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 openended 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 openended 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 )  010  10 20  2030  3040  4050  5060 
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 
010 1020 2030 3040 4050 5060  12 18 27 x 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)

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 (3020)
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+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 
Answer
Income (in Rs.)  Number of families f)  Mid values (x)  fx 
7585 8595 95105 105115 115125 125135 135145 145155  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  =  ∑fx 
= 17450/150 = 116.33
Thus, the average mean income for 150 families is Rs . 116.33.

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) 
0100 100200 200300 300400 400500  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 )  1014  1519  2024  2529  3034  3539 
Numbers of workers  5  10  15  20  10  5 
Answer.
Daily income(in Rs )  Class interval (X)  Number of workers (f)  Cumulative frequency (cf) 
1014 1519 2024 2529 3034 3539  9.514.5 14.519.5 19.524.5 24.529.5 29.534.5 34.539.5  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 Q_{1}.
First Quartile frequency = D/4 = 65/4 = 16.25
Q_{1} = L + (N/4 – c.f) x h
f
By substituting the values in the formula, we get
Q_{1} = 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 Q_{3}.
Third quartile frequency = 3(D/4) = 3 (65/4)= 48.75
Q_{3} = L + (3(N/4) – c.f) x h
f
By substituting the values in the formula, we get
Q_{3} = 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)  5053  5356  5659  5962  6265  6568  6871  7174  7477 
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 
5053 5356 5659 5962 6265 6568 6871 7174 7477  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.