# ICSE Solutions for Selina Concise Chapter 24 Measure of Central Tendency Class 10 Maths

**Exercise 24(A)**

**1. Find the mean of the following set of numbers:**

**(i) 6, 9, 11, 12 and 7**

**(ii) 11, 14, 23, 26, 10, 12, 18 and 6**

**Solution**

**(i)** By definition, we know

Mean = ∑x/n

Here, n = 5

Thus,

Mean = (6 + 9 + 11 + 12 + 7)/ 5 = 45/5 = 9

**(ii)** By definition, we know

Mean = ∑x/n

Here, n = 8

Thus,

Mean = (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/8

= 120/8

= 15

**2. Marks obtained (in mathematics) by 9 student are given below:**

**60, 67, 52, 76, 50, 51, 74, 45 and 56**

**(a) find the arithmetic mean**

**(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.**

**Solution**

**(a)** Mean = ∑x/n

Here, n = 9

Thus,

Mean = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/ 9 = 531/9 = 59

**(b)** If the marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

**3. Find the mean of the natural numbers from 3 to 12.**

**Solution**

The numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Here n = 10

Mean = ∑x/n

= (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/ 10 = 75/10 = 7.5

**4. (a) Find the mean of 7, 11, 6, 5, and 6**

**(b) If each number given in (a) is diminished by 2, find the new value of mean.**

**Solution**

**(a)** Mean = ∑x/n , here n = 5

= (7 + 11 + 6 + 5 + 6)/ 5 = 35/5 = 7

**(b)** If 2 is subtracted from each number, then the mean will he changed as 7 – 2 = 5

**5. If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’**

**Solution**

Given,

No. of terms (n) = 5

Mean = 8

Sum of all terms = 8×5 = 40 **…(i)**

But, sum of numbers = 6 + 4 + 7 + a + 10 = 27 + a** …(ii)**

On equating (i) and (ii), we get

27 + a = 40

Thus, a = 13

**6. The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.**

**Solution**

Given,

No. of terms (n) = 5 and mean = 8

So, the sum of all terms = 5×8 = 40 **…(i)**

but sum of numbers = 6 + y + 7 + x + 14 = 27 + y + x **…(ii)**

On equating (i) and (ii), we get

27 + y + x = 40

⇒ x + y = 13

Hence, y = 13 – x

**7. The ages of 40 students are given in the following table:**

Age( in yrs) |
12 |
13 |
14 |
15 |
16 |
17 |
18 |

Frequency |
2 |
4 |
6 |
9 |
8 |
7 |
4 |

**Find the arithmetic mean.**

**Solution**

Age in yrs (x_{i}) |
Frequency (f_{i}) |
f_{i}x_{i} |

12 | 2 | 24 |

13 | 4 | 52 |

14 | 6 | 84 |

15 | 9 | 135 |

16 | 8 | 128 |

17 | 7 | 119 |

18 | 4 | 72 |

Total | 40 | 614 |

