Measures of Central Tendency
NCERT Solutions • Class 11 Statistics • Chapter 5Conceptual Questions
1. Which average would be suitable in the following cases?
- (i) Average size of readymade garments: Mode. (The standard size in highest demand is needed).
- (ii) Average intelligence of students in a class: Median. (Intelligence is qualitative data; Mean cannot be calculated).
- (iii) Average production in a factory per shift: Arithmetic Mean. (Quantitative data involving total production).
- (iv) Average wage in an industrial concern: Arithmetic Mean.
- (v) When the sum of absolute deviations from average is least: Median. (Property of Median: $\sum |X – M|$ is minimum).
- (vi) When quantities of the variable are in ratios: Geometric Mean. (Best for ratios and percentages).
- (vii) In case of open-ended frequency distribution: Median. (Mean is affected by undefined limits; Mode may be indeterminate).
2. Indicate the most appropriate alternative from the multiple choices provided.
- (i) The most suitable average for qualitative measurement is:
(b) Median. (It can locate the central position for attributes like beauty, honesty, intelligence). - (ii) Which average is affected most by the presence of extreme items?
(c) Arithmetic Mean. (Since it uses every value, a single outlier can distort the average significantly). - (iii) The algebraic sum of deviation of a set of n values from A.M. is:
(b) 0. (Property of Mean: $\sum (X – \bar{X}) = 0$).
3. Comment whether the following statements are true or false.
- (i) The sum of deviation of items from median is zero.
False. (It is zero from the Mean. From Median, the sum of absolute deviations is minimum). - (ii) An average alone is not enough to compare series.
True. (We also need measures of dispersion to understand variability). - (iii) Arithmetic mean is a positional value.
False. (Mean is a mathematical average; Median and Mode are positional averages). - (iv) Upper quartile is the lowest value of top 25% of items.
True. ($Q_3$ separates the lower 75% from the top 25%). - (v) Median is unduly affected by extreme observations.
False. (Median depends on position, not magnitude of extremes).
Numerical Problems
4. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series.
Given: Mean ($\bar{X}$) = 28.
(a) Missing Frequency:
$\bar{X} = \frac{\sum fm}{N} \Rightarrow 28 = \frac{2100 + 35f_1}{80 + f_1}$
$28(80 + f_1) = 2100 + 35f_1$
$2240 + 28f_1 = 2100 + 35f_1$
$140 = 7f_1 \Rightarrow \mathbf{f_1 = 20}$
(b) Median:
Total $N = 80 + 20 = 100$. Median Class is $N/2 = 50^{th}$ item.
Cumulative Freq just greater than 50 is 57, so Median Class is 20-30.
$L=20, cf=30, f=27, h=10$.
$Median = L + \frac{\frac{N}{2} – cf}{f} \times h$
$= 20 + \frac{50 – 30}{27} \times 10 = 20 + \frac{200}{27} = 20 + 7.41 = \mathbf{27.41}$
| Class | Mid-Val (m) | Freq (f) | fm | Cum Freq (cf) |
|---|---|---|---|---|
| 0-10 | 5 | 12 | 60 | 12 |
| 10-20 | 15 | 18 | 270 | 30 |
| 20-30 | 25 | 27 | 675 | 57 |
| 30-40 | 35 | $f_1$ | $35f_1$ | $57+f_1$ |
| 40-50 | 45 | 17 | 765 | $74+f_1$ |
| 50-60 | 55 | 6 | 330 | $80+f_1$ |
| Total | $N = 80+f_1$ | $\sum fm = 2100+35f_1$ |
(a) Missing Frequency:
$\bar{X} = \frac{\sum fm}{N} \Rightarrow 28 = \frac{2100 + 35f_1}{80 + f_1}$
$28(80 + f_1) = 2100 + 35f_1$
$2240 + 28f_1 = 2100 + 35f_1$
$140 = 7f_1 \Rightarrow \mathbf{f_1 = 20}$
(b) Median:
Total $N = 80 + 20 = 100$. Median Class is $N/2 = 50^{th}$ item.
Cumulative Freq just greater than 50 is 57, so Median Class is 20-30.
$L=20, cf=30, f=27, h=10$.
$Median = L + \frac{\frac{N}{2} – cf}{f} \times h$
$= 20 + \frac{50 – 30}{27} \times 10 = 20 + \frac{200}{27} = 20 + 7.41 = \mathbf{27.41}$
5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
Sum of Incomes: $120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260 = 2400$.
Number of Workers (N): 10.
Mean: $\bar{X} = \frac{\sum X}{N} = \frac{2400}{10} = \mathbf{Rs\ 240}$.
Number of Workers (N): 10.
Mean: $\bar{X} = \frac{\sum X}{N} = \frac{2400}{10} = \mathbf{Rs\ 240}$.
6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
Note: The data is “More than” cumulative. We must convert it to simple frequency.
Mean: $\bar{X} = \frac{\sum fm}{N} = \frac{17450}{150} = \mathbf{Rs\ 116.33}$
| Income (Class) | Mid-Val (m) | Freq (f) | fm |
|---|---|---|---|
| 75-85 | 80 | $150-140=10$ | 800 |
| 85-95 | 90 | $140-115=25$ | 2250 |
| 95-105 | 100 | $115-95=20$ | 2000 |
| 105-115 | 110 | $95-70=25$ | 2750 |
| 115-125 | 120 | $70-60=10$ | 1200 |
| 125-135 | 130 | $60-40=20$ | 2600 |
| 135-145 | 140 | $40-25=15$ | 2100 |
| 145-155 | 150 | 25 | 3750 |
| Total | N=150 | $\sum fm = 17450$ |
Mean: $\bar{X} = \frac{\sum fm}{N} = \frac{17450}{150} = \mathbf{Rs\ 116.33}$
7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
Cumulative Frequency Table:
Calculation:
$N = 380$. Median Item = $N/2 = 190^{th}$.
