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#### 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 ₹): 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
$\overline{)X}=\frac{\Sigma X}{N}=\frac{420}{10}=42\phantom{\rule{0ex}{0ex}}$
Thus, mean marks is 42

(b)
 X 550 490 670 890 435 590 575 ΣX = 4200

$\overline{)\mathrm{X}}=\frac{\Sigma X}{N}=\frac{4200}{7}=600$
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

$\overline{)X}=\frac{\Sigma X}{N}=\frac{1184}{8}=148$
Thus, mean height is equal to 148 cm.

#### Question 2:

Batsman A, B , C and D  played three matches . Calculate the average runs scored by each batsman.

 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

${\overline{)X}}_{A}=\frac{\Sigma {X}_{\mathrm{A}}}{N}=\frac{256}{6}=42.67\phantom{\rule{0ex}{0ex}}{\overline{)\mathit{X}}}_{\mathrm{B}}=\frac{\Sigma {X}_{\mathrm{B}}}{N}=\frac{336}{6}=56\phantom{\rule{0ex}{0ex}}{\overline{)X}}_{\mathrm{C}}=\frac{\Sigma {X}_{\mathrm{C}}}{N}=\frac{376}{6}=62.67\phantom{\rule{0ex}{0ex}}{\overline{)X}}_{\mathrm{D}}=\frac{\Sigma {X}_{\mathrm{D}}}{N}=\frac{284}{6}=47.33$

#### 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

$\overline{)X}=\frac{\mathrm{\Sigma }fX}{\mathrm{\Sigma }f}=\frac{706}{100}=7.06$

Thus, mean of the series is 7.06.

#### 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 (₹ in '000) : 50 50 55 60 65 65 65 60 60 50 Expenses (₹ 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

$\overline{)\mathit{X}}=\frac{\Sigma X}{N}=\frac{580}{10}=58$
So, mean sales is Rs 58,000

$\overline{)Y}=\frac{\Sigma Y}{N}=\frac{140}{10}=14$
So, mean expenses is Rs 14,000

#### 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

$\overline{)\mathit{X}}=\frac{\Sigma fX}{\Sigma f}=\frac{75980}{1071}=70.94$
Thus, the mean for the given data is Rs 70.94.

#### 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

$\overline{)X}=\frac{\mathrm{\Sigma }fX}{\mathrm{\Sigma }f}=\frac{550}{50}=11$
Thus, the mean is 11.

#### Question 7:

Calculate arithmetic mean of the following data:

 Profit ( in ₹) : 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

$\overline{)\mathit{X}}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{3350}{110}=30.45$
Thus, the mean profit is Rs 30.45.

#### 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.

$\overline{)\mathit{X}}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{3000}{100}=30$
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

#### 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.

#### 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 ₹ 275 and ₹ 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

#### 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.

#### 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.

#### 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

#### 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.

$\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{2900}{70}=41.42$
Thus, the mean marks is 41.42.

#### 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.

#### 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.

#### Question 17:

The mean salary paid to 1000 workers of a factory was found to be ₹ 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 ΣXCXI + correct values - incorrect values
= ΣXI + 197 + 185 − 297 − 165
= 180400 + 197 + 185 − 297 − 165
= 180320

Thus, the correct mean wage is Rs 180.32.

#### 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 $\Sigma f$ = 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

$\overline{)\mathit{X}}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{1565}{365}=4.29°\mathrm{C}$

Thus, the mean temperature us 4.29oC.

#### 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.

#### 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

${\overline{)\mathit{X}}}_{\mathit{W}}=\frac{\Sigma WX}{\Sigma W}=\frac{4987.5}{785}=6.35$
Thus, the weighted mean is Rs 6.35.

#### 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

${\overline{)X}}_{{W}_{\mathit{1}}}=\frac{\Sigma {X}_{1}{W}_{1}}{\Sigma {W}_{1}}=\frac{1451}{20}=72.55\phantom{\rule{0ex}{0ex}}{\overline{)X}}_{{W}_{\mathit{2}}}=\frac{\Sigma {X}_{2}{W}_{2}}{\Sigma {W}_{2}}=\frac{1977}{28}=70.6\phantom{\rule{0ex}{0ex}}{\overline{)X}}_{{W}_{\mathit{3}}}=\frac{\Sigma {X}_{3}{W}_{3}}{\Sigma {W}_{3}}=\frac{1513}{21}=72.04$
Average score of Mumbai is more than Kolkata and Chennai. So performance of Mumbai is better.