Mean = ∑f_{i} x_{i}/∑f_{i} = 614/40 = 15.35

**Exercise 24(B) **

**1. The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.**

Age – Years |
16 – 18 |
18 – 20 |
20 – 22 |
22- 24 |
24-26 |

No. of Students |
2 |
7 |
21 |
17 |
3 |

**Solution**

Age in years (C.I.) |
x_{i} |
Number of students (f_{i}) |
x_{i}f_{i} |

16 – 18 | 17 | 2 | 34 |

18 – 20 | 19 | 7 | 133 |

20 – 22 | 21 | 21 | 441 |

22 – 24 | 23 | 17 | 391 |

24 – 26 | 25 | 3 | 75 |

Total | 50 | 1074 |

Mean = ∑f_{i} x_{i}/∑f_{i} = 1074/50 = 21.48

**2. The following table gives the weekly wages of workers in a factory.**

Weekly Wages (Rs) |
No. of Workers |

50-55 |
5 |

55-60 |
20 |

60-65 |
10 |

65-70 |
10 |

70-75 |
9 |

75-80 |
6 |

80-85 |
12 |

85-90 |
8 |

**Calculate the mean by using:**

**(i) Direct Method**

**(ii) Short – Cut Method**

**Solution**

**(i)** Direct Method

Weekly Wages (Rs) |
Mid-Value (x_{i}) |
No. of Workers (f_{i}) |
f_{i}x_{i} |

50-55 | 52.5 | 5 | 262.5 |

55-60 | 57.5 | 20 | 1150.0 |

60-65 | 62.5 | 10 | 625.0 |

65-70 | 67.5 | 10 | 675.0 |

70-75 | 72.5 | 9 | 652.5 |

75-80 | 77.5 | 6 | 465.0 |

80-85 | 82.5 | 12 | 990.0 |

85-90 | 87.5 | 8 | 700.0 |

Total | 80 | 5520.00 |

Mean = ∑f_{i} x_{i}/ ∑f_{i} = 5520/80 = 69

**(ii) **Short – cut method

Weekly wages (Rs) |
No. of workers (f_{i}) |
Mid-value (x_{i}) |
A = 72.5 (d_{i }= x – A) |
f_{i}d_{i} |

50-55 | 5 | 52.5 | -20 | -100 |

55-60 | 20 | 57.5 | -15 | -300 |

60-65 | 10 | 62.5 | -10 | -100 |

65-70 | 10 | 67.5 | -5 | -50 |

70-75 | 9 | A = 72.5 | 0 | 0 |

75-80 | 6 | 77.5 | 5 | 30 |

80-85 | 12 | 82.5 | 10 | 120 |

85-90 | 8 | 87.5 | 15 | 120 |

Total | 80 | -280 |

Here, A = 72.5

**3. The following are the marks obtained by 70 boys in a class test:**

Marks |
No. of boys |

30 – 40 |
10 |

40 – 50 |
12 |

50 – 60 |
14 |

60 – 70 |
12 |

70 – 80 |
9 |

80 – 90 |
7 |

90 – 100 |
6 |

**Calculate the mean by:**

**(i) Short – cut method**

**(ii) Step – deviation method**

**Solution**

(i) Short – cut method

Marks |
No. of boys (f_{i}) |
Mid-value x_{i} |
A = 65 (d_{i }= x – A) |
f_{i}d_{i} |

30 – 40 | 10 | 35 | -30 | -300 |

40 – 50 | 12 | 45 | -20 | -240 |

50 – 60 | 14 | 55 | -10 | -140 |

60 – 70 | 12 | A = 65 | 0 | 0 |

70 – 80 | 9 | 75 | 10 | 90 |

80 – 90 | 7 | 85 | 20 | 140 |

90 – 100 | 6 | 95 | 30 | 180 |

Total | 70 | -270 |

Here, A = 65

**(ii)** Step – deviation method

Marks |
No. of boys (f_{i}) |
Mid-value x_{i} |
A = 65 u _{i} = (x_{i} – A)/h |
f_{i}u_{i} |