Class interval with cf > 190 is 200-300.
$L=200, cf=129, f=148, h=100$.
$Median = 200 + \frac{190 – 129}{148} \times 100$
$= 200 + \frac{6100}{148} = 200 + 41.22 = \mathbf{241.22\ acres}$.
| Class | Freq (f) | Cum Freq (cf) |
|---|---|---|
| Less than 100 | 40 | 40 |
| 100 – 200 | 89 | 129 |
| 200 – 300 | 148 | 277 |
| 300 – 400 | 64 | 341 |
| 400 and above | 39 | 380 |
Calculation:
$N = 380$. Median Item = $N/2 = 190^{th}$.
Class interval with cf > 190 is 200-300.
$L=200, cf=129, f=148, h=100$.
$Median = 200 + \frac{190 – 129}{148} \times 100$
$= 200 + \frac{6100}{148} = 200 + 41.22 = \mathbf{241.22\ acres}$.
8. 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.
Interpretation:
(a) Highest of lowest 50% = Median.
(b) Min of top 25% = Upper Quartile ($Q_3$).
(c) Max of lowest 25% = Lower Quartile ($Q_1$).
Cumulative Frequency:
10-14 (5), 15-19 (15), 20-24 (30), 25-29 (50), 30-34 (60), 35-39 (65). N=65.
(a) Median ($N/2 = 32.5$): Class 25-29.
$M = 25 + \frac{32.5 – 30}{20} \times 5 = 25 + 0.625 = \mathbf{Rs\ 25.63}$
*(Note: NCERT Answer 25.11 suggests inclusive series adjustment. $L=24.5$. $24.5 + 0.625 = 25.125$)*.
(b) $Q_3$ ($3N/4 = 48.75$): Class 25-29.
$Q_3 = 24.5 + \frac{48.75 – 30}{20} \times 5 = 24.5 + 4.69 = \mathbf{Rs\ 29.19}$
(c) $Q_1$ ($N/4 = 16.25$): Class 20-24.
$Q_1 = 19.5 + \frac{16.25 – 15}{15} \times 5 = 19.5 + 0.42 = \mathbf{Rs\ 19.92}$
(a) Highest of lowest 50% = Median.
(b) Min of top 25% = Upper Quartile ($Q_3$).
(c) Max of lowest 25% = Lower Quartile ($Q_1$).
Cumulative Frequency:
10-14 (5), 15-19 (15), 20-24 (30), 25-29 (50), 30-34 (60), 35-39 (65). N=65.
(a) Median ($N/2 = 32.5$): Class 25-29.
$M = 25 + \frac{32.5 – 30}{20} \times 5 = 25 + 0.625 = \mathbf{Rs\ 25.63}$
*(Note: NCERT Answer 25.11 suggests inclusive series adjustment. $L=24.5$. $24.5 + 0.625 = 25.125$)*.
(b) $Q_3$ ($3N/4 = 48.75$): Class 25-29.
$Q_3 = 24.5 + \frac{48.75 – 30}{20} \times 5 = 24.5 + 4.69 = \mathbf{Rs\ 29.19}$
(c) $Q_1$ ($N/4 = 16.25$): Class 20-24.
$Q_1 = 19.5 + \frac{16.25 – 15}{15} \times 5 = 19.5 + 0.42 = \mathbf{Rs\ 19.92}$
9. Calculate the mean, median and mode values for production yield of wheat.
Mean Calculation: Using Step Deviation (Assumed Mean A=63.5, h=3).
$\bar{X} = A + \frac{\sum fu}{N} \times h$. $\sum fu = 16$. N=150.
$\bar{X} = 63.5 + \frac{16}{150} \times 3 = 63.5 + 0.32 = \mathbf{63.82\ kg/ha}$.
Median Calculation: $N/2 = 75$. Class 62-65 (cf=91).
$M = 62 + \frac{75 – 55}{36} \times 3 = 62 + 1.67 = \mathbf{63.67\ kg/ha}$.
Mode Calculation: Highest freq = 36 (Class 62-65).
$f_1=36, f_0=30, f_2=28, L=62, h=3$.
$Z = L + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h$
$= 62 + \frac{36 – 30}{72 – 30 – 28} \times 3 = 62 + \frac{6}{14} \times 3 = 62 + 1.29 = \mathbf{63.29\ kg/ha}$.
$\bar{X} = A + \frac{\sum fu}{N} \times h$. $\sum fu = 16$. N=150.
$\bar{X} = 63.5 + \frac{16}{150} \times 3 = 63.5 + 0.32 = \mathbf{63.82\ kg/ha}$.
Median Calculation: $N/2 = 75$. Class 62-65 (cf=91).
$M = 62 + \frac{75 – 55}{36} \times 3 = 62 + 1.67 = \mathbf{63.67\ kg/ha}$.
Mode Calculation: Highest freq = 36 (Class 62-65).
$f_1=36, f_0=30, f_2=28, L=62, h=3$.
$Z = L + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h$
$= 62 + \frac{36 – 30}{72 – 30 – 28} \times 3 = 62 + \frac{6}{14} \times 3 = 62 + 1.29 = \mathbf{63.29\ kg/ha}$.