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S Chand Class 10 ICSE Maths Solutions Chapter 18 Arithmetic Mean, Median, Mode and Quartiles Ex 18(a)
Question 1.
Find the number half-way between 0.2 and 0.02.
Solution:
Half-way number (average) between 0.2 and 0.02 = \(\frac{0.2+0.02}{2}\) = \(\frac{0.22}{2}\) = 0.11
Question 2.
Find the mean of the following sets of numbers :
(a) 4, 5, 7, 8
(b) 3, 5, 0, 2, 8
(c) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
(d) -6, -2, -1, 0, 1, 2, 5, 9
(e) First five prime numbers
(f) First eight even natural numbers
(g) All factors of 30
(h) First five multiples of 8
(i) x, x+1, x+2, x+3, x+4, x+5, x+6
Solution:
(a) 4, 5, 7, 8
Here n = 4
∴ Mean = \(\frac{\text { Sum }}{n}\) = \(\frac{4+5+7+8}{4}\) = \(\frac { 24 }{ 4 }\) = 6
(b) 3, 5, 0, 2, 8
Here n = 5
∴ Mean = \(\frac{3+5+0+2+8}{5}\) = \(\frac { 18 }{ 5 }\) = 3.6
(c) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
Here n = 8
∴ Mean = \(\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}\) = \(\frac { 21.0 }{ 7 }\) = 3.0
(d) -6,-2,-1,0, 1,2, 5, 9
Here n = 8
∴ Mean = \(\frac{-6-2-1+0+1+2+5+9}{8}\) = \(\frac{-9+17}{8}\) = \(\frac{8}{8}\) = 1
(e) First 5 prime numbers are 2, 3, 5,7, 11
∴ Mean = \(\frac{2+3+5+7+11}{8}\) = \(\frac{28}{5}\) = 5.6
(f) First 8 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16
∴ Mean = \(\frac{2+4+6+8+10+12+14+16}{8}\) = \(\frac{72}{8}\) = 9
(g) All factors of 30 are 1, 2. 3, 5, 6, 10, 15, 30
Here n = 8
∴ Mean = \(\frac{1+2+3+5+6+10+15+30}{8}\) = \(\frac{72}{8}\) = 9
(h) First 5 multiples of 8 are 8, 16, 24, 32, 40
∴ Mean = \(\frac{8+16+24+32+40}{5}\) = \(\frac{120}{5}\) = 24
(i) x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6
Here n = 7
Mean = \(\frac{x+x+1+x+2+x+3+x+4+x+5+x+6}{7}\) = \(\frac{7 x+21}{7}\) = \(\frac{7(x+3)}{7}\) = x + 3
Question 3.
The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
Solution:
6, y, 7, x, 14
Here n = 5 and Mean = 8
Total sum = \(n \bar{x}\) = 5 × 8 = 40
∴ 6 + y + 7 + x + 14 = 40 ⇒ 7 + x + 27 = 40 ⇒ 7 = 40 – 27 – x =13 – x
Question 4.
Nisha secured 73, 86, 78 and 75 marks in four tests. What is the least number of points she should secure in her next test if she has to have an average of 80?
Solution:
Number of total tests = 5
Average of 5 tests = 80
∴ Total points, Nisha secured = 5 × 80 = 400
But her score in 4 tests = 73 + 86 + 78 + 75 = 312
∴ Score in fifth test = 400 – 312 = 88
Question 5.
A class of 10 students was given a test in Mathematics. The marks, out of 50, secured by the students were as follows :
31, 36, 27, 38, 45, 39, 32, 29, 41, 38 Find the mean score.
Solution:
Here n = 10
and scores are 31, 36, 27, 38, 45, 39, 32, 29, 41, 38
∴ Sum = 31 + 36 + 27 + 38 + 45 + 39 + 32 + 29+ 41 + 38 = 356
∴ Mean = \(\frac{356}{10}\) = 35.6
Question 6.
