NCERT Solutions Class 11 Economics Chapter 5 Measures of Central Tendency

NCERT Solution for Class 11 Statistics for chapter 5 Measures of Central Tendency

Exercises

Q1. Which average would be suitable in following cases?

(i) Average size of readymade garments.

(ii) Average intelligence of students in a class.

(iii) Average production in a factory per shift.

(iv) Average wage in an industrial concern.

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

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

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

Answer.

(i) Average size of readymade garments. Mode

Explanation: Mode is suitable average for average size of readymade garments because it gives the most frequent occurring value.

(ii) Average intelligence of students in a class. Median

Explanation: Median is a suitable average in case of a qualitative nature of the data.

(iii) Average production in a factory per shift. Mean

Explanation: Production can be measured on a quantitative scale so Arithmetic mean is suitable in this case.

(iv) Average wage in an industrial concern. Mean

Explanation: Wage can be measured on a quantitative scale so arithmetic mean is suitable in this case.

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

Explanation: Mean shall be used because sum of deviations from mean is always zero or least than the other averages.

(vi) When quantities of the variable are in ratios. Mean

Explanation: Ratios are quantitative, so it is suitable to use arithmetic mean.

(vii) In case of open-ended frequency distribution. Median

Explanation: Median is used because there is no need to adjust class size or magnitude for using median.

Q2. Indicate the most appropriate alternative from the multiple choices provided against each question.

(i) The most suitable average for qualitative measurement is

(a) Arithmetic mean

(b) Median

(c) Mode

(d) Geometric mean

(e) None of the above

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

(a) Median

(b) Mode

(c) Arithmetic mean

(d) None of the above

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

(a) N

(b) 0

(c) 1

(d) none of the above

 

Answer.

(i) The most suitable average for qualitative measurement is Median.

(ii) Arithmetic mean is the average affected by the presence of the extreme values.

(iii) 0 is the sum of deviations of a set of n values from AM.

Q3. Comment whether the following statements are true or false.

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

(ii) An average alone is not enough to compare series. (iii) Arithmetic mean is a positional value.

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

(v) Median is unduly affected by extreme observations.

Answer.

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

Explanation: Generally, sum of deviations from mean is zero; but only in the case of symmetric distribution (mean=median=mode) above statement is true.

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

Explanation: Averages are very rigid values, they don’t say anything about the variability of the series, and thus they are not enough to compare series.

(iii) Arithmetic mean is a positional value. False

Explanation: Arithmetic mean is not a positional value because it is calculated on the basis of all the observations.

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

Explanation: Quartile refers to a quarter, so when the frequency is arranged in a ascending order the upper quartile refers to the first 25% of the items.

(v) Median is unduly affected by extreme observations. False

Explanation: Median doesn’t get affected by extreme observations because it only takes the median class to calculate it. It is mean which gets affected by extreme observations.

 

Q4. 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 Rs )	0-10	10 -20	20-30	30-40	40-50	50-60
Number of retail shops	12	18	27	-	17	6

Answer.

(a) Let us take the missing frequency as x

Profit per retail shop (in Rs .) (X)	Number of retail shops (f)	Mid values (m)	fm
0-10 10-20 20-30 30-40 40-50 50-60	12 18 27  x 17 6	5 15 25 35 45 55	60 270 675 35x 765 330
	∑f = 80 + x		∑Fm = 2100 + 35x

Mean = ∑fm/∑f Mean = 28

Substituting the values in the formula we get,

28 = 2100 + 35x
            80 + x

28 x (80 + x) = 2100 + 35x

2240 + 28x = 2100 + 35x

7x = 140 x = 20

Thus, the missing value frequency is 20.

(b)

Profit per retail shop(in Rs ) (X) Number of retail shops (f) Cumulative frequency (cf) 0-10 10-20 20-30 30-40 40-50 50-60 12 18 27 20 17 6 12 30 57 77 94 100   ∑f = 100

Formula of median is as follow: Median
= L + (N/2 – c.f) x h
                f

By substituting the value in the formula we get,

Median = 20 + (50 – 30) x (30-20)
                           27

Median = 20 + (20) x (10) = 20 + 200 = 20 + 7.41 = 27.41
                        27                        27

Thus the median value of the series is 27.41.

Q5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers	A	B	C	D	E	F	G	H	I	J
Daily Income (in Rs .)	120	150	180	200	250	300	220	350	370	260

Answer.

Formula of mean is as follow:

Mean = Sum of all the observations
                No. of observations

= 120+150+180+200+250+300+220+350+370+260
                                   10

= 2400 = 240

      10

Thus, average income of the workers is Rs 240.

Q6. Following information pertains to the daily income of 150 families

Calculate the arithmetic mean.

