NM Shah (2016) Solutions for Class 11 Humanities Economics Chapter 5 Measures Of Central Tendency are provided here with simple step-by-step explanations. These solutions for Measures Of Central Tendency are extremely popular among class 11 Humanities students for Economics Measures Of Central Tendency Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NM Shah (2016) Book of class 11 Humanities Economics Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NM Shah (2016) Solutions. All NM Shah (2016) Solutions for class 11 Humanities Economics are prepared by experts and are 100% accurate.
Page No 162:
Question 1:
Calculate arithmetic averages of the following information:
(a) Marks obtained by 10 students:
30, 62, 47, 25, 52, 39, 56, 66, 12, 31
(b) Income of 7 families (in â¹): Also show
550, 490, 670, 890, 435, 590, 575
(c) Height of 8 students ( in cm) :
140, 145, 147, 152, 148, 144, 150, 158
Answer:
(a)
X | 30 | 62 | 47 | 25 | 52 | 39 | 56 | 66 | 12 | 31 |
ΣX = 420
Thus, mean marks is 42
(b)
X |
550 490 670 890 435 590 575 |
ΣX = 4200 |
Thus, mean income is equal to Rs 600.
X | |
550 490 670 890 435 590 575 |
-50 -110 70 290 -165 -10 -25 |
(c)
S.No. | X |
1 2 3 4 5 6 7 8 |
140 145 147 152 148 144 150 158 |
ΣX = 1184 |
Thus, mean height is equal to 148 cm.
Page No 162:
Question 2:
Batsman A, B , C and D played three matches . Calculate the average runs scored by each batsman.
â
Name of batsman | Match I | Match II | Match III | |||
I | II | I | II | I | II | |
Inning | Inning | Inning | Inning | Inning | Inning | |
A | 60 | 20 | 26 | 10 | 100 | 40 |
B | 40 | 50 | 60 | 36 | 70 | 80 |
C | 100 | 10 | 8 | 18 | 100 | 140 |
D | 20 | 40 | 46 | 84 | 42 | 52 |
Answer:
The data given for the marks scored by the batsmen in different innings can be summarised as follows.
Inning | Score by batsman A (XA) |
Score by batsman B (XB) |
Score by batsman C (XC) |
Score by batsman D (XD) |
1 2 3 4 5 6 |
60 20 26 10 100 40 |
40 50 60 36 70 80 |
100 10 8 18 100 140 |
20 40 46 84 42 52 |
ΣXA = 256 | ΣXB = 336 | ΣXC = 376 | ΣXD = 284 |
Page No 162:
Question 3:
Calculate mean of the following series:
Size | : | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Frequency | : | 6 | 12 | 15 | 28 | 20 | 14 | 5 |
Answer:
X | f | fX |
4 5 6 7 8 9 10 |
6 12 15 28 20 14 5 |
24 60 90 196 160 126 50 |
Σf = 100 | ΣfX = 706 |
Thus, mean of the series is 7.06.
Page No 162:
Question 4:
Find out mean of sales and expenses of following 10 firms:
Firms | : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Sales (â¹ in '000) | : | 50 | 50 | 55 | 60 | 65 | 65 | 65 | 60 | 60 | 50 |
Expenses (â¹ in '000) | : | 11 | 13 | 14 | 16 | 16 | 15 | 15 | 14 | 13 | 13 |
Answer:
Firm | Sales (X) (Rs in '000) |
Expenses (Y) (Rs in '000) |
1 2 3 4 5 6 7 8 9 10 |
50 50 55 60 65 65 65 60 60 50 |
11 13 14 16 16 15 15 14 13 13 |
ΣX = 580 | ΣY = 140 |
So, mean sales is Rs 58,000
So, mean expenses is Rs 14,000
Page No 162:
Question 5:
Calculate mean of the following frequency distribution:
Values | : | 60 | 62 | 64 | 67 | 70 | 73 | 77 | 81 | 85 | 89 |
Frequency | : | 54 | 82 | 103 | 176 | 212 | 180 | 115 | 78 | 50 | 21 |
Answer:
X | Frequency (f) |
fX |
60 62 64 67 70 73 77 81 85 89 |
54 82 103 176 212 180 115 78 50 21 |
3240 5084 6592 11792 14840 13140 8855 6318 4250 1869 |
Σf = 1071 | ΣfX = 75980 |
Thus, the mean for the given data is Rs 70.94.
Page No 163:
Question 6:
Calculate average of the following series:
X | : | 4 | 6 | 10 | 14 | 18 | 22 |
f | : | 7 | 9 | 16 | 8 | 6 | 4 |
Answer:
X | f | fX |
4 6 10 14 18 22 |
7 9 16 8 6 4 |
28 54 160 112 108 88 |
Σf = 50 | ΣfX = 550 |
Thus, the mean is 11.