30 – 40 | 10 | 35 | -3 | -30 |

40 – 50 | 12 | 45 | -2 | -24 |

50 – 60 | 14 | 55 | -1 | -14 |

60 – 70 | 12 | A = 65 | 0 | 0 |

70 – 80 | 9 | 75 | 1 | 9 |

80 – 90 | 7 | 85 | 2 | 14 |

90 – 100 | 6 | 95 | 3 | 18 |

Total | 70 | -27 |

Here, A = 65 and h = 10

**4. Find mean by step – deviation method:**

C. I. |
63-70 |
70-77 |
77-84 |
84-91 |
91-98 |
98-105 |
105-112 |

Freq |
9 |
13 |
27 |
38 |
32 |
16 |
15 |

**Solution**

C. I. |
Frequency (f_{i}) |
Mid-value x_{i} |
A = 87.50u _{i} = (x_{i} – A)/h |
f_{i}u_{i} |

63 – 70 | 9 | 66.50 | -3 | -27 |

70 – 77 | 13 | 73.50 | -2 | -26 |

77 – 84 | 27 | 80.50 | -1 | -27 |

84 – 91 | 38 | A = 87.50 | 0 | 0 |

91 – 98 | 32 | 94.50 | 1 | 32 |

98 – 105 | 16 | 101.50 | 2 | 32 |

105 – 112 | 15 | 108.50 | 3 | 45 |

Total | 150 | 29 |

Here, A = 87.50 and h = 7

**5. The mean of the following frequency distribution is ****21(1/7)** **. Find the value of ‘f’.**

C. I. |
0 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |

frequency |
8 |
22 |
31 |
f |
2 |

**Solution**

Given,

C. I. |
frequency |
Mid-value (x_{i}) |
f_{i}x_{i} |

0-10 | 8 | 5 | 40 |

10-20 | 22 | 15 | 330 |

20-30 | 31 | 25 | 775 |

30-40 | f | 35 | 35f |

40-50 | 2 | 45 | 90 |

Total | 63 + f | 1235 + 35f |

**Exercise 24(C) **

**1. A student got the following marks in 9 questions of a question paper.**

**3, 5, 7, 3, 8, 0, 1, 4 and 6.**

**Find the median of these marks.**

**Solution**

Arranging the given data in descending order:

8, 7, 6, 5, 4, 3, 3, 1, 0

Clearly, the middle term is 4 which is the 5^{th} term.

Hence, median = 4

**2. The weights (in kg) of 10 students of a class are given below:**

**21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.**

**Find the median of their weights.**

**Solution**

Arranging the given data in descending order:

28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5

It’s seen that,

The middle terms are 24 and 24, 5^{th} and 6^{th} terms

Thus,

Median = (24 + 24)/2 = 48/2 = 24

**3. The marks obtained by 19 students of a class are given below:**

**27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.**

**Find:**

**(i) median**

**(ii) lower quartile**

**(iii) upper quartile**

**(iv) interquartile range**

**Solution**

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

**(i)** The middle term is 10^{th} term i.e. 29

Hence, median = 29

**(ii)** Lower quartile

(iii) Upper quartile =

(iv) Interquartile range = q_{3} – q_{1} =35 – 26 = 9

**4. From the following data, find:**

**(i) Median**

**(ii) Upper quartile**

**(iii) Inter-quartile range**

**25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83**

**Solution**

Arranging the given data in ascending order, we have:

0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

**(i)** Median is the mean of 8^{th} and 9^{th} term

Thus, median = (40 + 45)/ 2 = 85/2 = 42.5

**(ii) **Upper quartile =

**(iii)** Interquartile range is given by,

q_{1} = 16^{th}/4 term = 18; q_{3} = 65

Interquartile range = q_{3} – q_{1}

Thus, q_{3} – q_{1} = 65 – 18 = 47

**5. The ages of 37 students in a class are given in the following table:**

Age (in years) |
11 |
12 |
13 |
14 |
15 |
16 |

Frequency |
2 |
4 |
6 |
10 |
8 |
7 |

**Find the median.**

**Solution**

Age (in years) |
Frequency |
Cumulative Frequency |

11 | 2 | 2 |

12 | 4 | 6 |

13 | 6 | 12 |

14 | 10 | 22 |

15 | 8 | 30 |

16 | 7 | 37 |

Number of terms (n) = 37

Median =

And, the 19^{th} term is 14

∴ the median = 14

**Exercise 24(D)**

**1. Find the mode of the following data:**

**(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6**

**(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8**

**Solution**

**(i)** It’s seen that 7 occurs 4 times in the given data.

Hence, mode = 7

**(ii)** Mode = 11

As 11 occurs 4 times in the given data.

**2. The following table shows the frequency distribution of heights of 50 boys:**

Height (cm) |
120 |
121 |
122 |
123 |
124 |

Frequency |
5 |
8 |
18 |
10 |
9 |

**Find the mode of heights.**

**Solution**

Clearly, Mode is 122 cm because it has occurred the maximum number of times i.e. frequency is 18.

**3. Find the mode of following data, using a histogram:**

Class |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |

Frequency |
5 |
12 |
20 |
9 |
4 |

**Solution**

Clearly, Mode is in 20-30, because in this class there are 20 frequencies.

**4. The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:**

Expenditure (Rs) |
No. of students |

20-25 |
4 |

25-30 |
7 |

30-35 |
23 |

35-40 |
18 |

40-45 |
6 |

45-50 |
2 |

**Solution**

Clearly, Mode is in 30-35 because it has the maximum frequency.

