Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency

Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency are useful for students, as they help them in scoring high marks in the board examination. These solutions are prepared by subject-matter experts at BYJU’S, describing the complete and easy method of solving problems. It provides students with the foundation for higher studies.

Chapter 24 – Measure of Central Tendency: It is a single value that attempts to describe a set of data by identifying the central position within that set of data. Here, we calculate the Frequency Distribution by using the Frequency Table, and the presentation of data in groups results in the Grouping of Data. In Frank Solutions, many such exercise problems are given, which will help students better understand these concepts. They can download the PDF of Frank Solutions for Class 10 Maths Chapter 24 from the link given below.

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Access Answers to Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency

1. Find the mean of the first 12 even numbers.

Solution:-

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 + … + xn)/n

Where n is the total number,

n = 12

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

x̅ = 156/12

x̅ = 13

Hence, the mean of the first 12 even numbers is 13.

2. Find the mean of the first 10 prime numbers.

Solution:-

We know that the first 10 prime numbers are,

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

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total number,

n = 10

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

x̅ = 129/10

x̅ = 12.9

Hence, the mean of the first 10 prime numbers is 12.9.

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

Solution:-

All numbers from 7 to 17 are,

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

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total number,

n = 11

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

x̅ = 132/11

x̅ = 12

Hence, the 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.

Solution:-

All odd numbers from 5 to 20 are,

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

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total number,

n = 8

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

x̅ = 96/8

x̅ = 12

Hence, the 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, the mean of 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.

Solution:-

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 + … + xn)/n

Where n is the total number,

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, the 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, the 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.

Solution:-

From 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 + … + xn)/n

Where n is the total number,

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.

Solution:-

From 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 + … + xn)/n

Where n is the total number,

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.

Solution:-

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 + … + xn)/n

₹ 8,670 = ∑xn/10

∑xn = 8,670 × 10

∑xn = ₹ 86,700

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

So, ∑xn = ₹ (86,700 + 9,000)

∑xn = ₹ 95,700

Then, n = 11

x̅ = (x1 + x2 + x3 + … + xn)/n

x̅ = ₹ 95,700/11

x̅ = ₹ 8,700

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

9. The heights 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.

Solution:-

From the question, it is given that,

The heights 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 + … + xn)/n

Where n is the total number,

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

Solution:-

So, now we have to prepare the frequency distribution table

Class Interval xi fi fixi
0 – 10 5 4 20
10 – 20 15 7 105
20 – 30 25 6 150
30 – 40 35 3 105
40 – 50 45 5 225
Total 25 605

we know that,

x̅ = ∑fixi/∑fi

x̅ = 605/25

x̅ = 24.2

Therefore, the mean is 24.2.

(ii)

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

Solution:-

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

Class Interval xi fi fixi
0 – 10 5 4 20
10 – 20 15 4 60
20 – 30 25 7 175
30 – 40 35 10 350
40 – 50 45 12 540
50 – 60 55 8 440
60 – 70 65 5 325
Total 50 1910

we know that,

x̅ = ∑fixi/∑fi

x̅ = 1910/50

x̅ = 38.2

Therefore, the mean is 38.2.

(iii)

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

Solution:-

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

Class Interval xi fi fixi
0 – 6 3 7 21
6 – 12 9 5 45
12 – 18 15 10 150
18 – 24 21 12 252
24 – 30 27 6 162
Total 40 630

we know that,

x̅ = ∑fixi/∑fi

x̅ = 630/40

x̅ = 15.75

Therefore, the mean is 15.75.

(iv)

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

Solution:-

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

Class Interval xi fi fixi
25 – 35 30 6 180
35 – 45 40 10 400
45 – 55 50 8 400
55 – 65 60 12 720
65 – 75 70 4 280
Total 40 1980

we know that,

x̅ = ∑fixi/∑fi

x̅ = 1980/40

x̅ = 49.5

Therefore, the mean is 49.5.

(v)

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

Solution:-

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

Class Interval xi fi fixi
50 – 60 55 8 440
60 – 70 65 6 390
70 – 80 75 12 900
80 – 90 85 11 935
90 – 100 95 13 1235
Total 50 3900

we know that,

x̅ = ∑fixi/∑fi

x̅ = 3900/50

x̅ = 78

Therefore, the mean is 78.

(vi)

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

Solution:-

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

Class Interval xi fi fixi
1 – 10 5.5 9 49.5
11 – 20 15.5 12 186
21 – 30 25.5 15 382.5
31 – 40 35.5 10 355
41 – 50 45.5 14 637
Total 60 1610

we know that,

x̅ = ∑fixi/∑fi

x̅ = 1610/60

x̅ = 26.83

Therefore, the mean is 26.83.

(vi)

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

Solution:-

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

Class Interval xi fi fixi
101 – 110 105.5 11 1160.5
111 – 120 115.5 16 1848
121 – 130 125.5 20 2510
131 – 140 135.5 30 4065
141 – 150 145.5 14 2037
151 – 160 155.5 9 1399.5
Total 100 13020

we know that,

x̅ = ∑fixi/∑fi

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

Solution:-

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

Class Interval xi fi fixi
0 – 10 5 7 35
10 – 20 15 x 15x
20 – 30 25 15 375
30 – 40 35 y 35y
40 – 50 45 10 450
Total 50 860 + 15x + 35y

We know that,

∑fi = x1 + x2 + … + xn

50 = 7 + x + 15 + y + 10

x + y + 32 = 50

x + y = 18 … (i)

Also, we know that, x̅ = ∑fixi/∑fi

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 sides 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 the 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.

11. Find the mean of the following frequency distribution by the shortcut method.

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

Solution:-

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

Class Interval xi fi A = 25

d = x – A

fid
0 – 10 5 9 – 20 -180
10 – 20 15 12 – 10 -120
20 – 30 A = 25 15 0 0
30 – 40 35 10 10 100
40 – 50 45 14 20 280
Total 60 80

we know that,

x̅ = A + ∑fid/∑fi

x̅ = 25 + 80/60

x̅ = 25 + 1.33

x̅ = 26.33

Therefore, the value of the mean is 26.33.

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

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

Solution:-

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

Class Interval xi fi A = 25

d = x – A

fid
1 – 10 5.5 7 – 30 -210
11 – 20 15.5 10 20 -200
21 – 30 25.5 14 -10 -140
31 – 40 35.5 17 0 0
41 – 50 45.5 15 10 150
51 – 60 55.5 11 20 220
61 – 70 65.5 6 30 180
Total 80 0

we know that,

x̅ = A + ∑fid/∑fi

x̅ = 35.5 + 0/80

x̅ = 35.5 + 0

x̅ = 35.5

Therefore, the value of the 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

Solution:-

Class Interval xi fi A = 125

u = (x – A)/hi

fiu
100 – 110 105 15 – 2 -30
110 – 120 115 18 -1 -18
120 – 130 A = 125 32 0 0
130 – 140 315 25 1 25
140 – 150 145 10 2 20
Total 100 -3

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

we know that,

x̅ = A + h × ∑fiu/∑fi

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

Solution:-

Class Interval xi fi A = 125

u = (x – A)/hi

fiu
0 – 20 10 12 – 3 -36
20 – 40 30 24 -2 -48
40 – 60 50 52 -1 -52
60 – 80 A = 70 88 0 0
80 – 100 90 66 1 66
100 – 120 110 42 2 84
120 – 140 130 16 3 48
Total 300 62

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

we know that,

x̅ = A + h × ∑fiu/∑fi

x̅ = 70 + 20 × (62/300)

x̅ = 70 + 4.13

x̅ = 74.13

Therefore, the mean is 74.13.

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