}\|$ =1128.8

$N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and

$\sum f_i|x_i - \overline{x}|$ = 1128.8

$\therefore$ $M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|$

$= \frac{1128.8}{100} = 11.29$

Hence, the mean deviation about the mean is 11.29

Question:11. Find the mean deviation about median for the following data :

Marks	$\small 0-10$	$\small 10-20$	$\small 20-30$	$\small 30-40$	$\small 40-50$	$\small 50-60$
Number of girls	$\small 6$	$\small 8$	$\small 14$	$\small 16$	$\small 4$	$\small 2$

Answer:

Marks	Number of Girls $f_i$	Cumulative Frequency c.f.	Mid Points $x_i$	$\|x_i - M\|$	$f_i\|x_i - M\|$
0-10	6	6	5	22.85	137.1
10-20	8	14	15	12.85	102.8
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
					$\sum f_i\|x_i - M\|$ =517.1

Now, N = 50, which is even.

The class interval containing $\left (\frac{N}{2} \right)^{th}$ or $25^{th}$ item is 20-30. Therefore, 20-30 is the median class.

We know,

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

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median $= 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and

$\sum f_i|x_i - M|$ = 517.1

$\therefore$ $M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|$

$= \frac{517.1}{50} = 10.34$

Hence, the mean deviation about the median is 10.34

Question:12 Calculate the mean deviation about median age for the age distribution of $\small 100$ persons given below:

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

[ 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]

Answer:

Age (in years)	Number $f_i$	Cumulative Frequency c.f.	Mid Points $x_i$	$\|x_i - M\|$	$f_i\|x_i - M\|$
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
					$\sum f_i\|x_i - M\|$ =735

Now, N = 100, which is even.

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

We know,

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

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

Therefore, Median $= 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and

$\sum f_i|x_i - M|$ = 735

$\therefore$ $M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|$

$= \frac{735}{100} = 7.35$

Hence, the mean deviation about the median is 7.35

Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1

The Class 11 Maths chapter 15 exercise is prepared by expert faculties with rich experience.
Exercise 15.1 Class 11 Maths can help one enhance their marks in the exam as one question of 5 marks is mostly asked from this.
Class 11 Maths chapter 15 exercise 15.1 solutions provided here are comprehensive in manner.

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Frequently Asked Questions (FAQs)

1. Which topics are covered in Exercise 15.1 Class 11 Maths?

Exercise 15.1 Class 11 Maths includes mean deviation about mean and median.

2. Is this chapter 15 related to other chapters ?

No, mostly concepts are new and can be understood if other chapters are not read.

3. What is the difficulty level of the questions asked in this chapter ?

Questions are easier to moderate level of difficulty if concepts are memorized.

4. Is it necessary to remember the formulas ?

Yes, some basic formulas must be remembered to solve the questions.

5. How many questions are there in the Exercise 15.1 Class 11 Maths ?

Total 12 questions are discussed in this exercise.

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