Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency are useful for students as it helps them in scoring high marks in the examination. These solutions are prepared by subject matter experts at BYJUâ€™S, describing the complete method of solving problems. Learners can download the pdf of Frank Solutions for Class 10 Maths Chapter 24 from the below links. It also provides the foundation for higher studies.
Chapter 24 – Measure of Central Tendency, 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 Frequency table, presentation of data in groups result in Grouping of Data. In the Frank Solutions, many such exercise problems are given which helps in better learning.
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1. Find the mean of 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, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
n = 12
xÌ… = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24)/12
xÌ… = 156/12
xÌ… = 13
Hence, mean of first 12 even numbers is 13.
2. Find the mean of 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, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
n = 10
xÌ… = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10
xÌ… = 129/10
xÌ… = 12.9
Hence, mean of 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, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
n = 11
xÌ… = (7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17)/11
xÌ… = 132/11
xÌ… = 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.
Solution:-
All odd numbers from 5 to 20 are,
5, 7, 9, 11, 13, 15, 17, 19
Then, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
n = 8
xÌ… = (5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/11
xÌ… = 96/8
xÌ… = 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
xÌ… = (20 + 28 + 36 + 44 + 52 + 60 + 68 + 76)/8
xÌ… = 384/8
xÌ… = 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.
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, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
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, 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, 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:-
Form the question it is given that, 8, 14, 20, x, 12
Mean = 13
We have to find the value of x,
Then, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
n = 5
13 = (8 + 14 + 20 + x + 12)/5
13 Ã— 5 = (54 + x)
65 = 54 + x
x = 65 â€“ 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:-
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, xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
n = 8
15 = (11 + 14 + p + 26 + 10 + 12 + 18 + 6)/8
15 Ã— 8 = (97 + p)
120 = 97 + p
p = 120 â€“ 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 jons the group, find the new monthly income.
Solution:-
From the question it is given that,
The mean monthly income of 10 persons is â‚¹ 8,670.
Number of persons, n = 10
We know that,
xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
â‚¹ 8,670 = âˆ‘x_{n}/10
âˆ‘x_{n} = 8,670 Ã— 10
âˆ‘x_{n} = â‚¹ 86,700
Also it is given that, a new member with a monthly income of â‚¹ 9,000.
So, âˆ‘x_{n} = â‚¹ (86,700 + 9,000)
âˆ‘x_{n} = â‚¹ 95,700
Then, n = 11
xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
xÌ… = â‚¹ 95,700/11
xÌ… = â‚¹ 8,700
Therefore, the new mean monthly income is â‚¹ 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.
Solution:-
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,
xÌ… = (x_{1} + x_{2} + x_{3} + â€¦ + x_{n})/n
Where n is the total numbers,
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 â€“ 10 |
10 â€“ 20 |
20 â€“ 30 |
30 â€“ 40 |
40 â€“ 50 |
Frequency |
4 |
7 |
6 |
3 |
5 |
Solution:-
So, now we have to prepare the frequency distribution table
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
0 â€“ 10 |
5 |
4 |
20 |
10 â€“ 20 |
15 |
7 |
105 |
20 â€“ 30 |
25 |
6 |
150 |
30 â€“ 40 |
35 |
3 |
105 |
40 â€“ 50 |
45 |
5 |
225 |
Total |
25 |
605 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
xÌ… = 605/25
xÌ… = 24.2
Therefore, the mean is 24.2.
(ii)
Class |
0 â€“ 10 |
10 â€“ 20 |
20 â€“ 30 |
30 â€“ 40 |
40 â€“ 50 |
50 – 60 |
60 â€“ 70 |
Frequency |
4 |
4 |
7 |
10 |
12 |
8 |
5 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
0 â€“ 10 |
5 |
4 |
20 |
10 â€“ 20 |
15 |
4 |
60 |
20 â€“ 30 |
25 |
7 |
175 |
30 â€“ 40 |
35 |
10 |
350 |
40 â€“ 50 |
45 |
12 |
540 |
50 â€“ 60 |
55 |
8 |
440 |
60 â€“ 70 |
65 |
5 |
325 |
Total |
50 |
1910 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
xÌ… = 1910/50
xÌ… = 38.2
Therefore, the mean is 38.2.
