# NCERT Solution Class 11 Statistics Measures of Central Tendency

NCERT Solution Class 11 Statistics Measures of Central Tendency with answers available in Pdf for free download. The NCERT Solutions for Class 11 Economics with answers have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Standard 11 by CBSE, NCERT and KVS. Solutions to questions given in NCERT book for Class 11 Economics are an important part of exams for Grade 11 Economics and if practiced properly can help you to get higher marks. Refer to more Chapter-wise Solutions for NCERT Class 11 Economics and also download more latest study material for all subjects

## Measures of Central Tendency Class 11 NCERT Solutions

Class 11 Economics students should refer to the following NCERT questions with answers for Measures of Central Tendency in standard 11. These NCERT Solutions with answers for Grade 11 Economics will come in exams and help you to score good marks

### Measures of Central Tendency NCERT Solutions Class 11

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

Exercises

Q1. Which average would be suitable in following cases?

(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.

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

(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.

(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

(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 0-1010-2020-3030-4040-5050-60 121827 x176 51525354555 6027067535x765330 ∑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) 0-1010-2020-3030-4040-5050-60 12182720176 1230577794100 ∑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

Formula of mean is as follow:

Mean = Sum of all the observations
No. of observations

= 120+150+180+200+25+300+220+35+370+26
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

 Income (in Rs.) Number of families  f) Mid values (x) fx 75-8585-9595-105105-115115-125125-135135-145145-155 1025202510201525 8090100110120130140150 8002250200027501200260021003750 ∑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 than100 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.

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) 0-100100-200200-300300-400400-500 40891486439 40129277341380 ∑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

 Daily income(in Rs ) Class interval (X) Number of workers  (f) Cumulative frequency (cf) 10-1415-1920-2425-2930-3435-39 9.5-14.514.5-19.519.5-24.524.5-29.529.5-34.534.5-39.5 5101520105 51530506065 ∑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

To calculate mean, median and mode values

 Production yield (kg per hectare) (X) Number of farms (f) Mid values(m) Cumulative Frequency (cf) fm 50-5353-5656-5959-6262-6565-6868-7171-7474-77 381430362816105 51.554.557.560.563.566.569.572.575.5 311255591119135145150 154.54368051815228618621112725377.5 ∑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.

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