Find the mean of the following frequency distributions:
(a)
| Weight (kg) | 30 | 31 | 32 | 33 | 34 |
| No. of students | 8 | 10 | 15 | 8 | 9 |
(b)
| Marks | 20 | 25 | 30 | 35 | 40 |
| No. of students | 5 | 10 | 12 | 8 | 5 |
(c)
| X | 2 | 5 | 7 | 8 |
| y | 2 | 4 | 6 | 3 |
(d)
| X | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
| f | 30 | 60 | 20 | 40 | 10 | 50 |
Solution:
(a)
| Weight (in kg) (x) | No. of students (f) | f × x |
| 30
31 32 33 34 |
8
10 15 8 9 |
240
310 480 264 306 |
| Total | 50 | 1600 |
Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1600 }{ 50 }\) = 32 kg
(b)
| Marks (x) | No. of students (f) | f × x |
| 20 | 5 | 100 |
| 25 | 10 | 250 |
| 30 | 12 | 360 |
| 35 | 8 | 280 |
| 40 | 5 | 200 |
| Total | 40 | 1190 |
∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1190 }{ 40 }\) = 29.75
(c)
| x | f | f × x |
| 2 | 2 | 4 |
| 5 | 4 | 20 |
| 7 | 6 | 42 |
| 8 | 3 | 24 |
| Total | 15 | 90 |
∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 90 }{ 15 }\) = 6
(d)
| x | f | f × x |
| 0.1
0.2 0.3 0.4 0.5 0.6 |
30
60 20 40 10 50 |
3
12 6 16 5 30 |
| Total | 210 | 72 |
∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 72 }{ 210 }\) = \(\frac { 12 }{ 35 }\) = 0.34
Question 7.
Fill in the blanks :
While calculating the mean of the grouped data, we make the assumption that frequency in any class is centred at its
Solution:
Mean of grouped data in any class is called its class mark.
Question 8.
The frequency distribution of marks obtained by 40 students of a class is as under :
Calculate the Arithmetic mean.
| Marks | 0-8 | 8-16 | 16-24 | 24-32 | 32-40 | 40-48 |
| Students | 5 | 3 | 10 | 16 | 4 | 2 |
Solution:
| Marks | Class Mark x | Frequency (f) | f × x |
| 0-8
8-16 16-24 24-32 32-40 40-48 |
4
12 20 28 36 44 |
5
3 10 16 4 2 |
20
36 200 448 144 88 |
| Total | 40 | 936 |
Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 936 }{ 40 }\) = 23.4 marks
Question 9.
Find the mean of the following data:
(a)
| Marks | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 |
| No. of students | 4 | 6 | 12 | 5 | 3 |
(b)
| Class | 0-10 | 11-20 | 21-30 | 31-40 | 41-50 |
| Frequency | 3 | 4 | 2 | 5 | 6 |
Solution:
(a)
| Marks | Class Mark x |
Frequency (f) | f × x |
| 10-14
15-19 20-24 25-29 30-34 |
12
17 22 27 32 |
4
6 12 5 3 |
48
102 264 135 96 |
| Total | 30 | 645 |
Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 645 }{ 30 }\) = 21.5 marks
(b)
| Class | Class Mark x |
Frequency (f) | f × x |
| 0-10
11-20 21-30 31-40 41-50 |
5
15.5 25.5 35.5 45.5 |
3
4 2 5 6 |
15.0
62.0 51.0 177.5 273.0 |
| Total | 20 | 578.5 |
Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 578.5 }{ 20 }\) = 28.925
Question 10.
In a class of 60 boys the marks obtained in a monthly test were as under :
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Students | 10 | 25 | 12 | 08 | 05 |
Find the mean marks of the class.
Solution:
| Marks | Class Mark (x) | Frequency (f) | f × x |
| 10-20 | 15 | 10 | 150 |
| 20-30 | 25 | 25 | 625 |
| 30-40 | 35 | 12 | 420 |
| 40-50 | 45 | 08 | 360 |
| 50-60 | 55 | 05 | 275 |
| Total | 60 | 1830 |
Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1830 }{ 60 }\) = 30.5
Question 11.
Compute the mean of the following frequency table by:
(i) direct-method and
(ii) short-cut method
| Class | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 |
| Frequency | 10 | 6 | 4 | 12 | 8 | 4 | 2 | 1 | 3 |
Solution:
| Class | Class Marks (x) |
Frequency (f) | f × x | A = 27.5 d = x – A |
f × d |
| 5-10
10-15 15-20 20-25 25-30 30-35 35-40 40-45 45-50 |
7.5
12.5 17.5 22.5 A=27.5 32.5 37.5 42.5 47.5 |
10
6 4 12 8 4 2 1 3 |
75.0
75.0 70.0 270.0 220.0 130.0 75.0 42.5 142.5 |
-20
-15 -10 -5 0 5 10 15 20 |
-200
-90 -40 -60 0 20 20 15 60 |
| Total | 50 | 1100.0 | -275 |
(i) Direct method — Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1100 }{ 50 }\) = 22
(ii) Short-cut method— Mean = A + \(\frac{\sum f d}{\sum f}\) = 27.5 + \(\frac{-275}{50}\) = 27.5 – 5.5 = 22
Question 12.