(iii)
Class |
0 â€“ 6 |
6 â€“ 12 |
12 â€“ 18 |
18 â€“ 24 |
24 â€“ 30 |
Frequency |
7 |
5 |
10 |
12 |
6 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
0 â€“ 6 |
3 |
7 |
21 |
6 â€“ 12 |
9 |
5 |
45 |
12 â€“ 18 |
15 |
10 |
150 |
18 â€“ 24 |
21 |
12 |
252 |
24 â€“ 30 |
27 |
6 |
162 |
Total |
40 |
630 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
xÌ… = 630/40
xÌ… = 15.75
Therefore, the mean is 15.75.
(iv)
Class |
25 â€“ 35 |
35 â€“ 45 |
45 â€“ 55 |
55 â€“ 65 |
65 â€“ 75 |
Frequency |
6 |
10 |
8 |
12 |
4 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
25 â€“ 35 |
30 |
6 |
180 |
35 â€“ 45 |
40 |
10 |
400 |
45 â€“ 55 |
50 |
8 |
400 |
55 â€“ 65 |
60 |
12 |
720 |
65 – 75 |
70 |
4 |
280 |
Total |
40 |
1980 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
xÌ… = 1980/40
xÌ… = 49.5
Therefore, the mean is 49.5.
(v)
Class |
50 â€“ 60 |
60 â€“ 70 |
70 â€“ 80 |
80 â€“ 90 |
90 â€“ 100 |
Frequency |
8 |
6 |
12 |
11 |
13 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
50 â€“ 60 |
55 |
8 |
440 |
60 â€“ 70 |
65 |
6 |
390 |
70 â€“ 80 |
75 |
12 |
900 |
80 â€“ 90 |
85 |
11 |
935 |
90 – 100 |
95 |
13 |
1235 |
Total |
50 |
3900 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
xÌ… = 3900/50
xÌ… = 78
Therefore, the mean is 78.
(vi)
Class |
1 â€“ 10 |
11 â€“ 20 |
21 â€“ 30 |
31 â€“ 40 |
41 â€“ 50 |
Frequency |
9 |
12 |
15 |
10 |
14 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
1 â€“ 10 |
5.5 |
9 |
49.5 |
11 â€“ 20 |
15.5 |
12 |
186 |
21 â€“ 30 |
25.5 |
15 |
382.5 |
31 â€“ 40 |
35.5 |
10 |
355 |
41 – 50 |
45.5 |
14 |
637 |
Total |
60 |
1610 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
xÌ… = 1610/60
xÌ… = 26.83
Therefore, the mean is 26.83.
(vi)
Class |
101 â€“ 110 |
111 â€“ 120 |
121 â€“ 130 |
131 â€“ 140 |
141 â€“ 150 |
151 – 160 |
Frequency |
9 |
12 |
15 |
10 |
14 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
101 â€“ 110 |
105.5 |
11 |
1160.5 |
111 â€“ 120 |
115.5 |
16 |
1848 |
121 â€“ 130 |
125.5 |
20 |
2510 |
131 â€“ 140 |
135.5 |
30 |
4065 |
141 – 150 |
145.5 |
14 |
2037 |
151 – 160 |
155.5 |
9 |
1399.5 |
Total |
100 |
13020 |
we know that,
xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
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 â€“ 20 |
20 â€“ 30 |
30 â€“ 40 |
40 â€“ 50 |
Frequency |
7 |
x |
15 |
y |
10 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
f_{i}x_{i} |
0 â€“ 10 |
5 |
7 |
35 |
10 â€“ 20 |
15 |
x |
15x |
20 â€“ 30 |
25 |
15 |
375 |
30 â€“ 40 |
35 |
y |
35y |
40 – 50 |
45 |
10 |
450 |
Total |
50 |
860 + 15x + 35y |
We know that,
âˆ‘f_{i} = x_{1} + x_{2} + â€¦ + x_{n}
50 = 7 + x + 15 + y + 10
x + y + 32 = 50
x + y = 18 â€¦ (i)
Also we know that, xÌ… = âˆ‘f_{i}x_{i}/âˆ‘f_{i}
So,
25.8 = (860 + 15x + 35y)/50
By cross multiplication we get,
15x + 35y + 860 = 1290
15x + 35y = 1290 â€“ 860
15x + 35y = 430 â€¦ [divide both side by 5]
3x + 7y = 86 â€¦ (ii)
Now multiplying equation (i) by 3 we get,
3x + 3y = 54 â€¦ (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 â€“ 8
x = 10
Hence the value of x = 10 and y = 8.
