NCERT Class 11 Economics Chapter 5 Measures of Central Tendency

NCERT Class 11 Economics Chapter 5 Measures of Central Tendency Solutions to each chapter is provided in the list so that you can easily browse through different chapters NCERT Class 11 Economics Chapter 5 Measures of Central Tendency Question Answer and select need one. NCERT Class 11 Economics Chapter 5 Measures of Central Tendency Textual Solutions Download PDF. CBSE Class 11 Statistics For Economics Textbook Solutions.

NCERT Class 11 Economics Chapter 5 Measures of Central Tendency

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 11 Economics Chapter 5 Measures of Central Tendency Solutions are part of All Subject Solutions. Here we have given NCERT Class 11 Economics Textual Question and Answer, CBSE Solutions For Class 11 Economics Solutions for All Chapters, You can practice these here.

Chapter: 5

PART – (A) STATISTICS FOR ECONOMICS

TEXTUAL QUESTION ANSWERS

1. Which average would be suitable in the following cases?

(a) Average size of readymade garments.

Ans: Mode.

(b) Average intelligence of students in a class.

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Ans: Median.

(c) Average production in a factory per shift.

Ans: Arithmetic Mean.

(d) Average wage in an industrial concern.

Ans: (d) Arithmetic Mean.

(e) When the sum of absolute deviations from average is least.

Ans: Arithmetic Mean.

(f) When quantities of the variable are in ratios.

Ans: Geometric mean.

(g) In case of open-ended frequency distribution.

Ans: Median or Mode.

2. Indicate the most appropriate alternative from the multiple choices provided against each question.

(a) The most suitable average for qualitative measurement is:

(i) Arithmetic means.

(ii) Median.

(iii) Mode.

(iv) Geometric mean.

(v) None of the above.

Ans: (ii) Median.

(b) Which average is affected most by the presence of extreme items?

(i) Median.

(ii) Mode.

(iii) Arithmetic mean.

(iv) None of the above.

Ans: (iii) Arithmetic mean.

(c) The algebraic sum of deviation of a set of n values from A.M. is:

(i) n

(ii) 0

(iii) 1

(iv) None of the above.

Ans: (ii) 0

3. Comment whether the following statements are true or false.

(a) The sum of deviation of items from median is zero.

Ans: False.

(b) An average alone is not enough to compare series.

Ans: True.

(c) Arithmetic mean is a positional value.

Ans: False.

(d) Upper quartile is the lowest value of the top 25% of items.

Ans: True.

(e) Median is unduly affected by extreme observations.

Ans: False.

4. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:

Profit per retail shop (in ₹)0-1010-2020-3030-4040-5050-60
Number of retail shops121827176

Ans: (a) Let the missing frequency be f₁. 

Arithmetic Mean = 28

Profit Per Retail Shop (in₹) Class IntervalNumber of Retail Shops (f)Mid Value (m)fm
0-1012560
10-201815270
20-302725675
30-40f₁3535f₁
40-501745765
50-60655330
∑f = 80 + f₁∑fm = 2100 + 35 f₁

2240 + 28f₁ = 2100 + 35f₁

2240 – 2100 = 35f₁ = 28f₁

140 = 7f₁

f₁ = 20

Hence, the missing frequency is 20.

(b) 

Class IntervalFrequencyCumulative Frequency
0-101212
10-201830
20-302757
30-402077
40-501794
50-6061000
∑f = 100

50th item lies in the 57th cumulative frequency and the corresponding class interval is 20-30.

5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

WorkersABCDEFGHIJ
Daily Income (in ₹)120150180200250300220350370260

Ans:

WorkersDaily Income (in ₹) (x)
A120
B150
C180
D200
E250
F300
G220
H350
I370
J260
∑X = 2,400

N = 10

∴ Arithmetic Mean = ₹ 240

6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Income (in₹)Number of families
More than 75150
More than 85140
More than 95115
More than 10595
More than 11570
More than 12560
More than 13540
More than 14525

Ans: 

Income Class IntervalNumber of Families (c.f.)Frequency (f)Mid Value (m)fm
75 – 85150150 – 140 = 1080800
85 – 95140140 – 115 = 25902,250
95 – 105115115 – 95 = 201002,000
105 – 1159595 – 70 = 251102,750
115 – 1257070 – 60 = 101201,200
125 – 1356060 – 40 = 201302,600
135 – 1454040 – 25 = 151402,100
145 – 15525251503,750
Σf = 150Σfm = 17,450

7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Size of Land Holdings (in acres)Number of families
Less than 10040
100-20089
200-300148
300-40064
400 and above39

Ans: 

Size of Land Holdings Class IntervalNumber of Families (f)Cumulative Frequency (c.f.)
0 – 1004040
100 – 20089129
200 – 300148277
300 – 40064341
400 – 50039380
Σf = 380

190ᵗʰ item lies in The 129ᵗʰcumulative frequency and the corresponding class interval is 200-300.

= 200 + 41.22 = 241.22

Median size of land holdings = 241.22 acres.

8. The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers.

Daily Income (in₹)10-1415-1920-2425-2930-3435-39
Number of workers5101520105

Ans: 

Daily Income (in) Class IntervalNumber of Workers(f)Cumulative Frequency (c.f.)
9.5 – 14.555
14.5 – 19.51015
19.5 – 24.51530
24.5 – 29.52050
29.5 – 34.51060
34.5 – 39.5565
Σf = 65

(a) Highest income of lowest 50% workers will be given by the Median. Σf = N = 65

32.5ᵗʰ item he’s in the 50% cumulative frequency and the corresponding class interval is 24.5-29.5.

= ₹ 25.13

(b) Minimum income earned by top 25% workers will be given by the lower quartile Q₁.

= 16.25ᵗʰ item

16.25ᵗʰ item lies in the 30ᵗʰ cumulative frequency and the corresponding class interval is 19.5-24.5

=  ₹19.92

(c) Maximum income earned by lowest 25% workers will be given by the upper quartile Q₃.

= 3 × 16.25ᵗʰ item = 48.75ᵗʰ item

48.75ᵗʰ item lines in 50ᵗʰ item and the corresponding class interval is 24.5-29.5

=  ₹ 29.19

9. The following table gives production yield in kg per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.

Production yield (kg per hectare)Number of farms
50-533
53-568
56-5914
59-6230
62-6536
65-6828
68-7116
71-7410
74-775

Ans: 

Production yield (kg per hectare)fc.f.M.V.(x)dxdx’fdx’
50-533351.5-12-4-12
53-5681154.5-9-3-24
56-59142557.5-6-2-28
59-62305560.5-3-1-30
62-65369163.5000
65-682811966.53128
68-711613569.56232
71-741014572.59330
74-77515075.512420
Σf = 150Σfdx’ = 16

By interpretation:

By inspection we find that mode lies in the class 62-65, because this class has maximum frequency, i.e., 36.

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