**Exercise 24(E)**

**1. The following distribution represents the height of 160 students of a school.**

Height (in cm) |
No. of Students |

140 – 145 |
12 |

145 – 150 |
20 |

150 – 155 |
30 |

155 – 160 |
38 |

160 – 165 |
24 |

165 – 170 |
16 |

170 – 175 |
12 |

175 – 180 |
8 |

**Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:**

**i. The median height.**

**ii. The interquartile range.**

**iii. The number of students whose height is above 172 cm.**

**Solution**

Height (in cm) |
No. of Students |
Cumulative frequency |

140 – 145 | 12 | 12 |

145 – 150 | 20 | 32 |

150 – 155 | 30 | 62 |

155 – 160 | 38 | 100 |

160 – 165 | 24 | 124 |

165 – 170 | 16 | 140 |

170 – 175 | 12 | 152 |

175 – 180 | 8 | 160 |

N = 160 |

Now, let’s draw an ogive taking height of student along x-axis and cumulative frequency along y-axis.

**(i)** So, Median = 160/2 = 80^{th} term

Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5.

**(ii)** As, the number of terms = 160

Lower quartile (Q_{1}) = (160/4) = 40^{th} term = 152

Upper quartile (Q_{3}) = (3 x 160/4) = 120^{th} term = 164

Inner Quartile range = Q_{3} – Q_{1}

= 164 – 152

= 12

**(iii)** Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145.

Now, The number of students whose height is above 172 cm

= 160 – 144 = 16

**2. Draw ogive for the data given below and from the graph determine: (i) the median marks.**

**(ii) the number of students who obtained more than 75% marks.**

Marks |
10 – 19 |
20 -29 |
30 – 39 |
40 – 49 |
50 – 59 |
60 – 69 |
70 – 79 |
80 – 89 |
90 – 99 |

No. of students |
14 |
16 |
22 |
26 |
18 |
11 |
6 |
4 |
3 |

**Solution**

Marks |
No. of students |
Cumulative frequency |

9.5 – 19.5 | 14 | 14 |

19.5 – 29.5 | 16 | 30 |

29.5 – 39.5 | 22 | 52 |

39.5 – 49.5 | 26 | 78 |

49.5 – 59.5 | 18 | 96 |

59.5 – 69.5 | 11 | 107 |

69.5 – 79.5 | 6 | 113 |

79.5 – 89.5 | 4 | 117 |

89.5 – 99.5 | 3 | 120 |

Scale:

1cm = 10 marks on X axis

1cm = 20 students on Y axis

**(i)** So, the median = 120/ 2 = 60^{th} term

Through mark 60, draw a parallel line to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

**(ii)** Total marks = 100

75% of total marks = 75/100 ×100 = 75 marks

Hence, the number of students getting more than 75% marks = 120 – 111 = 9 students.

**3. The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.**

**Solution**

Mean of 1, 7, 5, 3, 4 and 4 = (1 + 7 + 5 + 3 + 4 + 4)/6

= 24/6

= 4

So, m = 4

Now, given that

The mean of 3, 2, 4, 2, 3, 3 and p = m -1 = 4 – 1 = 3

Thus, 17 + p = 3 x n …. ,where n = 7

⇒ 17 + p = 21

⇒ p = 4

Arranging the terms in ascending order, we have:

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4^{th} term = 3

Hence, q = 3

**4. In a malaria epidemic, the number of cases diagnosed were as follows:**

Date (July) |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |

Number |
5 |
12 |
20 |
27 |
46 |
30 |
31 |
18 |
11 |
5 |
0 |
1 |

**On what days do the mode and upper and lower quartiles occur?**

**Solution**

Date |
Number |
C.f. |

1 | 5 | 5 |

2 | 12 | 17 |

3 | 20 | 37 |

4 | 27 | 64 |

5 | 46 | 110 |

6 | 30 | 140 |

7 | 31 | 171 |

8 | 18 | 189 |

9 | 11 | 200 |

10 | 5 | 205 |

11 | 0 | 205 |

12 | 1 | 206 |

**(i)** Mode = 5^{th} July as it has maximum frequencies.

**(ii) **Total number of terms = 206

Upper quartile = 206 ×(3/4) = 154.5^{th} = 7^{th} July

Lower quartile = 206 ×(1/4) = 51.5^{th} = 4^{th} July

**5. The income of the parents of 100 students in a class in a certain university are tabulated below.**

Income (in thousand Rs) |
0 – 8 |
8 – 16 |
16 – 24 |
24 – 32 |
32 – 40 |

No. of students |
8 |
35 |
35 |
14 |
8 |

** (i) Draw a cumulative frequency curve to estimate the median income.**

**(ii) If 15% of the students are given freeships on the basis of the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.**