The following table gives the classification of 100 cows of a dairy farm, according to the amount of milk given by each in a dairy.
| Amount of milk in | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 | 14-16 | 16-18 |
| No. of cows | 4 | 14 | 17 | 20 | 10 | 13 | 12 | 10 | 10 |
Calculate the mean correct to first place of decimal.
Solution:
| Amount of milk in kg | Class Mark (x) | No. of Cows (f) | f × x |
| 0-2 | 1 | 4 | 4 |
| 2-4 | 3 | 14 | 42 |
| 4-6 | 5 | 17 | 85 |
| 6-8 | 7 | 20 | 140 |
| 8-10 | 9 | 10 | 90 |
| 10-12 | 11 | 13 | 143 |
| 12-14 | 13 | 12 | 156 |
| 14-16 | 15 | 10 | 150 |
| 16-18 | 17 | 10 | 170 |
| Total | 110 | 980 |
∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{980}{110}\) = 8.9 kg
Question 13.
The weights (in grams) of 50 apples picked out at random from a consignment are given below : 82, 118, 80,110, 104, 84, 106,107, 76, 82,109,107,115, 93,187, 95,123,125, 111, 92, 86, 70,126, 78,130, 129,139,119,115,128,100,186,84, 99, 113,204, 111, 141,136,123, 90, 115, 98, 110, 78, 90, 107, 81, 131, 75.
(i) What is the range of the weights ?
(ii) Form a frequency distribution with class intervals 70-89, 90-109, and so on.
(iii) Use your frequency distribution to calculate the mean.
Solution:
(i) Maximum weight = 204 g
Minimum weight = 70 g
∴ Range = 204 – 70 = 134 g
(ii) Now forming the frequency distribution table, taking class intervals such as 70-89, 90-109, so on
| Class interval | Class Marks (x) |
Frequency (f) |
A = 139.5 d = x – A |
f × d |
| 70-89
90-109 110-129 130-149 150-169 170-189 190-209 |
79.5
99.5 119.5 A = 139.5 159.5 179.5 199.5 |
12
14 16 5 0 2 1 |
-60
-40 -20 0 20 40 60 |
-720
-560 -320 0 0 80 60 |
| Total | Class Marks (x) | 50 | -1460 |
Let A = 139.5
(iii) Mean = A + \(\frac{\sum f d}{\sum f}\) = 139.5 + \(\frac{-1460}{50}\) = 139.5 – 29.2 = 110.3
Question 14.
The following table gives the weekly wages of workers in a factory.
| Weekly wages (in Rs.) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 | 80-85 | 85-90 |
| No. of workers | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
Calculate :
(i) the mean,
(ii) the number of workers getting weekly wages below Rs. 80, and
(iii) the number of workers getting Rs. 65 and more but less than Rs. 85 as weekly wages.
Solution:
| Weekly wages | Class Marks (x) |
Frequency (f) |
A=67.5 d=x – A |
f × d |
| 50-55 | 52.5 | 5 | -15 | -75 |
| 55-60 | 57.5 | 20 | -10 | -200 |
| 60-65 | 62.5 | 10 | -5 | -50 |
| 65-70 | A = 67.5 | 10 | 0 | 0 |
| 70-75 | 72.5 | 9 | 5 | 45 |
| 75-80 | 77.5 | 6 | 10 | 60 |
| 80-85 | 82.5 | 12 | 15 | 180 |
| 85-90 | 87.5 | 8 | 20 | 160 |
| Total | 80 | 120 |
(i) Mean = A + \(\frac{\sum f d}{\sum f}\) = 67.5 + \(\frac{120}{80}\) = 67.5 + 1.5 = 69
(ii) Number of workers getting wages below Rs: 80 = 60
(iii) Number of workers getting more than Rs. 65 but less them 85 = 10 + 9 + 6 + 12 = 37
Question 15.
