Class 11 Economics (Statistics for Economics) Chapter 5 Measures of Central Tendency

Key Concept: Frequency Identification

c) 8

[Solution Description] The mode of a data set is the value that appears most frequently. In the given set 6, 7, 8, 8, 9, 10, the number 8 occurs twice, while all other numbers appear only once. Therefore, the mode is 8.

Key Concept: Mode Definition

c) The most frequently occurring value in the dataset

[Solution Description] The mode is defined as the value that appears most frequently in a dataset.

Key Concept: Median in Discrete Series

b) 10

[Solution Description] To find the median in a discrete series, we first calculate the cumulative frequency. The cumulative frequencies are: 3, 10 (3+7), 14 (10+4), and 20 (14+6). The total number of observations $N = 20$. The position of the median is given by $\frac{(N+1)}{2} = \frac{21}{2} = 10.5$. The median lies between the 10th and 11th observation, which falls in the second class interval as the cumulative frequency up to this point is 10. Therefore, the median score is 10.

Key Concept: Mode Formula Application

b) 32.67

[Solution Description] Using the mode formula $\text{Mode} = L + \left( \frac{D1}{D1 + D2} \right) \times h$, we substitute the given values:

Mode $= 30 + \left( \frac{10}{10 + 5} \right) \times 4$

Mode $= 30 + \left( \frac{10}{15} \right) \times 4$

Mode $= 30 + \left( \frac{2}{3} \right) \times 4$

Mode $= 30 + \frac{8}{3}$

Mode $= 30 + 2.67$

Mode $\approx 32.67$

Key Concept: Basic Calculation

b) 30

[Solution Description] The arithmetic mean is calculated by taking the sum of all values and dividing it by the number of values.

So, $\text{Arithmetic Mean} = \frac{10 + 20 + 30 + 40 + 50}{5} = \frac{150}{5} = 30$.

Key Concept: Median as Percentile

b) $P_{50}$

[Solution Description] The median is the value that divides the data set into two equal halves, meaning 50% of the data is below it and 50% is above it. This corresponds to the 50th percentile, $P_{50}$.

Key Concept: Definition Recall

b) The most frequently occurring value

[Solution Description]

The mode is defined as the most frequently observed data value in a series. It is the value that appears most often.

Key Concept: Quartile Identification

b) Q2

[Solution Description]

The median of a dataset, which divides it into two equal parts, is also referred to as the second quartile or Q2.

Key Concept: Quartile Calculation

c) 14

[Solution Description] For a data set with 9 values, $Q3$ can be found at position $\left(\frac{3(n+1)}{4}\right)$ where $n = 9$. This gives position $\frac{3(9+1)}{4} = \frac{30}{4} = 7.5$. To find $Q3$, we average the 7th and 8th values, which are 13 and 15. So, $Q3 = \frac{13 + 15}{2} = 14$.

Key Concept: Purpose of Central Tendency

c) To summarize data with a single value

[Solution Description] Measures of central tendency are used to summarize a large set of data points with a single value that represents the center or typical value, helping simplify and understand the data.

Key Concept: Comparison with Arithmetic Mean

c) The weighted mean is higher than the arithmetic mean

[Solution Description] First, compute the simple arithmetic mean:

$\text{Simple Arithmetic Mean} = \frac{70 + 80 + 90}{3} = 80$

Now calculate the weighted arithmetic mean using weights 1, 2, and 3:

$\text{Weighted Mean} &= \frac{(1 \times 70) + (2 \times 80) + (3 \times 90)}{1 + 2 + 3}$

$= \frac{70 + 160 + 270}{6}$

$= \frac{500}{6} \approx 83.33$

Thus, the weighted arithmetic mean is higher than the simple arithmetic mean.

Key Concept: Deviation Calculation

b) -150

[Solution Description] The deviation $d$ is calculated as $d = X – A$, where $X = 700$ is the individual observation and $A = 850$ is the assumed mean. Calculating gives $d = 700 – 850 = -150$.

Key Concept: Assumed Mean Method

b) 246.59

[Solution Description]

To calculate the mean using the assumed mean method, we first calculate the deviations from the assumed mean $A = 250$. The formula is:

$X = A + \frac{\Sigma f d}{\Sigma f}$

where:

– $d = X_i – A$ are the deviations,

– $f$ is the frequency.

