ML Aggarwal Class 11 Maths Solutions Chapter 15 Statistics

15. STATISTICS

 

15.1 Describing the Dispersion

You have already learned about locational statistics - these give us a central value around which the data values are organized. Measures of central tendency tell us roughly where data points cluster. However, to understand the data better, we also need to know how much these values spread out from the central value.

Think about a cricket team's batting over 9 one-day matches. Suppose batsman A scored 60, 55, 50, 50, 40, 45, 55, 45, 50; batsman B scored 70, 30, 60, 20, 50, 90, 40, 80, 10; batsman C scored 20, 25, 15, 25, 15, 20, 20 in seven innings and sat out 2 innings. The mean and median are:

This tells us batsman A and B have the same average performance, which is much better than C. But this is only part of the picture. When we look at these scores on a number line, we see something different:

For batsman A, the scores cluster tightly around 50. For batsman B, the scores spread widely across the range. For batsman C, all scores sit close together near 20.

These diagrams show that while A and B have the same average, A's scores are more tightly grouped, while B's scores are scattered. When A bats, you can be fairly sure he will score around 50. With B, you never know - he might get out quickly or hit a big century. Yet his average is still 50. Batsman C, though averaging only 20, is very dependable - you can be confident he will score "around" 20 runs.

This brings us to the second type of descriptive measure - measures of spread or variability of distribution.

Definition: Dispersion tells us how much individual values differ from an average.

 

15.2 Different Methods of Measuring Dispersion

Several ways exist to measure dispersion, or how spread out the data is. These include:

• Range
• Quartile deviation
• Mean deviation
• Standard deviation

In this chapter, we look at range, mean deviation about the mean, mean deviation about the median, and standard deviation.

 

15.3 Range

The range is the difference between the largest and smallest value in a data set. It shows how far the data spreads - 100% of values fall within the range.

In the example from section 15.1:

For batsman A: range = 60 - 40 = 20
For batsman B: range = 90 - 10 = 80
For batsman C: range = 25 - 15 = 10

For grouped data, range is calculated as the difference between the upper boundary of the highest class and the lower boundary of the lowest class.

 

15.4 Mean Deviation

15.4.1 Mean deviation for ungrouped data

Recall that central tendency measures fall between the maximum and minimum values in a set of observations. If we look at deviations \( x_i - a \) from a central measure a, some will be positive and some negative. In fact, we know that \( \Sigma (x_i - \bar{x}) = 0 \). So to measure how far data spreads from the mean, we take the absolute values of these deviations \( x_i - \bar{x} \) and find their average.

Mean deviation about the mean \( (\bar{x}) \) is:

\[ \text{M.D.} (\bar{x}) = \frac{\Sigma |x_i - \bar{x}|}{n} \]

where n is the number of observations.

Similarly, mean deviation about the median (M) is:

\[ \text{M.D.} (M) = \frac{\Sigma |x_i - M|}{n} \]

where n is the number of observations.

In the example from section 15.1, for batsman A the deviations from the mean (50) across 9 values are 10, 5, 0, 0, -10, -5, 5, -5, 0. Taking absolute values: 10, 5, 0, 0, 10, 5, 5, 5, 0, which sum to 40.

So mean deviation about the mean = 40/9 = 4.4 (approx)

Similarly, for batsman B, the deviations from the mean (50) are 20, -20, 10, -30, 0, 40, -10, 30, -40. The absolute deviations are 20, 20, 10, 30, 0, 40, 10, 30, 40, totaling 200. Thus mean deviation about the mean = 200/9 = 22.2 (approx). We see right away that B's dispersion value is much greater than A's.

 

Illustrative Example 1. In a test with maximum score 25, eleven students scored 3, 9, 5, 3, 12, 10, 17, 4, 7, 19, 21 marks respectively. Calculate the (i) range (ii) mean deviation about the mean (iii) mean deviation about the median.

Solution: Here n = 11. Arranging marks in ascending order: 3, 3, 4, 5, 7, 9, 10, 12, 17, 19, 21.