Find the mean of the following data:
| Marks obtained | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 |
| No. of students | 7 | 19 | 32 | 42 | 50 |
Solution:
| Marks obtained | Class Mark no. (x) | No. of workers (c.f.) | (f) | fx |
| 0-10 | 5 | 7 | 7 | 35 |
| 10-20 | 15 | 19 | 12 | 180 |
| 20-30 | 25 | 32 | 13 | 325 |
| 30-40 | 35 | 42 | 10 | 350 |
| 40-50 | 45 | 50 | 8 | 360 |
| Total | 50 | 1250 |
Mean = \(\frac{\sum f x}{\sum f^{\prime}}\) = \(\frac{1250}{50}\) = 25
Question 16.
Recast the following cumulative table into the form of an ordinary frequency distribution and determine the value of the mean :
Solution:
| Marks | Class Marks | c.f | f | A=27.5 d = x – A |
f × d |
| 0-5 | 2.5 | 14 | 14 | -25 | -350 |
| 5-10 | 7.5 | 25 | 11 | -20 | -220 |
| 10-15 | 12.5 | 37 | 12 | -15 | -180 |
| 15-20 | 17.5 | 56 | 19 | -10 | -190 |
| 20-25 | 22.5 | 94 | 38 | -5 | -190 |
| 25-30 | A = 27.5 | 112 | 18 | 0 | 0 |
| 30-35 | 32.5 | 130 | 18 | 5 | 90 |
| 35-40 | 37.5 | 144 | 14 | 10 | 140 |
| 40-45 | 42.5 | 148 | 4 | 15 | 60 |
| 45-50 | 47.5 | 150 | 2 | 20 | 40 |
| Total | 150 | -800 |
Mean = A + \(\frac{\sum f d}{\sum f}\) = 27.5 + \(\frac{-800}{150}\) = 27.5 – 5.3 = 22.2
Question 17.
Find the missing frequency in the following data if arithmetic mean is 19.92.
| Class | 4-8 | 8-12 | 12-16 | 16-20 | 20-24 | 24-28 | 28-32 | 32-36 | 36-40 |
| Frequency | 11 | 13 | 16 | 14 | – | 9 | 17 | 6 | 4 |
Solution:
Let the missing frequency be p, then
| Class | Class Marks (x) | Frequency (f) | f × x |
| 4-8 | 6 | 11 | 66 |
| 8-12 | 10 | 13 | 130 |
| 12-16 | 14 | 16 | 224 |
| 16-20 | 18 | 14 | 252 |
| 20-24 | 22 | p | 22 p |
| 24-28 | 26 | 9 | 234 |
| 28-32 | 30 | 17 | 510 |
| 32-36 | 34 | 6 | 204 |
| 36-40 | 38 | 4 | 152 |
| Total | 90+p | 1772+22 p |
Mean is given = 19.92
∴ Mean = \(\frac{\sum f x}{\sum f}\) ⇒ 19.92 = \(\frac{1772+22 p}{90+p}\) ⇒ 1772 + 22p = 1792.8 + 19.92p
22p = 19.92p = 1792.8 – 1772
2.08p = 20.8
∴ p = \(\frac{20.8}{2.08}\) = \(\frac{208 \times 100}{10 \times 208}\) = 10
∴ Missing Frequency = 10
Question 18.
Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution of marks obtained in an Arithmetic test.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| No. of students | 2 | 5 | 20 | 8 | 7 |
Solution:
| Marks | Class Marks (x) | Frequency (f) | f × x |
| 0-10 | 5 | 2 | 10 |
| 10-20 | 15 | 5 | 75 |
| 20-30 | 25 | 20 | 500 |
| 30-40 | 35 | 8 | 280 |
| 40-50 | 45 | 7 | 315 |
| Total | 42 | 1180 |
Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{1180}{42}\) = 28.095 = 28.1 (approx)
Question 19.
The following table gives the marks scored by students in the examination:
| Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
| No. of students | 3 | 7 | 15 | 24 | 16 | 8 | 5 | 2 |
Calculate the mean mark, correct to two decimal places.
Solution:
| Marks (Class) | Class Marks (x) | Frequency (f) | Let A = 17.5 d = x – A |
f × d |
| 0-5
5-10 10-15 15-20 20-25 25-30 30-35 35-40 |
2.5
7.5 12.5 A = 17.5 22.5 27.5 32.5 37.5 |
3
7 15 24 16 8 5 2 |
-15
-10 -5 0 5 10 15 20 |
-45
-70 -75 0 80 80 75 40 |
| Total | 80 | 85 |
A = 17.5
Mean = A + \(\frac{\sum f d}{\sum f}\) = 17.5 + \(\frac { 85 }{ 80 }\) = 17.5 + 1.06 = 18.56