Income (in Rs)	Number of families
More than 75	150
More than 85	140
More than 95	115
More than 105	95
More than 115	70
More than 125	60
More than 135	40
More than 145	25

Answer 

Income (in Rs.)	Number of families  f)	Mid values (x)	fx
75-85 85-95 95-105 105-115 115-125 125-135 135-145 145-155	10 25 20 25 10 20 15 25	80 90 100 110 120 130 140 150	800 2250 2000 2750 1200 2600 2100 3750
	∑f = 150		∑fx = 17450

Formula of mean is as follow:
Mean = Sum of all the observations                     No. of observations	=	∑fx  ∑f

= 17450/150 = 116.33

Thus, the average mean income for 150 families is Rs . 116.33.

Size of Land Holdings (in acres) Less than 100 100- 200 200- 300 300- 400 400 and above Number of families 40 89 148 64 39

Q7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Answer.

To calculate the mean size of holding, first calculate the cumulative frequency.

Size of land holdings (in acres) (X)	Number of families  (f)	Cumulative  frequency (cf)
0-100 100-200 200-300 300-400 400-500	40 89 148 64 39	40 129 277 341 380
	∑f =380

Then, find the median frequency

Median frequency = N/2 = 380/2 = 190

Formula of median is as follow: Median = L + (N/2 – c.f) x h
                                                                         f

By substituting the value in the formula we get,

= 200 + (190 – 129) x 100 = 241.21
                   148

Thus, the median size of land holding is 241.21 acres.

 

Q8. The following series relates to the daily income of workers employed in affirm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by 25% workers.

Daily Income (in Rs )	10-14	15-19	20-24	25-29	30-34	35-39
Numbers of workers	5	10	15	20	10	5

Answer.

Daily income(in Rs )	Class interval (X)	Number of workers  (f)	Cumulative frequency (cf)
10-14 15-19 20-24 25-29 30-34 35-39	9.5-14.5 14.5-19.5 19.5-24.5 24.5-29.5 29.5-34.5 34.5-39.5	5 10 15 20 10 5	5 15 30 50 60 65
		∑f = 65

a) To compute highest income of lowest 50% we need to calculate median

Median frequency = N/2 = 65/2 = 32.5

Formula for median is s follows

Median = L + (N/2 – c.f) x h
                          f

By substituting the values in the formula, we get

Median = 24.5 + (32.5 –30) x 5
                               20

= 24.5 + 0.625 = 25.125

Highest income of lowest 50% workers is Rs 25.125.

b) For minimum income of 25% of workers we need to calculate Q₁.

First Quartile frequency = D/4 = 65/4 = 16.25

Q₁ = L + (N/4 – c.f) x h
                    f

By substituting the values in the formula, we get

Q₁ = 19.5 + (16.25 – 15) x 5 = 19.9166
                            15

Minimum income earned by the top 25% workers is Rs 19.92.

c) For maximum income of 25% of workers we need to calculate Q₃.

Third quartile frequency = 3(D/4) = 3 (65/4)= 48.75

Q₃ = L + (3(N/4) – c.f) x h
                       f

By substituting the values in the formula, we get

Q₃ = 24.5 + (48.75 – 30) x 5
                           20

= 24.5 + 4.6875 = 29.1875

Maximum income earned by 25% workers is Rs 29.19.

Q9. 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)	50-53	53-56	56-59	59-62	62-65	65-68	68-71	71-74	74-77
Number of farms	3	8	14	30	36	28	16	10	5

Answer.

To calculate mean, median and mode values

Production yield (kg per hectare) (X)	Number of farms (f)	Mid values (m)	Cumulative Frequency (cf)	fm
50-53 53-56 56-59 59-62 62-65 65-68 68-71 71-74 74-77	3 8 14 30 36 28 16 10 5	51.5 54.5 57.5 60.5 63.5 66.5 69.5 72.5 75.5	3 11 25 55 91 119 135 145 150	154.5 436 805 1815 2286 1862 1112 725 377.5
	∑f =150			∑fm= 9573

Formula of mean is as follows:

Mean = ∑fm = 9573/150 = 63.82 kg/hectare
             ∑f

Formula of median is as follows: Median = L + (N/2 – cf) x h
                                                                           f

By substituting the value in the formula we get,

= 62 + (75 – 55) x 3
                  36

= 62 + 1.67 = 63.67 kg/hectare

Formula of mode is as follows: Mode = L + d1       x h
                                                            d1 + d2

By substituting the value in the formula we get,

= 62 + 6/6+8 x 3 = 62 + 18/14 = 62 + 1.28

= 63.28 kg/hectare

Mean, median and mode values are 63.82 kg/hectare, 63.67 kg/hectare and 63.28 kg/hectare.