The numerical expressions which represent the characteristics of a group (a large collection of numerical data) are called Measures of Central Tendency. In this chapter, students will learn and solve problems based on three types of statistical averages, namely Mean, Median and Mode. Further in the chapter, different methods of finding these central tendencies are studied and tested under the exercise problems. Students can refer to the Selina Solutions for Class 10 Mathematics for clarifying any doubts regarding solving problems. It also helps students for their exam preparations and mainly improves problem-solving skills, which is vital. The solutions of the Selina Solutions for Class 10 Mathematics Chapter 24 Measures of Central Tendency exercises are available in the links given below.
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Exercise 24(A) Page No: 356
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 = âˆ‘x/ n
Here, n = 5
Thus,
Mean = (6 + 9 + 11 + 12 + 7)/ 5 = 45/5 = 9
(ii) By definition, we know
Mean = âˆ‘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 = âˆ‘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 = âˆ‘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 = âˆ‘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 â€“ 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 x 5 = 40 â€¦â€¦ (i)
But, sum of numbers = 6 + 4 + 7 + a + 10 = 27 + a â€¦.. (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 x 8 = 40 â€¦â€¦. (i)
but sum of numbers = 6 + y + 7 + x + 14 = 27 + y + x â€¦â€¦. (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 = âˆ‘f_{i} x_{i}/ âˆ‘f_{i} = 614/40 = 15.35
Exercise 24(B) Page No: 361
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 = âˆ‘f_{i} x_{i}/ âˆ‘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 = âˆ‘f_{i} x_{i}/ âˆ‘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 â€“ 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 â€“ 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} â€“ 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 = 87.50
u_{i} = (x_{i} â€“ 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. Find the value of ‘f’.
C. I. | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
freq | 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 |
9324 + 148f = 8645 + 245f
245f â€“ 148f = 9324 â€“ 8645
f = 679/97
Thus, f = 7
Exercise 24(C) Page No: 372
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}Â 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
Itâ€™s seen that,
The middle terms are 24 and 24, 5^{th}Â and 6^{th}Â 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}Â term i.e. 29
Hence, median = 29
(ii) Lower quartile
(iii) Upper quartile =
(iv) Interquartile range = q_{3}Â – q_{1}Â =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}Â and 9^{th}Â 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} â€“ q_{1}
Thus,
q_{3} â€“ q_{1} = 65 â€“ 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
Therefore, the median = 14
Exercise 24(D) Page No: 374
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) Itâ€™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) Page No: 375
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 â€“ 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 â€“ 160 | 38 | 100 |
160 – 165 | 24 | 124 |
165 – 170 | 16 | 140 |
170 – 175 | 12 | 152 |
175 – 180 | 8 | 160 |
N = 160 |
Now, letâ€™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} â€“ Q_{1}
= 164 â€“ 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. DrawÂ ogiveÂ for the data given below and from the graph determine: (i) the median marks.
(ii)Â theÂ 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 =Â 75/100 x 100 = 75Â 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 â€¦. ,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}Â 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}Â July as it has maximum frequencies.
(ii) Total number of terms = 206
Upper quartile =Â 206 x (3/4) = 154.5^{th} = 7^{th} July
Lower quartile = 206 x (1/4) = 51.5^{th} = 4^{th} July
5. TheÂ income of the parents of 100 students in a class in a certain university areÂ tabulated below.
Income (in thousand Rs) | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |
No. of students | 8 | 35 | 35 | 14 | 8 |
Â (i)Â Draw a cumulative frequency curve to estimate the median income.
(ii)Â If 15% of the students are givenÂ freeshipsÂ on the basis of the basis of the income of their parents, find the annual income of parents, below which theÂ freeshipsÂ will be awarded.
(iii)Â 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 |
âˆ‘f = 100 | âˆ‘ 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 x 100)/ 100 = 15
From c.f. 15, draw a horizontal line which intersects the curve at P.
From P, draw a perpendicular to x â€“ 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 = âˆ‘ fx/ âˆ‘ 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
âˆ‘ x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257
(i) Mean = âˆ‘x/ âˆ‘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)Â 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}Â = 120 x (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) x 100 = 78 %
(ii) 30% of students = (30 x 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 =Â 171/25 = 6.84
(ii)Â 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 = âˆ‘ fx/ âˆ‘ f = (1940 + 35f)/ (36 + f) â€¦â€¦ (i)
But, given mean = 52 â€¦. (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 x (3/4) = 240^{th} term = 10.3 thousands = Rs 10,300
The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.