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

# 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.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

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 Class 11 Chapter 15-Exercise: 15.1

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.

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.

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.

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.

 $x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 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.

 $x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 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.

 $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 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 $\dpi{100} 13^{th}$ and $\dpi{100} 14^{th}$ observations.

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

Therefore, Median, M $\dpi{100} = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7$

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

 $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 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 $\dpi{100} 16^{th}$ observations.

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

Therefore, Median, M $\dpi{100} = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30$

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 Number of persons

 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 Number of person

 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

 Marks Number of girls

 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 $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $\dpi{100} 25^{th}$ item is 20-30. Therefore, 20-30 is the median class.

We know,

Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$

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

Therefore, Median $\dpi{100} = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$

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

 Age (in years) Number

[ 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]

 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 $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$

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

Therefore, Median $\dpi{100} = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$

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

## More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1

The NCERT Class 11 Maths chapter Statistics is very scoring chapters and questions are asked on the expected concepts only. It does not have much application in other chapters hence if one is not good at other chapters can strengthen this chapter to score well in the exam. Exercise 15.1 Class 11 Maths Mainly discusses the mean deviation about mean and median. Hence NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 can be seen as a scoring exercise from the exam perspective.

## 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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## Subject Wise NCERT Exemplar solutions

Happy learning!!!

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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