Calculate $d$ and $fd$ for each group:

$\text{For } X = 150, \quad d = 150 – 250 = -100, \quad fd = 7 \times (-100) = -700$

$\text{For } X = 200, \quad d = 200 – 250 = -50, \quad fd = 10 \times (-50) = -500$

$\text{For } X = 250, \quad d = 250 – 250 = 0, \quad fd = 12 \times 0 = 0$

$\text{For } X = 300, \quad d = 300 – 250 = 50, \quad fd = 9 \times 50 = 450$

$\text{For } X = 350, \quad d = 350 – 250 = 100, \quad fd = 6 \times 100 = 600$

The sum of $fd$ is:

$\Sigma fd = -700 – 500 + 0 + 450 + 600 = -150$

Total frequency $\Sigma f = 7 + 10 + 12 + 9 + 6 = 44$.

Now apply these values in the formula:

$X = 250 + \frac{-150}{44} = 250 – 3.41 = 246.59$

Therefore, the mean is approximately 246.59.

Key Concept: Median Basics

b) 9

[Solution Description] To find the median, arrange the numbers in order and identify the middle one. The dataset $4, 7, 9, 10, 12$ is already ordered. As there are 5 data points (an odd number), the median is the third value, which is 9.

Key Concept: Basic Definitions

d) Arithmetic mean is greater than the median which is greater than the mode in positive skew.

[Solution Description] In a positively skewed distribution, the relationship is typically $Me > Mi > Mo$.

Key Concept: Basic Definition

d) Median

[Solution Description] The measures of central tendency include the Arithmetic Mean, Median, and Mode. These are used to summarize data with a single number. Range and Variance are measures of spread, not central tendency.

Key Concept: Extreme Values

b) When there are outliers present

[Solution Description] The median is preferred when dealing with data that has outliers or extreme values, as it provides a central value that is not skewed by these extremes.

Key Concept: Median Calculation

c) 14

[Solution Description]

To find the median, arrange the numbers in ascending order and locate the central number. For an odd number of observations, the median is the middle value.

Arrange the data in order: 10, 12, 14, 16, 18

Since there are 5 numbers, the position of the median is given by:

$\frac{N+1}{2} = \frac{5+1}{2} = 3$

Thus, the median is the third number in the ordered list, which is 14.

Key Concept: Median Definition

d) The middle value of an ordered dataset

[Solution Description] In a discrete series, the median is defined as the middle value when all data values are ordered from smallest to largest. It divides the dataset into two equal halves.

Key Concept: Basic Formula Application

b) 291.67

[Solution Description] To find the arithmetic mean using the direct method, we use the formula $X = \frac{\Sigma fX}{\Sigma f}$. Here, the calculation steps are:

Calculate $\Sigma fX = 250 \times 10 + 300 \times 15 + 350 \times 5 = 2500 + 4500 + 1750 = 8750$.

Calculate $\Sigma f = 10 + 15 + 5 = 30$.

Thus, the arithmetic mean is

$X = \frac{8750}{30} = 291.67.$

Therefore, the arithmetic mean is approximately 291.67.

Key Concept: Assumed Mean Method Formula

a) $X = A + \frac{\Sigma fd}{\Sigma f}$

[Solution Description]

In the assumed mean method, the formula used to calculate the arithmetic mean $X$ in a discrete series is:

$X = A + \frac{\Sigma fd}{\Sigma f}$

Here, $A$ is the assumed mean, $\Sigma fd$ is the sum of the products of frequencies and deviations from the assumed mean, and $\Sigma f$ is the total sum of frequencies.

Key Concept: Median Position

b) 6th item

[Solution Description]

To find the median position in a discrete series, we use the formula:

$\text{Median Position} = \frac{N+1}{2}$

where $N$ is the number of items.

Substituting $N = 11$, we get:

$\text{Median Position} = \frac{11+1}{2} = 6$

Thus, the median position is the 6th item.

Key Concept: Definition Understanding

d) The sum of all values divided by the number of values

[Solution Description]

An arithmetic mean is described as the sum of all observations divided by the number of observations.

It provides a measure of central tendency that represents the average value in a dataset.

Key Concept: Definition Recall

b) The left and right sides are mirror images.

[Solution Description] A symmetric distribution is characterized by the property that its left and right sides are mirror images of each other.

