T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 10 - Measures of Central Tendency- Median and Mode

T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 10 – Measures of Central Tendency- Median and Mode is regarded as an important concept to be studied thoroughly by the students. Here, we have provided T.R. Jain and V.K. Ohri Solutions for Class 11.

Board

CBSE

Class

Class 11

Subject

Statistics for Economics

Chapter

Chapter 10

Chapter Name

Measures of Central Tendency- Median and Mode

Number of questions solved

04

Category

T.R. Jain and V.K. Ohri

Chapter 10 -Measures of Central Tendency- Median and Mode covers the below-mentioned concepts:

  • Median
  • How to find the median value?
  • Missing frequency
  • Graphic determination of the median
  • Partition value: Quartile

T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 10 – Measures of Central Tendency- Median and Mode

Question 1

The following series show marks in statistics of 9 students in class 11. Find the median marks.

Marks

22

16

18

13

15

19

17

20

23

Solution:

The data is first arranged in the ascending order:

S.no.

Marks

1

13

2

15

3

16

4

17

5

18

6

19

7

20

8

22

9

23

N = 9

\(M\, =\, Size\, of\,\left (\frac{N\,+ 1}{2} \right)th\, item\) \(=\, Size\, of\, \left (\frac{9\, +\, 1}{2} \right)th\, item\)

= Size of 5th item = 18

Hence, Median = 18

Question 2

The following table gives the marks obtained by some students. Calculate the median marks obtained by the students.

Marks

0-5

5-10

10-15

15-20

20-25

25-30

30-35

35-40

40-45

Number of students

6

12

17

30

10

10

8

5

2

Solution:

Marks

Frequency (f)

Cumulative Frequency

0-5

6

6

5-10

12

18

10-15

17

35(c.f.)

(l1)15-20

30 (f)

65

20-25

10

75

25-30

10

85

30-35

8

93

35-40

5

98

40-45

2

100

= N = 100

M = Size of \(\left (\frac{N}{2}\right)th\, item\)

= Size of \(\left (\frac{100}{2}\right)th\, item\) = Size of 50th item

50th item lies in 65th cumulative frequency and the corresponding median class is 15-20.

\(M = l1\, +\, \frac{N/2\, -\, c.f.}{f}\times i\)

= \(=\, 15\, +\, \frac{100/2\, -\, 35\times 5}{30}\)

= \(=\, 15\, +\, \frac{15-35}{30}\times 5\, =\, 15\, +\, \frac{15}{30\times 5}\)

= 15 + 2.5 = 17.5

Median = 17.5 marks

Question 3

Find the mode from the following data:

8, 10, 5, 8, 12, 7, 8, 9, 11, 7

Solution:

Arrange the series in an ascending order as:

5, 7, 7, 8, 8, 8, 9, 10, 11, 12

An inspection of the series shows that the values 8 occurs most frequency in the series.

Hence, Mode (Z) = 8

Question 4

Calculate the mode from the following data:

Wages (₹)

0-5

5-10

10-15

15-20

20-25

25-30

30-35

Number of workers

3

7

15

30

20

10

5

Solution:

Wages (₹)

Frequency (f)

0-5

3

5-10

7

10-15

15 (f0)

(ll) 15-20

30 (f1)

20-25

20 (f2)

25-30

10

30-35

5

Since the series is regular, we may not do grouping for the location of the model group. By inspection, the modal class is 15-20.

Z = l1 \(+\, \frac{f1-f0}{2f1\, -\, f0\,- f2}\times i\)

Here, l1 = 15, f1 = 30, f0 = 15, f2 = 20, i = 5

Substituting the values, we get,

\(Z\, =\, 15\, +\, \frac{30\, 15}{2\left(30 \right)-15-20}\, \times 5\) \(=\, 15+\frac{15}{60-35}\times 5\) \(=\, 15+\frac{15}{25}\times 5\)

= 15+3=18

Thus, Mode = 18

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