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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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
xÌ… = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24)/12
xÌ… = 156/12
xÌ… = 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
xÌ… = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10
xÌ… = 129/10
xÌ… = 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
xÌ… = (7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17)/11
xÌ… = 132/11
xÌ… = 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
xÌ… = (5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/11
xÌ… = 96/8
xÌ… = 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
xÌ… = (20 + 28 + 36 + 44 + 52 + 60 + 68 + 76)/8
xÌ… = 384/8
xÌ… = 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
xÌ… = (32 + 33 +34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46)/15
xÌ… = 585/15
xÌ… = 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
xÌ… = (27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41)/15
xÌ… = 510/15
xÌ… = 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
xÌ… = (142 + 158 + 152 + 143 + 139 + 144 + 146 + 148 + 151)/9
xÌ… = 1323/9
xÌ… = 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
xÌ… = 605/25
xÌ… = 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
xÌ… = 1910/50
xÌ… = 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
xÌ… = 630/40
xÌ… = 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
xÌ… = 1980/40
xÌ… = 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
xÌ… = 3900/50
xÌ… = 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
xÌ… = 1610/60
xÌ… = 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
xÌ… = 13020/100
xÌ… = 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
xÌ… = 25 + 80/60
xÌ… = 25 + 1.33
xÌ… = 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
xÌ… = 35.5 + 0/80
xÌ… = 35.5 + 0
xÌ… = 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
xÌ… = 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)
xÌ… = 70 + 4.13
xÌ… = 74.13
Therefore, the mean is 74.13.
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