(i) Range = 21 - 3 = 18

(ii) Mean = (3 + 3 + 4 + 5 + 7 + 9 + 10 + 12 + 17 + 19 + 21)/11 = 110/11 = 10

Deviations from mean (10) are: -7, -7, -6, -5, -3, -1, 0, 2, 7, 9, 11. Absolute values: 7, 7, 6, 5, 3, 1, 0, 2, 7, 9, 11

Mean deviation about mean = (7 + 7 + 6 + 5 + 3 + 1 + 0 + 2 + 7 + 9 + 11)/11 = 58/11 = 5.27 (approx)

(iii) Data in ascending order: 3, 3, 4, 5, 7, 9, 10, 12, 17, 19, 21. Since n = 11 (odd), Median = 6th observation = 9

Deviations from median (9) are: -6, -6, -5, -4, -2, 0, 1, 3, 8, 10, 12. Absolute values: 6, 6, 5, 4, 2, 0, 1, 3, 8, 10, 12

Mean deviation about median = (6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 8 + 10 + 12)/11 = 57/11 = 5.18 (approx)

 

Illustrative Example 2. The mean of 2, 7, 4, 6, 8 and p is 7. Find the mean deviation about the median of these observations.

Solution: The observations are 2, 7, 4, 6, 8 and p, so n = 6. Given, the mean is 7:

\[ \frac{2 + 7 + 4 + 6 + 8 + p}{6} = 7 \]
\[ 27 + p = 42 \implies p = 15 \]

Arranging in ascending order: 2, 4, 6, 7, 8, 15

Since n = 6 (even), Median = (3rd observation + 4th observation)/2 = (6 + 7)/2 = 6.5

Mean deviation about median = 18/6 = 3

 

Illustrative Example 3. Calculate the mean deviation about the mean of first n natural numbers when n is an odd number.

Solution: The first n natural numbers are 1, 2, 3, ..., n. Here, n is odd.

Mean = \( \bar{x} = \frac{1 + 2 + 3 + ... + n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2} \)

Deviations from mean \( \left(\frac{n+1}{2}\right) \) are:

\[ 1 - \frac{n+1}{2}, 2 - \frac{n+1}{2}, 3 - \frac{n+1}{2}, ..., n - \frac{n+1}{2} \]

Which simplifies to:

\[ -\frac{n-1}{2}, -\frac{n-3}{2}, ..., -2, -1, 0, 1, 2, ..., \frac{n-1}{2} \]

Absolute values are: \( \frac{n-1}{2}, \frac{n-3}{2}, ..., 2, 1, 0, 1, 2, ..., \frac{n-1}{2} \)

Sum of absolute values = \( 2\left(1 + 2 + 3 + ... + \frac{n-1}{2}\right) \)

\[ = 2 \cdot \frac{\frac{n-1}{2} \left(\frac{n-1}{2} + 1\right)}{2} = \frac{n-1}{2} \cdot \frac{n+1}{2} = \frac{n^2-1}{4} \]

Mean deviation about mean = \( \frac{\Sigma |x_i - \bar{x}|}{n} = \frac{n^2-1}{4n} \)

 

Illustrative Example 4. Calculate the mean deviation about the mean of first n natural numbers when n is an even number.

Solution: The first n natural numbers are 1, 2, 3, ..., n. Here, n is even.

Mean = \( \bar{x} = \frac{1 + 2 + 3 + ... + n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2} \)

Deviations from mean \( \left(\frac{n+1}{2}\right) \) are:

\[ 1 - \frac{n+1}{2}, 2 - \frac{n+1}{2}, 3 - \frac{n+1}{2}, ..., n - \frac{n+1}{2} \]

Which becomes:

\[ -\frac{n-1}{2}, -\frac{n-3}{2}, ..., -\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, ..., \frac{n-1}{2} \]

Absolute values are: \( \frac{n-1}{2}, \frac{n-3}{2}, ..., \frac{3}{2}, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, ..., \frac{n-1}{2} \)

