Frank Chapter 24 Measure of Central Tendency Solutions Class 10 Maths

Frank Solutions for Chapter 24 Measure of Central Tendency Class 10 ICSE Mathematics

Exercise 24.1

1. Find the mean of first 12 even numbers.

Answer

We know that, the first 12 even numbers are,

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

where n is the total numbers,

n = 12

x̅ = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24)/12

⇒ x̅ = 156/12

⇒ x̅ = 13

Hence, mean of first 12 even numbers is 13.

2. Find the mean of first 10 prime numbers.

Answer

We know that, the first 10 prime numbers are,

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 10

x̅ = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10

⇒ x̅ = 129/10

⇒ x̅ = 12.9

Hence, mean of first 10 prime numbers is 12.9.

3. Find the mean of all numbers from 7 to 17.

Answer

All numbers from 7 to 17 are,

7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 11

x̅ = (7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17)/11

⇒ x̅ = 132/11

⇒ x̅ = 12

Hence, mean of all numbers from 7 to 17.

4. Find the mean of all odd numbers from 5 to 20. Find the new mean when each number is multiplied by 4.

Answer

All odd numbers from 5 to 20 are,

5, 7, 9, 11, 13, 15, 17, 19

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 8

x̅ = (5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/11

⇒ x̅ = 96/8

⇒ x̅ = 12

Hence, mean of all odd numbers from 5 to 20 is 12.

Then, all odd numbers from 5 to 20 multiplied by 4 are,

20, 28, 36, 44, 52, 60, 68, 76

n = 8

x̅ = (20 + 28 + 36 + 44 + 52 + 60 + 68 + 76)/8

⇒ x̅ = 384/8

⇒ x̅ = 48

Hence, mean all odd numbers from 5 to 20 multiplied by 4 is 48.

5. Find the mean of all natural numbers from 32 to 46. Find the new mean when each number is diminished by 5.

Answer

All natural numbers from 32 to 46 are,

32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46.

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 15

x̅ = (32 + 33 +34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46)/15

⇒ x̅ = 585/15

⇒ x̅ = 39

Hence, mean of all natural numbers from 32 to 46 is 39.

Then, all natural numbers from 32 to 46 diminished by 5 are,

27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41

n = 15

x̅ = (27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41)/15

⇒ x̅ = 510/15

⇒ x̅ = 34

Hence, mean all natural numbers from 32 to 46 diminished by 5 is 34.

6. If the mean of 8, 14, 20, x and 12 is 13, find x.

Answer

Form the question it is given that, 8, 14, 20, x, 12

Mean = 13

We have to find the value of x,

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 5

13 = (8 + 14 + 20 + x + 12)/5

⇒ 13 × 5 = (54 + x)

⇒ 65 = 54 + x

⇒ x = 65 – 54

⇒ x = 11

Therefore, the value of x is 11.

7. If the mean of 11, 14, p, 26, 10, 12, 18 and 6 is 15, find p.

Answer

Form the question it is given that, 11, 14, p, 26, 10, 12, 18 and 6.

Mean = 15

We have to find the value of p,

Then, x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 8

15 = (11 + 14 + p + 26 + 10 + 12 + 18 + 6)/8

⇒ 15 × 8 = (97 + p)

⇒ 120 = 97 + p

⇒ p = 120 – 97

⇒ p = 23

Therefore, the value of p is 23.

8. The mean monthly income of 10 persons is Rs 8,670. If a new member with a monthly income of Rs 9,000 joins the group, find the new monthly income.

Answer

From the question it is given that,

The mean monthly income of 10 persons is ₹8,670.

Number of persons, n = 10

We know that,

x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

₹8,670 = ∑x_n/10

⇒ ∑x_n = 8,670 × 10

⇒ ∑x_n = ₹86,700

Also it is given that, a new member with a monthly income of ₹ 9,000.

So, ∑x_n = ₹ (86,700 + 9,000)

⇒ ∑x_n = ₹ 95,700

Then, n = 11

x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

⇒ x̅ = ₹ 95,700/11

⇒ x̅ = ₹ 8,700

Therefore, the new mean monthly income is ₹ 8,700.

9. The height of 9 persons are 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 148 cm and 151 cm. Find the mean height.

Answer

From the question it is given that,

The height of 9 persons are, 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 148 cm and 151 cm.