11. Find the mean of the following frequency distribution by the short cut method.
Class |
0 – 10 |
10 â€“ 20 |
20 â€“ 30 |
30 â€“ 40 |
40 â€“ 50 |
Frequency |
9 |
12 |
15 |
10 |
14 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
A = 25 d = x â€“ A |
f_{i}d |
0 â€“ 10 |
5 |
9 |
– 20 |
-180 |
10 â€“ 20 |
15 |
12 |
– 10 |
-120 |
20 â€“ 30 |
A = 25 |
15 |
0 |
0 |
30 â€“ 40 |
35 |
10 |
10 |
100 |
40 – 50 |
45 |
14 |
20 |
280 |
Total |
60 |
80 |
we know that,
xÌ… = A + âˆ‘f_{i}d/âˆ‘f_{i}
xÌ… = 25 + 80/60
xÌ… = 25 + 1.33
xÌ… = 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 â€“ 20 |
21 â€“ 30 |
31 â€“ 40 |
41 â€“ 50 |
51 – 60 |
61 â€“ 70 |
Frequency |
7 |
10 |
14 |
17 |
15 |
11 |
6 |
Solution:-
So, now we have to prepare the frequency distribution table,
Class Interval |
x_{i} |
f_{i} |
A = 25 d = x â€“ A |
f_{i}d |
1 â€“ 10 |
5.5 |
7 |
– 30 |
-210 |
11 â€“ 20 |
15.5 |
10 |
20 |
-200 |
21 â€“ 30 |
25.5 |
14 |
-10 |
-140 |
31 â€“ 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,
xÌ… = A + âˆ‘f_{i}d/âˆ‘f_{i}
xÌ… = 35.5 + 0/80
xÌ… = 35.5 + 0
xÌ… = 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 â€“ 20 |
21 â€“ 30 |
31 â€“ 40 |
41 â€“ 50 |
51 – 60 |
61 â€“ 70 |
Frequency |
7 |
10 |
14 |
17 |
15 |
11 |
6 |
Solution:-
Class Interval |
x_{i} |
f_{i} |
A = 125 u = (x â€“ A)/h_{i} |
f_{i}u |
100 â€“ 110 |
105 |
15 |
– 2 |
-30 |
110 â€“ 120 |
115 |
18 |
-1 |
-18 |
120 â€“ 130 |
A = 125 |
32 |
0 |
0 |
130 â€“ 140 |
315 |
25 |
1 |
25 |
140 – 150 |
145 |
10 |
2 |
20 |
Total |
100 |
-3 |
So, from the table A = 125 and h_{i} = 10
we know that,
xÌ… = A + h Ã— âˆ‘f_{i}u/âˆ‘f_{i}
xÌ… = 125 + 10 Ã— (-3/100)
xÌ… = 125 â€“ 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 â€“ 40 |
40 â€“ 60 |
60 â€“ 80 |
80 â€“ 100 |
100 – 120 |
120 â€“ 140 |
Frequency |
12 |
24 |
52 |
88 |
66 |
42 |
16 |
Solution:-
Class Interval |
x_{i} |
f_{i} |
A = 125 u = (x â€“ A)/h_{i} |
f_{i}u |
0 â€“ 20 |
10 |
12 |
– 3 |
-36 |
20 â€“ 40 |
30 |
24 |
-2 |
-48 |
40 â€“ 60 |
50 |
52 |
-1 |
-52 |
60 â€“ 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 h_{i} = 20
we know that,
xÌ… = A + h Ã— âˆ‘f_{i}u/âˆ‘f_{i}
xÌ… = 70 + 20 Ã— (62/300)
xÌ… = 70 + 4.13
xÌ… = 74.13
Therefore, the mean is 74.13.