#### 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.

#### 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

$\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{15000}{500}=30$
Thus, the mean is 30 years.

#### 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

$\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{3000}{100}=30$
Thus, the mean is 30.

#### Question 25:

Find the missing value , if mean is 23.

 X 5 15 ? 35 45 Frequency 4 8 10 6 2

#### Answer:

 X Frequency (f) fX 5 15 ? (x) 35 45 4 8 10 6 2 20 120 10x 210 90 Σf = 30 ΣfX = 440+10x

$\overline{)X}=23\phantom{\rule{0ex}{0ex}}\overline{)X}=\frac{\mathrm{\Sigma }fX}{\mathrm{\Sigma }f}\phantom{\rule{0ex}{0ex}}23=\frac{440+10x}{30}\phantom{\rule{0ex}{0ex}}690=440+10x\phantom{\rule{0ex}{0ex}}250=10x\phantom{\rule{0ex}{0ex}}x=\frac{250}{10}=25$
Thus, the missing value of X is 25.

#### Question 26:

Find the missing frequency, if mean is 62.

 C.I 0-40 40-80 80-120 120-160 Frequency 6 9 ? 2

#### Answer:

 C.I. Mid-Point (m) Frequency (f) fm 0 − 40 40 − 80 80 − 120 120 − 160 20 60 100 140 6 9 ? (f) 2 120 540 100f 280 Σf = 17+f Σfm = 940+100f

$\overline{)X}=62\phantom{\rule{0ex}{0ex}}\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}\phantom{\rule{0ex}{0ex}}62=\frac{940+100f}{17+f}\phantom{\rule{0ex}{0ex}}1054+62f=940+100f\phantom{\rule{0ex}{0ex}}38f=114\phantom{\rule{0ex}{0ex}}f=\frac{114}{38}=3$
Thus, the missing freuency is 3.

#### Question 27:

Calculate arithmetic mean:

 Marks (More than) 0 10 20 30 40 50 No. of Students 100 88 70 43 23 6

#### Answer:

 Marks Mid Value (m) f fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 5 15 25 35 45 55 100 − 88 = 12 88 − 70 = 18 70 − 43 = 27 43 − 23 = 20 23 − 6 = 17 6 60 270 675 700 765 330 Σf = 100 Σfm = 2800

For the calculation of mean, first the given more than series is converted in continuous series as above.

$\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{2800}{100}=28$
Thus, the mean marks is 28.

#### Question 28:

Calculate mean of the following data:

 Values below 10 20 20 40 50 Frequency 20 44 84 120 140

#### Answer:

 Marks Mid Value (m) f fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 5 15 25 35 45 20 44 − 20 = 24 84 − 44 = 40 120 − 84 = 36 140 − 120 = 20 100 360 1000 1260 900 Σf = 140 Σfm = 3620

For the calculation of mean, first the given less than series is converted in continuous series as above.

$\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{3620}{140}=25.86$
Thus, the mean is 25.86.

#### Question 29:

Calculate missing frequency, if mean of the distribution is 28.

 C.I 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 12 ? 27 20 17 6

#### Answer:

 C.I. Mid-Point (m) Frequency (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 5 15 25 35 45 55 12 ? (f) 27 20 17 6 60 15f 675 700 765 330 Σf = 82+f Σfm = 2530+15f

$\overline{)X}=28\phantom{\rule{0ex}{0ex}}\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}\phantom{\rule{0ex}{0ex}}28=\frac{2530+15f}{82+f}\phantom{\rule{0ex}{0ex}}2296+28f=2530+15f\phantom{\rule{0ex}{0ex}}13f=234\phantom{\rule{0ex}{0ex}}f=\frac{234}{13}=18$
Thus, the missing freuency is 18.

#### Question 30:

Calculate mean from the information given below.

 C.I 5-6 8-10 11-13 14-16 17-19 Frequency 10 8 7 3 2

#### Answer:

 C.I. Mid-Point (m) Frequency (f) fm 4.5 − 7.5 7.5 − 10.5 10.5 − 13.5 13.5 − 16.0 16.5 − 19.50 6 9 12 15 18 10 8 7 3 2 60 72 84 45 36 Σf = 30 Σfm = 297

$\overline{)X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}\phantom{\rule{0ex}{0ex}}\overline{)X}=\frac{297}{30}=9.9$
Thus, the mean is 9.9.

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