Measures of Central TendencyClass 11 Statistics For Economics NCERT Solutions

The suitable averages for the given cases are as follows:

(i) Average size of readymade garments: Mode would be the most suitable average. A manufacturer is interested in the size that is most frequently demanded to plan production accordingly. The mode identifies the most frequently occurring value in a dataset.

(ii) Average intelligence of students in a class: Median would be suitable. Intelligence is a qualitative attribute that can be ranked but not measured precisely in numerical terms. The median is a positional average that is appropriate for ranked data.

(iii) Average production in a factory per shift: Arithmetic Mean would be the most suitable average. Production is a quantitative variable, and the arithmetic mean provides a single value representing the average output per shift, considering the total production over a period.

(iv) Average wage in an industrial concern: Median is often more suitable than the arithmetic mean. While the mean can be used, wage data often includes a few very high or very low values (extreme values). The arithmetic mean is heavily influenced by these extremes, whereas the median, being a positional average, is not and gives a better representation of the typical wage.

(v) When the sum of absolute deviations from average is least: Median is the average for which the sum of absolute deviations is the least. This is a mathematical property of the median.

(vi) When quantities of the variable are in ratios: Geometric Mean is the most suitable average. It is specifically used for averaging ratios, percentages, or growth rates. The chapter mentions this type of average as being suitable for certain situations.

(vii) In case of open-ended frequency distribution: Median and Mode are suitable. The arithmetic mean cannot be calculated for open-ended distributions because the mid-point of the open-ended class cannot be determined. Both median and mode can be calculated as their computation does not depend on the extreme values of the distribution.

(i) The most suitable average for qualitative measurement is (b) median. Qualitative data can be ranked (e.g., poor, average, good). The median is a positional average suitable for such ranked data. The mode is also used for qualitative data, especially when it is categorical (e.g., colors), to find the most frequent category.

(ii) Which average is affected most by the presence of extreme items? (c) arithmetic mean. The arithmetic mean is calculated using all the values in the dataset. Therefore, a very large or very small value (an extreme item) can significantly pull the mean up or down. The chapter explicitly states this as a property of the arithmetic mean.

(iii) The algebraic sum of deviation of a set of n values from A.M. is (b) 0. This is a fundamental property of the arithmetic mean (A.M.). The chapter states, "the sum of deviations of items about arithmetic mean is always equal to zero. Symbolically, Σ(X - X̄) = 0."

(i) The sum of deviation of items from median is zero. False. The sum of deviations of items from the arithmetic mean is always zero. This property does not hold true for the median.

(ii) An average alone is not enough to compare series. True. An average provides a measure of central tendency but gives no information about the spread or dispersion of the data within the series. Two series can have the same average but vastly different distributions of values. Therefore, measures of dispersion are also needed for a meaningful comparison.

(iii) Arithmetic mean is a positional value. False. The median is a positional value because it is determined by its position in an ordered dataset (the middle value). The arithmetic mean is a calculated value based on the magnitude of all observations in the series.

(iv) Upper quartile is the lowest value of top 25% of items. True. The upper quartile (Q3) is the value that separates the lowest 75% of the data from the highest 25%. Therefore, it represents the minimum or lowest value for the top 25% of the items.

(v) Median is unduly affected by extreme observations. False. The median is not affected by extreme observations. Since it is a positional average, changing the value of the largest or smallest item does not change the position of the middle item, and thus the median remains the same.

Let the missing frequency for the class interval 30-40 be 'f'. To find its value, we use the formula for the arithmetic mean for grouped data: Mean (X̄) = Σfm / Σf.

We can construct the following table:

Given that the arithmetic mean (X̄) is 28.

28 = (2100 + 35f) / (80 + f) 28 * (80 + f) = 2100 + 35f 2240 + 28f = 2100 + 35f 2240 - 2100 = 35f - 28f 140 = 7f f = 140 / 7 f = 20

So, the missing frequency is 20.

Now that we have the complete frequency distribution, we can calculate the median. First, we find the cumulative frequency (c.f.). Total frequency (N) = Σf = 80 + 20 = 100.

Median item = (N/2)th item = (100/2)th item = 50th item. The 50th item lies in the class interval 20-30, as its cumulative frequency (57) is the first to be greater than 50. Thus, the median class is 20-30.