Key Concept: Median Position

b) 5th

[Solution Description] To find the position of the median in a set of $N$ items, use the formula $\frac{(N+1)}{2}$. Here, $N = 9$, so: $\text{Position of Median} = \frac{(9+1)}{2} = 5.$ Therefore, the median is at the 5th position.

Key Concept: Suitable Average

c) Mode

[Solution Description]

To find the most common value or item, the mode is used as it represents the value that occurs with the highest frequency.

Key Concept: Definition Understanding

d) The most frequently observed data value

[Solution Description] Mode is defined as the most frequently observed data value in a data set.

Key Concept: Basic Calculation

b) 35

[Solution Description] To find the arithmetic mean, sum all the observations and divide by the number of observations.

$X = \frac{\Sigma X}{N} = \frac{15 + 25 + 35 + 45 + 55}{5} = \frac{175}{5} = 35$

So, the arithmetic mean is 35.

Key Concept: Assumed Mean Method

c) 30

[Solution Description] To find the arithmetic mean using the assumed mean method, we first calculate the deviations $(d)$ from the assumed mean $A = 30$.

$d = X – A$

For each observation:

– For $X = 20$, $d = 20 – 30 = -10$

– For $X = 25$, $d = 25 – 30 = -5$

– For $X = 30$, $d = 30 – 30 = 0$

– For $X = 35$, $d = 35 – 30 = +5$

– For $X = 40$, $d = 40 – 30 = +10$

Sum of deviations, $\Sigma d = (-10) + (-5) + 0 + 5 + 10 = 0$

Now apply the formula:

$\bar{X} = A + \frac{\Sigma d}{N} = 30 + \frac{0}{5} = 30$

Thus, the arithmetic mean is 30.

Key Concept: Median Class Identification

c) The class with cumulative frequency of 55

[Solution Description]

To find the median class, calculate $\frac{N}{2} = \frac{100}{2} = 50$. Locate where this value falls in the cumulative frequency distribution. Here, 50 is between 30 and 55, which corresponds to the class interval with cumulative frequency 55. Thus, the median class is the one associated with cumulative frequency 55.

Key Concept: Basic Calculation

c) 18

[Solution Description] To calculate the arithmetic mean, sum all the numbers and divide by the count of numbers.

Given numbers are: 12, 15, 20, 25

Step 1: Calculate the sum of these numbers:

$12 + 15 + 20 + 25 = 72$

Step 2: Count the number of observations:

There are 4 numbers.

Step 3: Divide the sum by the number of observations:

$\text{Arithmetic Mean} = \frac{72}{4} = 18$

Key Concept: Quartile Calculation

b) 6.5

[Solution Description] To find $Q1$, we use the formula $Q1 = \text{size of } \frac{(N + 1)}{4}\text{th item}$. Here, $N = 5$ (since there are 5 numbers). Thus,

$Q1 = \frac{(5 + 1)}{4} = \frac{6}{4} = 1.5^{th}$

The 1st position value is 5 and the 2nd position value is 8. Therefore, the $1.5^{th}$ value will be a midpoint between 5 and 8. We calculate it as:

$Q1  = 5 + 0.5 \times (8 – 5)$

$= 5 + 1.5 = 6.5$

So, the lower quartile $Q1 = 6.5$.

Key Concept: Basic Definition

c) 15

[Solution Description] To find the arithmetic mean, sum all the observations and divide by the number of observations. The formula is $\bar{X} = \frac{\Sigma X}{N}$. Here, $\Sigma X = 5 + 10 + 15 + 20 + 25 = 75$ and $N = 5$. Thus, $\bar{X} = \frac{75}{5} = 15$.

Key Concept: Mode Calculation

c) 36.67

[Solution Description] The formula for calculating the mode is:

$Mo = L + \left( \frac{D1}{D1 + D2} \right) \times h$

Substituting the given values,

$Mo = 30 + \left( \frac{8}{8 + 4} \right) \times 10$

$Mo = 30 + \left( \frac{8}{12} \right) \times 10$

$Mo = 30 + \left( \frac{2}{3} \right) \times 10$

$Mo = 30 + 6.67$

Therefore, the mode is approximately 36.67.