Page No 163:
Question 7:
Calculate arithmetic mean of the following data:
Profit ( in â¹) | : | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
No of shops | : | 12 | 18 | 27 | 20 | 17 | 16 |
Answer:
Profit (in Rs) |
Mid Value (m) |
No. of Shops Frequency (f) |
fm |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 |
5 15 25 35 45 55 |
12 18 27 20 17 16 |
60 270 675 700 765 880 |
Σf = 110 | Σfm = 3350 |
Thus, the mean profit is Rs 30.45.
Page No 163:
Question 8:
Calculate the arithmetic mean from the following data:
Marks | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 |
No. of Students | 5 | 15 | 55 | 75 | 100 |
Answer:
Marks | Mid Value (m) |
No. of student (f) |
fm |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 |
5 15 25 35 45 |
5 10 40 20 25 |
25 150 1000 700 1125 |
Σf = 100 | Σfm = 3000 |
For the calculation of mean first the given less than series is converted in continuous class intervals as above.
Thus, mean is 30 marks.
Marks | Mid Value (m) |
No. of student (f) |
||
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 |
5 15 25 35 45 |
5 10 40 20 25 |
-25 -15 -5 5 15 |
-125 -150 -200 100 375 |
Σf = 100 | =0 |
Page No 163:
Question 9:
A candidate obtains 46 marks in English , 67 in Mathematics , 53 in Hindi , 72 in History and 58 in Economics. It is agreed to give triple weights to marks in English and Double weights to marks in Mathematics as compared to other subjects . Calculate Weighted Mean. Also, compare it with simple Arithmetic Mean.
Answer:
Subject | Mark (%) (X) |
Weight (W) |
WX |
English Mathematics Hindi History Economics |
46 67 53 72 58 |
3 2 1 1 1 |
138 134 53 72 58 |
ΣX = 296 | ΣW = 8 | ΣWX = 455 |
The given information can be summarised as above.
Thus, the weighted arithmetic mean is greater than the simple arithmetic mean.
Page No 163:
Question 10:
There are two branches of an establishment employing 100 and 80 persons respectively. If the Arithmetic Means of the monthly salaries by the two branches are â¹ 275 and â¹ 225 respectively, find out the arithmetic mean of the salaries of the employees of the establishment as a whole.
Answer:
The given information can be summarised as follows.
Thus, the combined mean salary is Rs 252.8
Page No 163:
Question 11:
The mean marks obtained in an examination by a group of 100 students were found to be 49.46. The mean marks obtained in the same examination by another group of 200 students were 52.32. Find out the mean of marks obtained by both the groups of students taken together.
Answer:
The given information can be summarised as follows.
Thus, the mean marks of both the groups taken together is 51.37.
Page No 163:
Question 12:
The mean marks of 100 students were found to be 40 . Later on it was discovered that a score of 53 was misread as 83. Find the corrected mean corresponding to the corrected score .
Answer:
Given:
Incorrect Mean = 40
Number of students = 100
Correct value = 53
Incorrect value = 83
Thus, the corrected mean marks is 39.7.
Page No 164:
Question 13:
The mean weight of 25 boys in group A of a class is 61 Kg and the mean weight of 35 boys in group B of the same class is 58 Kg. Find the mean weight of 60 boys.
Answer:
The given information can be summarised as follows:
Thus, the combined mean weight is 59.25 kg
Page No 164:
Question 14:
Calculate mean of the following data:
Marks Below | : | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
No. of Students | : | 5 | 9 | 17 | 29 | 45 | 60 | 70 |
Answer:
Marks | Mid Value (m) |
f | fm |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 |
5 15 25 35 45 55 65 |
5 9 − 5 = 4 17 − 9 = 8 29 − 17 = 12 45 − 29 = 16 60 − 45 = 15 70 − 60 = 10 |
25 60 200 420 720 825 650 |
Σf = 70 | Σfm = 2900 |
For the calculation of mean, first the given less than series is converted in continuous series as above.
Thus, the mean marks is 41.42.
Page No 164:
Question 15:
Calculate Combined Mean:
Section | Mean Marks | No. of Students |
A | 75 | 50 |
B | 60 | 60 |
C | 55 | 50 |
Answer:
Thus, the combined mean marks is 63.125.
Page No 164:
Question 16:
The average marks for statistics in a class of 30 were 52. The top six students had an average of 31 marks . What were the average marks of the other students?
Answer:
The given information can be summarised as follows:
Combined average marks = 52
So, average marks of the remaining 24 students is 57.25.
Page No 164:
Question 17:
The mean salary paid to 1000 workers of a factory was found to be â¹ 180.4 . Later on it was discovered that the wages of two workers were wrongly taken as 297 and 165 instead of 197 and 185. Find the correct mean.
Answer:
Given:
Number of workers = 1000
Incorrect values = 297, 165
Correct values = 197, 185
Incorrect mean = 180.4
Correct ΣXC =ΣXI + correct values - incorrect values
= ΣXI + 197 + 185 − 297 − 165
= 180400 + 197 + 185 − 297 − 165
= 180320
Thus, the correct mean wage is Rs 180.32.