Using the median formula: Median = L + [((N/2) - c.f.) / f] * h Where:

Median = 20 + [(50 - 30) / 27] * 10 Median = 20 + (20 / 27) * 10 Median = 20 + 200 / 27 Median = 20 + 7.4074 Median = 27.41 (approx)

Thus, the value of the missing frequency is 20 and the median of the series is Rs 27.41.

To find the arithmetic mean of the daily income of the ten workers, we will use the direct method for ungrouped data.

The formula for the arithmetic mean (X̄) is: X̄ = ΣX / N

Where:

The daily incomes (X) are: 120, 150, 180, 200, 250, 300, 220, 350, 370, 260. The number of workers (N) is 10.

First, we calculate the sum of the incomes (ΣX): ΣX = 120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260 ΣX = 2400

Next, we divide the sum by the number of workers: X̄ = 2400 / 10 X̄ = 240

Therefore, the arithmetic mean of the daily income of the workers is Rs 240.

First, we need to convert the 'more than' cumulative frequency distribution into a simple frequency distribution with class intervals.

*The last class is assumed to have the same width as the others.

Now, we can calculate the arithmetic mean using the formula X̄ = Σfm / Σf.

Arithmetic Mean (X̄) = Σfm / Σf X̄ = 17500 / 150 X̄ = 1750 / 15 X̄ = 116.67

Thus, the arithmetic mean daily income is Rs 116.67. (Note: The textbook answer is Rs 116.3. This small difference may arise from a different assumption for the open-ended final class interval.)

To find the median size of land holdings, we first need to create a cumulative frequency table.

Total number of families (N) = 380.

Median item = (N/2)th item = (380/2)th item = 190th item.

Looking at the cumulative frequency column, the 190th item falls in the class interval 200-300, as its cumulative frequency (277) is the first to be greater than 190. Therefore, the median class is 200-300.

We use the formula for the median of a continuous series: Median = L + [((N/2) - c.f.) / f] * h

Where:

Median = 200 + [(190 - 129) / 148] * 100 Median = 200 + (61 / 148) * 100 Median = 200 + 6100 / 148 Median = 200 + 41.216 Median = 241.22 (approx)

Therefore, the median size of land holdings is 241.22 acres.

This problem requires us to compute the Median (for part a), the Upper Quartile (Q3, for part b), and the Lower Quartile (Q1, for part c).

First, we must convert the inclusive class intervals to exclusive class intervals and prepare a cumulative frequency table.

Total number of workers (N) = 75. Class width (h) = 5.

(a) Highest income of lowest 50% workers (Median or Q2): Median item = (N/2)th item = (75/2)th item = 37.5th item. This lies in the class 24.5 - 29.5.

(b) Minimum income earned by the top 25% workers (Upper Quartile or Q3): Q3 item = (3N/4)th item = (3 * 75 / 4)th item = 56.25th item. This lies in the class 29.5 - 34.5.

(c) Maximum income earned by lowest 25% workers (Lower Quartile or Q1): Q1 item = (N/4)th item = (75/4)th item = 18.75th item. This lies in the class 19.5 - 24.5.

We will calculate the mean, median, and mode for the given frequency distribution.

1. Calculation of the Mean (using Step-Deviation Method) Let Assumed Mean (A) = 63.5 and class interval (h) = 3.

Mean (X̄) = A + (Σfd' / Σf) * h X̄ = 63.5 + (16 / 150) * 3 X̄ = 63.5 + 16 / 50 X̄ = 63.5 + 0.32 = 63.82 kg. per hectare.

2. Calculation of the Median First, we find the cumulative frequency (c.f.).

N = 150. Median item = (N/2)th item = 75th item. The 75th item lies in the class 62-65.

3. Calculation of the Mode The highest frequency is 36, which corresponds to the class interval 62-65. This is the modal class.

Mode (Mₒ) = L + [D₁ / (D₁ + D₂)] * h Where:

Mode = 62 + [6 / (6 + 8)] * 3 Mode = 62 + (6 / 14) * 3 Mode = 62 + 18 / 14 Mode = 62 + 1.2857 = 63.29 kg. per hectare (approx).