**(iii) Calculate the Arithmetic mean.**

**Solution**

**(i) **Cumulative Frequency Curve

Income (in thousand Rs.) |
No. of students (f) |
Cumulative Frequency |
Class mark (x) |
fx |

0 – 8 | 8 | 8 | 4 | 32 |

8 – 16 | 35 | 43 | 12 | 420 |

16 – 24 | 35 | 78 | 20 | 700 |

24 – 32 | 14 | 92 | 28 | 392 |

32 – 40 | 8 | 100 | 36 | 288 |

∑f = 100 | ∑ fx = 1832 |

We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:

Here, N = 100

N/2 = 50

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at B.

Through B, a vertical line is drawn which meets OX at M.

OM = 17.6 units

Hence, median income = 17.6 thousands

**(ii)** 15% of 100 students = (15×100)/100 = 15

From c.f. 15, draw a horizontal line which intersects the curve at P.

From P, draw a perpendicular to x–axis meeting it at Q which is equal to 9.6

Thus, freeship will be awarded to students provided annual income of their parents is upto 9.6 thousands.

**(ii)** Mean = ∑ fx/∑ f = 1832/100 = 18.32

**6. The marks of 20 students in a test were as follows:**

**2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.**

**Calculate:**

**(i) the mean**

**(ii) the median**

**(iii) the mode**

**Solution**

Arranging the terms in ascending order:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20

Number of terms = 20

∑x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257

**(i)** Mean = ∑x/∑n = 257/ 20 = 12.85

**(ii)** Median = (10^{th} term + 11^{th} term)/2 = (13 + 14)/ 2 = 27/2 = 13.5

**(iii)** Mode = 15 since it has maximum frequencies i.e. 3

**7. The marks obtained by 120 students in a mathematics test is given below:**

Marks |
No. of students |

0-10 |
5 |

10-20 |
9 |

20-30 |
16 |

30-40 |
22 |

40-50 |
26 |

50-60 |
18 |

60-70 |
11 |

70-80 |
6 |

80-90 |
4 |

90-100 |
3 |

**Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:**

**(i) the median**

**(ii) the number of students who obtained more than 75% in test.**

**(iii) the number of students who did not pass in the test if the pass percentage was 40.**

**(iv) the lower quartile**

**Solution**

Marks |
No. of students |
c.f. |

0-10 | 5 | 5 |

10-20 | 9 | 14 |

20-30 | 16 | 30 |

30-40 | 22 | 52 |

40-50 | 26 | 78 |

50-60 | 18 | 96 |

60-70 | 11 | 107 |

70-80 | 6 | 113 |

80-90 | 4 | 117 |

90-100 | 3 | 120 |

**(i)** Median = (120 + 1)/2 = 60.5^{th} term

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A. From A draw a perpendicular to x-axis meeting it at B.