We know that,

x̅ = (x₁ + x₂ + x₃ + … + x_n)/n

Where n is the total numbers,

n = 9

x̅ = (142 + 158 + 152 + 143 + 139 + 144 + 146 + 148 + 151)/9

⇒ x̅ = 1323/9

⇒ x̅ = 147 cm

Therefore, the mean height is 147 cm.

10. Find the mean of the following frequency distribution:

(i)

Class	0 - 10	10 – 20	20 - 30	30 – 40	40 - 50
Frequency	4	7	6	3	5

 

(ii)

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
Frequency	4	4	7	10	12	8	5

(iii)

Class	0 - 6	6 - 12	12 – 18	18 - 24	24 - 30
Frequency	7	5	10	12	6

 

(iv) 

Class	25 - 35	35 - 45	45 - 55	55 - 65	65 – 75
Frequency	6	10	8	12	4

 

(v)

Class	50 – 60	60 - 70	70 - 80	80 – 90	90 - 100
Frequency	8	6	12	11	13

(vi) 

Class	1 - 10	11 - 20	21 - 30	31 - 40	41 - 50
Frequency	9	12	15	10	14

 

(vii) 

Class	101 - 110	111 - 120	121 – 130	131 – 140	141 - 150	151 - 160
Frequency	9	12	15	10	14

Answer

(i)  So, now we have to prepare the frequency distribution table

we know that,

x̅ = ∑f_ix_i/∑f_i

⇒ x̅ = 605/25

⇒ x̅ = 24.2

Therefore, the mean is 24.2.

(ii) So, now we have to prepare the frequency distribution table,

x̅ = ∑f_ix_i/∑f_i

we know that,

x̅ = 1910/50

⇒ x̅ = 38.2

Therefore, the mean is 38.2.

(iii) So, now we have to prepare the frequency distribution table,

x̅ = ∑f_ix_i/∑f_i

we know that,

x̅ = 630/40

⇒ x̅ = 15.75

Therefore, the mean is 15.75.

(iv) So, now we have to prepare the frequency distribution table,

So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑f_ix_i/∑f_i

⇒ x̅ = 1980/40

⇒ x̅ = 49.5

Therefore, the mean is 49.5.

(v) So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑f_ix_i/∑f_i

⇒ x̅ = 3900/50

⇒ x̅ = 78

Therefore, the mean is 78.

(vi) So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑f_ix_i/∑f_i

⇒ x̅ = 1610/60

⇒ x̅ = 26.83

Therefore, the mean is 26.83.

(vii) So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑f_ix_i/∑f_i

⇒ x̅ = 13020/100

⇒ x̅ = 130.2

Therefore, the mean is 130.2.

11. The mean of the following frequency distribution is 25.8 and the sum of all the frequencies is 50. Find x and y.

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	7	x	15	y	10

Answer

So, now we have to prepare the frequency distribution table

We know that,

∑f_i = x₁ + x₂ + … + x_n

⇒ 50 = 7 + x + 15 + y + 10

⇒ x + y + 32 = 50

⇒ x + y = 18 …(i)

Also we know that, x̅ = ∑f_ix_i/∑f_i

So,

25.8 = (860 + 15x + 35y)/50

By cross multiplication we get,

15x + 35y + 860 = 1290

⇒ 15x + 35y = 1290 – 860

⇒ 15x + 35y = 430  [divide both side by 5]

⇒ 3x + 7y = 86 …(ii)

Now multiplying equation (i) by 3 we get,

3x + 3y = 54 …(iii)

Subtract equation (ii) from equation (iii) we get,

4y = 32

⇒ y = 32/4

⇒ y = 8

Substitute value of y in equation (i) to get the value of x,

x + y = 18

⇒ x + 8 = 18

⇒ x = 18 – 8

⇒ x = 10

Hence, the value of x = 10 and y = 8.

12. Find the mean of the following frequency distribution by the short cut method.

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	9	12	15	10	14

Answer

So, now we have to prepare the frequency distribution table, 

we know that,

x̅ = A + ∑f_id/∑f_i

⇒ x̅ = 25 + 80/60

⇒ x̅ = 25 + 1.33

⇒ x̅ = 26.33

Therefore, the value of mean is 26.33.