Key Concept: Assumed Mean Basics

c) 52

[Solution Description] Using the assumed mean formula $X = A + \frac{\Sigma fd}{\Sigma f}$:

1. Substitute the given values into the formula:

$X = 50 + \frac{24}{12}$

2. Calculate $\frac{24}{12} = 2$.

3. Therefore, $X = 50 + 2 = 52$.

Hence, the arithmetic mean is 52.

Key Concept: Qualitative Data

c) Mode

[Solution Description] The mode is typically used to describe qualitative data because it identifies the most frequently occurring category within the dataset.

Key Concept: Median in Continuous Series

c) 30

[Solution Description] The cumulative frequency up to each class is calculated as follows:

– Cumulative frequency for 0-10 = 3,

– Cumulative frequency for 10-20 = 3 + 7 = 10,

– Cumulative frequency for 20-30 = 10 + 5 = 15,

– Cumulative frequency for 30-40 = 15 + 15 = 30.

Total number of observations $N = 30$.

Median position is at $N/2 = 30/2 = 15$. Therefore, the median class is 20-30 because the cumulative frequency just exceeds 15 there.

Using the formula:

$\text{Median} = L + \left(\frac{\frac{N}{2} – \text{c.f.}}{f}\right) \times h$

where:

$L  = 20$,

$\text{c.f.}  = 10 \text{ (cumulative frequency before the median class)}$,

$f  = 5 \text{ (frequency of the median class)}$,

$h  = 10 \text{ (width of the class interval)}$.

Substitute these values in:

$\text{Median} = 20 + \left(\frac{15 – 10}{5}\right) \times 10 = 20 + \left(\frac{5}{5}\right) \times 10 = 20 + 10 = 30.$

So, the median is 30.

Key Concept: Basic Calculation

b) 10

[Solution Description] To find the arithmetic mean for ungrouped data using the direct method, we sum all observations and divide by the number of observations. Here, the observations are 5, 10, and 15.

First, add the numbers:

$5 + 10 + 15 = 30$

Next, divide the total by the number of observations, which is 3:

$\text{Arithmetic Mean} = \frac{30}{3} = 10$

Key Concept: Quartile Definition

b) 25%

[Solution Description] The first quartile (Q1) represents the point below which 25% of the data falls.

Key Concept: Central Tendency Choice

c) Median

[Solution Description] The median is the most suitable measure of central tendency when there are extreme values, such as a few very high salaries, because it is not affected by outliers.

Key Concept: Basic Formula Application

a) $X = A + \frac{\Sigma d’}{N} \times c$

[Solution Description] The arithmetic mean using the Step Deviation Method is calculated using the formula $X = A + \frac{\Sigma d’}{N} \times c$. Here, $A$ is the assumed mean, $\Sigma d’$ is the sum of step deviations, $N$ is the number of observations, and $c$ is the common factor.

Key Concept: Arithmetic Mean Calculation

c) 30

[Solution Description]

To calculate the arithmetic mean using the direct method, we first find the mid-point ($x_i$) of each class interval and then multiply it by its frequency ($f_i$). Sum these products to get $\sum f_ix_i$. Divide this sum by the total frequency ($\sum f_i$).

Steps:

1. Mid-points ($x_i$):

– For 10–20: $x_1 = \frac{10 + 20}{2} = 15$

– For 20–30: $x_2 = \frac{20 + 30}{2} = 25$

– For 30–40: $x_3 = \frac{30 + 40}{2} = 35$

– For 40–50: $x_4 = \frac{40 + 50}{2} = 45$

2. Products of mid-points and frequencies:

– $f_1x_1 = 7 \times 15 = 105$

– $f_2x_2 = 12 \times 25 = 300$

– $f_3x_3 = 15 \times 35 = 525$

– $f_4x_4 = 6 \times 45 = 270$

3. Sum of products: $\sum f_ix_i = 105 + 300 + 525 + 270 = 1200$

4. Total frequency: $\sum f = 7 + 12 + 15 + 6 = 40$

5. Arithmetic Mean: $\bar{x} = \frac{\sum f_ix_i}{\sum f} = \frac{1200}{40} = 30$

Key Concept: Application of Sum of Deviations

c) 3

[Solution Description] The deviation of each number from the mean is calculated as follows:

For 14: $14 – 15 = -1$

For 16: $16 – 15 = 1$

For 18: $18 – 15 = 3$

The sum of deviations is $-1 + 1 + 3 = 3$. Therefore, the sum of deviations is 3.