Then, the value of point B is the median = 43

**(ii)** Number of students who obtained up to 75% marks in the test = 110

Number of students who obtained more than 75% marks in the test = 120 – 110 = 10

**(iii)** Number of students who obtained less than 40% marks in the test = 52 (from the graph; x = 40, y = 52)

(iv) Lower quartile = Q_{1} = 120×(1/4) = 30^{th} term = 30

**8. Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.**

Weight |
Frequency |

40 – 45 |
5 |

45 – 50 |
17 |

50 – 55 |
22 |

55 – 60 |
45 |

60 – 65 |
51 |

65 – 70 |
31 |

70 – 75 |
20 |

75 – 80 |
9 |

**Use your ogive to estimate the following:**

**(i) The percentage of students weighing 55 kg or more**

**(ii) The weight above which the heaviest 30% of the student fall**

**(iii) The number of students who are (a) underweight (b) overweight, if 55.70 kg is considered as standard weight.**

**Solution**

Weight |
Frequency |
c. f. |

40-45 | 5 | 5 |

45-50 | 17 | 22 |

50-55 | 22 | 44 |

55-60 | 45 | 89 |

60-65 | 51 | 140 |

65-70 | 31 | 171 |

70-75 | 20 | 191 |

75-80 | 9 | 200 |

(i) The number of students weighing more than 55 kg = 200 – 44 = 156

Thus, the percentage of students weighing 55 kg or more = (156/200)×100 = 78 %

(ii) 30% of students = (30×200)/100 = 60

Heaviest 60students in weight = 9 + 21 + 30 = 60

Weight = 65 kg (From table)

(iii) (a) underweight students when 55.70 kg is standard = 46 (approx.) from graph

(b) overweight students when 55.70 kg is standard = 200 – 55.70 = 154 (approx.) from graph

**9. The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.**

Marks obtained |
5 |
6 |
7 |
8 |
9 |
10 |

No. of students |
3 |
9 |
6 |
4 |
2 |
1 |

**Solution**

Marks obtained(x) |
No. of students (f) |
c.f. |
fx |

5 | 3 | 3 | 15 |

6 | 9 | 12 | 54 |

7 | 6 | 18 | 42 |

8 | 4 | 22 | 32 |

9 | 2 | 24 | 18 |

10 | 1 | 25 | 10 |

Total | 25 | 171 |

Number of terms = 25

(i) Mean = 171/25 = 6.84

(ii) Median = (25 + 1)/2 ^{th} = 13^{th} term = 7

(iii) Mode = 6 since it has the maximum frequency i.e. 6

**10. The mean of the following distribution is 52 and the frequency of class interval 30 – 40 is ‘f’. Find f.**

Class Interval |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
50 – 60 |
60 – 70 |
70 – 80 |

Frequency |
5 |
3 |
f |
7 |
2 |
6 |
13 |

**Solution**

C.I. |
Frequency(f) |
Mid value (x) |
fx |

10-20 | 5 | 15 | 75 |

20-30 | 3 | 25 | 75 |

30-40 | f | 35 | 35f |

40-50 | 7 | 45 | 315 |

50-60 | 2 | 55 | 110 |

60-70 | 6 | 65 | 390 |

70-80 | 13 | 75 | 975 |

Total | 36 + f | 1940 + 35f |

Mean = ∑ fx/∑ f = (1940 + 35f)/(36 + f) **...(i)**

But, given mean = 52 **…(ii)**

From (i) and (ii), we have

(1940 + 35f)/ (36 + f) = 52

⇒ 1940 + 35f = 1872 + 52f

⇒ 17f = 68

Thus, f = 4

**11. The monthly income of a group of 320 employees in a company is given below:**

Monthly Income (thousands) |
No. of employees |

6 – 7 |
20 |

7 – 8 |
45 |

8 – 9 |
65 |

9 – 10 |
95 |

10 – 11 |
60 |

11 – 12 |
30 |

12 – 13 |
5 |

**Draw an ogive of the given distribution on a graph paper taking 2 cm = Rs 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine:**

**(i) the median wage.**

**(ii) number of employees whose income is below Rs 8500.**

**(iii) if salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.**

**(iv) the upper quartile.**

**Solution**

Monthly Income (thousands) |
No. of employees (f) |
Cumulative frequency |

6-7 | 20 | 20 |

7-8 | 45 | 65 |

8-9 | 65 | 130 |

9-10 | 95 | 225 |

10-11 | 60 | 285 |

11-12 | 30 | 315 |

12-13 | 5 | 320 |

Total | 320 |

Number of employees = 320

**(i)** Median = 320/2 = 160^{th} term

Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = Rs 9.3 thousands

**(ii)** The number of employees with income below Rs 8,500 = 95 (approx from the graph)

**(iii)** Number of employees with income below Rs 11,500 = 305 (approx from the graph)

Thus, the number of employees (senior employees) = 320 – 305 = 15

**(iv)** Upper quartile = Q_{3} = 320×(3/4) = 240^{th} term = 10.3 thousands = Rs 10,300