13. Find the mean of the following frequency distribution by the short cut method:

Class	1 - 10	11 - 20	21 – 30	31 - 40	41 - 50	51 – 60	61 – 70
Frequency	7	10	14	17	15	11	6

Answer

So, now we have to prepare the frequency distribution table,

we know that,

x̅ = A + ∑f_id/∑f_i 

⇒ x̅ = 35.5 + 0/80

⇒ x̅ = 35.5 + 0

⇒ x̅ = 35.5

Therefore, the value of mean is 35.5.

14. Find the mean of the following frequency distribution by the step deviation method:

Class	1 - 10	11 - 20	21 - 30	31 – 40	41 – 50	51 – 60	61 – 70
Frequency	7	10	14	17	15	11	6

Answer

So, from the table A = 125 and h_i = 10

We know that,

x̅ = A + h × ∑f_iu/∑f_i

⇒ x̅ = 125 + 10 × (-3/100)

⇒ x̅ = 125 – 0.3

⇒ x̅ = 124.70

Therefore, the mean is 124.70.

15. Find the mean of the following frequency distribution by the step deviation method:

Class	0 - 20	20 – 40	40 – 60	60 – 80	80 - 100	100 - 120	120 – 140
Frequency	12	24	52	88	66	42	16

Answer

So, from the table A = 70 and h_i = 20

we know that, 

x̅ = A + h × ∑f_iu/∑f_i

⇒ x̅ = 70 + 20 × (62/300)

⇒ x̅ = 70 + 4.13

⇒ x̅ = 74.13

Therefore, the mean is 74.13.

Exercise 24.2

1. The weights of 11 students in a class are 36 kg, 45 kg, 44 kg, 37 kg, 36 kg, 41 kg, 45 kg, 43 kg, 39 kg, 42 kg and 40 kg. Find the median of their weights.

Answer

Arranging the given data in descending order:

45 kg, 45 kg, 44 kg, 43 kg, 42 kg, 41 kg, 40 kg, 39 kg, 37 kg, 36 kg, 36 kg

The middle term is 41 kg which is 6^th term

Therefore, Median of weights = 41 kg.

2. The percentage marks obtained in 10 subjects by a student are 84, 88, 72, 91, 68, 75, 98, 96, 79 and 86. Find the median of the marks obtained.

Answer

Arranging the given data in descending order:

98, 96, 91, 88, 86, 84, 79, 75, 72, 68

The middle term are 86 and 84 which are 5^th and 6^th terms

Median = (86 + 84)/2 = 170/2 = 85

Therefore, Median of marks = 85

3. Find the median of the first 15 whole numbers.

Answer

The first 15 whole numbers are :

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

Arranging the given data in descending order:

14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0

The middle term is 7 which is 8th term.

Therefore, Median of numbers = 7

4. Find the median of all prime numbers between 20 and 50.

Answer

The prime number between 20 and 50 are:

23, 29, 31, 37, 41, 43, 47

Arranging the given data in descending order:

47, 43, 41, 37, 31, 29, 23

The middle term is 37 which is 4^th term

Therefore, Median of numbers = 37.

5. Find the median of the following:

(i) 11, 8, 15, 5, 9, 4, 19, 6, 18

(ii) 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

(iii) 3x, x + 5, x + 7, x + 9, x + 11, x + 13

Answer

(i) 11, 8, 15, 5, 9, 4, 19, 6, 18

Arranging the given data in descending order:

19, 18, 15, 11, 9, 8, 6, 5, 4

The middle term is 9 which is 5^th term

Therefore, Median of numbers = 9.

(ii) 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Arranging the given data in descending order:

35, 34, 32, 31, 29, 26, 25, 23, 22, 20

The middle terms are 29 and 26 which are 5^th and 6^th terms

Median = (29 + 26)/2 = 55/2 = 27.5

Therefore, Median of numbers = 27.5

(iii) 3x, x + 5, x + 7, x + 9, x + 11, x + 13

Arranging the given data in descending order:

x + 13, x + 11, x + 9, x + 7, x + 5, 3x

The middle terms are x + 9 and x + 7 which are 3^rd and 4^th terms

median = (x + 9 + x + 7)/2

= (2x + 16)/2

= x + 8

Therefore, Median = x + 8.