Page No 164:
Question 18:
Calculate arithmetic measure from the following data:
Temp.( 0C) | No. of days (f) |
−40 to −30 | 10 |
−30 to −20 | 28 |
−20 to −10 | 30 |
−10 to 0 | 42 |
0 to 10 | 65 |
10 to 20 | 180 |
20 to 30 | 10 |
= 365 |
Answer:
Temprature | Mid Value (m) |
No. of Days (f) |
fm |
(−40) − (−30) (−30) − (−20) (−20) − (−10) (−10) − 0 0 − 10 10 − 20 20 − 30 |
−35 −25 −15 −5 5 15 25 |
10 28 30 42 65 180 10 |
−350 −700 −450 −210 325 2700 250 |
Σf = 365 | Σfm = 1565 |
Thus, the mean temperature us 4.29oC.
Page No 164:
Question 19:
A candidate obtains the following percentage of marks: Sanskrit 75, Mathematics 84, Economics 56, English 78, Politics 57, History 54, Geography 47,. It is agreed to give double weights to marks in English , Mathematics and Sanskrit . What is the weighted and simple arithmetic mean?
Answer:
Subject | Marks (X) |
Weight (W) |
WX |
Sanskrit Mathematics Economics English Politics History Geography |
75 84 56 78 57 54 47 |
2 2 1 2 1 1 1 |
150 168 56 156 57 54 47 |
ΣX = 451 | ΣW = 10 | ΣWX = 688 |
Thus, the weighted mean and simple arithmetic mean are 68.8 and 64.43 respectively.
Page No 164:
Question 20:
Calculate weighted mean by weighting each price by the quantity consumed:
Food items | Quantity Consumed | Price in Rupees (per Kg) |
Flour | 500 kg | 1.25 |
Ghee | 200 kg | 20.00 |
Sugar | 30 kg | 4.50 |
Potato | 15 kg | 0.50 |
Oil | 40 kg | 5.50 |
Answer:
Food Item | Price (RS) (X) |
Quantity Consumed (W) |
WX |
Flour Ghee Sugar Potato Oil |
1.25 20 4.50 0.50 5.50 |
500 kg 200 kg 30 kg 15 kg 40 kg |
625 4000 135 7.5 220 |
ΣW = 785 | ΣWX = 4987.5 |
Thus, the weighted mean is Rs 6.35.
Page No 164:
Question 21:
Comment on the performance of the students of three universities given below using weighted mean:
Courses of study | Mumbai | Kolkata | Chennai | |||
% pass |
No. of students | % pass |
No. of Students | % pass |
No. of Students | |
M.A | 71 | 3 | 82 | 2 | 81 | 2 |
M.COM. | 83 | 4 | 76 | 3 | 76 | 3.5 |
B.A | 73 | 5 | 73 | 6 | 74 | 4.5 |
B.COM. | 74 | 2 | 76 | 7 | 58 | 2 |
B.Sc | 65 | 3 | 65 | 3 | 70 | 7 |
M.Sc | 66 | 3 | 60 | 7 | 73 | 2 |
Answer:
Courses | Mumbai | Kolkata | Chennai | ||||||
X1 | W1 | X1W1 | X2 | W2 | X2W2 | X3 | W3 | X3W3 | |
M.A M.Com B.A B.Com B.Sc M.Sc |
71 83 73 74 65 66 |
3 4 5 2 3 3 |
213 332 365 148 195 198 |
82 76 73 76 65 60 |
2 3 6 7 3 7 |
164 228 438 532 195 420 |
81 76 74 58 70 73 |
2 3.5 4.5 2 7 2 |
162 266 333 116 490 146 |
ΣW1 = 20 | ΣX1W1 = 1451 | ΣW2 = 28 | ΣX2W2 = 1977 | ΣW3 = 21 | ΣX3W3 = 1513 |
Average score of Mumbai is more than Kolkata and Chennai. So performance of Mumbai is better.
Page No 165:
Question 22:
A distribution consists of three components with total frequencies of 200, 250 and 300 having means of 25, 10 and 15 respectively . Find out the mean of combined distribution.
Answer:
Thus, mean of combined distribution is 16.
Page No 165:
Question 23:
Find the average age of 500 people in a town.
Age (in Years) | 0-10 | 10-20 | 20-30 | 30-60 | 60-90 |
No. of people | 50 | 90 | 200 | 120 | 40 |
Answer:
Age (in years) |
Mid Value (m) |
No. of People Frequency (f) |
fm |
0 − 10 10 − 20 20 − 30 30 − 60 60 − 90 |
5 15 25 45 75 |
50 90 200 120 40 |
250 1350 5000 5400 3000 |
Σf = 500 | Σfm = 15000 |
Thus, the mean is 30 years.
Page No 165:
Question 24:
Calculate mean of the following data:
Values more than | 0 | 10 | 20 | 30 | 40 |
Frequency | 100 | 95 | 85 | 45 | 25 |
Answer:
Class Intervals |
c.f. | Mid Value (m) |
Frequency (f) |
fm |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 |
100 95 85 45 25 |
5 15 25 35 45 |
5 10 40 20 25 |
25 150 1000 700 1125 |
Σf = 100 | Σfm = 3000 |
Thus, the mean is 30.
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