6. Find the median of the following frequency distribution:

(i)

Weight (kg)	36	38	40	42	44
No. of Students	11	26	29	24	10

(ii)

Weight (kg)	3500	4000	4500	5000	5500	6000
No. of people	9	17	23	15	6	5

Answer

(i)

(ii)

7. The frequency distribution table below shows the height of 50 students of grade 10.

Heights (in cm)	138	139	140	141	142
Frequency	6	11	16	10	7

Find the median, the upper quartile and the lower quartile of the heights.

Answer

8. Find the lower quartile, the upper quartile, the interquartile range and the semi-interquartile range for the following frequency distributions:

(i)

Shoe size	5	6	7	8	9	10	11
Frequency	8	1	7	14	11	5	4

(ii)

Marks	25	30	35	40	45	50
No. of students	6	15	12	10	18	9

(iii)

Variate	10	11	12	13	14	15	16	17	18	19	20
Frequency	1	2	3	1	2	4	2	1	1	2	1

 Answer

(i)

(ii)

(iii)

9. Estimate the median, the lower quartile and the upper quartile of the following frequency distributions by drawing an ogive:

(i)

Class Interval	0 - 10	10 – 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
Frequency	4	12	21	18	15	7	3

(ii)

Marks	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of boys	10	12	14	12	9	7	6

(iii)

Marks (less than)	10	20	30	40	50	60	70	80
No. of Students	5	15	30	54	72	86	94	100

(iv)

Age (in yrs)	Under 10	Under 20	Under 30	Under 40	Under 50	Under 60
No. of males	6	10	25	32	43	50

(v)

Marks (more than)	90	80	70	60	50	40	30	20	10	0
No. of students	6	13	22	34	48	60	70	78	80	80

Answer

(i)

(ii) 

(iii)

(iv) 

(v)

10. The marks obtained by 200 students in an examination are given below:

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 – 60	60 – 70	70 – 80	80 - 90	90 - 100
No. of students	5	10	11	20	27	38	40	29	14	6

Using a graph paper, draw an ogive for the above distribution and estimate

(i) the median

(ii) the lower quartile

(iii) the number of students who obtained more than 80% marks in the examination.

Answer

(i)

(ii)

(iii)

11. The marks of 200 students in a test is given below:

Marks %	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79	80 - 89
No. of Students	7	11	20	46	57	37	15	7

Draw an ogive and find the

(i) the median

(ii) the number of students who scored more than 35% marks

Answer

(i) 

(ii) 

Exercise- 24.3

1. Find the mode of the following:

(i) 6, 7, 1, 8, 6, 5, 9, 4, 6, 7, 1, 3, 2, 6, 7, 8

(ii) 21, 22, 28, 23, 24, 21, 26, 22, 29, 27, 21, 21, 26, 24, 23

(iii) 3, 4, 5, 7, 6, 3, 5, 4, 3, 5, 6, 4, 7, 5, 4, 5, 4, 3, 4, 5, 7, 6, 5, 6, 6, 7

(iv) 15, 17, 16, 17, 10, 12, 14, 16, 19, 12, 16, 15, 16

(iv) 20, 20, 30, 30, 30, 30, 35, 40, 40, 45, 45, 45, 50, 55, 60, 60, 60, 65, 70, 70, 70

Answer

2. Find the mode of the following frequency distribution:

(i)

Variate	20	21	22	23	24	25	26
Frequency	21	20	26	35	22	13	10

(ii)

Pocket money per week in Rs	25	50	75	100	125	150
No. of students	4	7	13	18	6	2

(iii)

Hrs. Spent daily in studies	3	3.5	4	4.5	5	5.5	6	6.5
No. of students	8	7	3	5	10	6	3	4

Answer

(i)

(ii)

(iii)

3. Draw a histogram for the following distribution and estimate the mode:

(i)

Marks	0 - 10	10 - 20	20 - 30	30 – 40	40 - 50	50 - 60	60 - 70	70 - 80
No. of students	3	7	15	24	16	8	5	2

(ii)

I. Q. Score	80 - 100	100 - 120	120 - 140	140 - 160	160 – 180	180 - 200
No. of Students	6	9	16	13	4	2

(iii)

Mangoes per tree	0 - 9	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59
No. of trees	10	16	20	14	6	4

(iv)

Marks %	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79	80 - 89	90 - 99
No. of students	14	26	40	92	114	78	36

Answer

(i)

(ii)

(iii)

(iv)