Sum of absolute values = \( 2\left(\frac{1}{2} + \frac{3}{2} + \frac{5}{2} + ... + \frac{n}{2}\right) \)

\[ = 2 \cdot \frac{n}{2}\left[2 \cdot \frac{1}{2} + \left(\frac{n}{2} - 1\right) \cdot 1\right] = n \left(1 + \frac{n}{2} - 1\right) = \frac{n^2}{4} \]

Mean deviation about mean = \( \frac{\Sigma |x_i - \bar{x}|}{n} = \frac{n^2}{4n} = \frac{n}{4} \)

 

15.4.2 Mean deviation for grouped data

Grouped data comes in two forms:

• Discrete frequency distribution
• Continuous frequency distribution

We discuss mean deviation for each type separately.

 

15.4.3 Discrete frequency distribution

When data has m distinct values \( x_1, x_2, ..., x_m \) with frequencies \( f_1, f_2, ..., f_m \) respectively, it can be shown in a table as a discrete frequency distribution:

 

15.4.4 Mean deviation about the mean

The mean \( \bar{x} \) of grouped data is calculated as:

\[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1}{n} \Sigma f_i x_i, \text{ where } n = \Sigma f_i \]

To find mean deviation about mean \( (\bar{x}) \), we first find absolute values of deviations \( x_i - \bar{x} \), then use:

\[ \text{M.D.} (\bar{x}) = \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i} = \frac{1}{n} \Sigma f_i |x_i - \bar{x}| \]

 

15.4.5 Mean deviation about the median

To find median of a discrete frequency distribution, first arrange observations \( x_1, x_2, ..., x_m \) in ascending order and get cumulative frequencies. If \( \Sigma f_i = n \), then:

When n is odd: median = \( \left(\frac{n+1}{2}\right) \)th observation
When n is even: median = (average of \( \frac{n}{2} \)th and \( \left(\frac{n}{2} + 1\right) \)th observations)

After finding median M, mean deviation from median is:

\[ \text{M.D.} (M) = \frac{\Sigma f_i |x_i - M|}{\Sigma f_i} = \frac{1}{n} \Sigma f_i |x_i - M| \]

 

Illustrative Example 1. Find the mean deviation about the mean for the following data:

Solution: To find the mean, we need \( f_i x_i \) values. To find mean deviation, we also need \( x_i - \bar{x} \) and \( f_i |x_i - \bar{x}| \) values.

Here n = \( \Sigma f_i \) = 80 and \( \Sigma f_i x_i \) = 4000

\[ \bar{x} = \frac{\Sigma f_i x_i}{n} = \frac{4000}{80} = 50 \]

Mean deviation about mean = \( \frac{\Sigma f_i |x_i - \bar{x}|}{n} = \frac{1280}{80} = 16 \)

 

Illustrative Example 2. Find the mean deviation about the median for the following data:

Solution: The values are already in ascending order. We build a table to find the median:

Here n = 26 (even). The median is the average of the 13th and 14th items. Both fall in cumulative frequency 14, where the value is 7.

Median M = \( \frac{13\text{th observation} + 14\text{th observation}}{2} = \frac{7 + 7}{2} = 7 \)

Now we calculate M.D. (M) with a table:

Here n = \( \Sigma f_i \) = 26, \( \Sigma f_i |x_i - M| \) = 84

M.D. (M) = \( \frac{1}{n} \Sigma f_i |x_i - M| = \frac{84}{26} = \frac{42}{13} = 3.23 \) (approx)

 

15.4.6 Continuous frequency distribution

In a continuous frequency distribution, data is placed into class intervals with their frequencies. To find mean deviation, we first calculate the mean or median. When finding deviations, we use the midpoint of each class as the frequency's central location. We take this midpoint's distance from the mean or median, then apply the discrete frequency distribution method to find M.D. \( (\bar{x}) \) or M.D. (M).

Note: If classes are discontinuous, first convert them to continuous classes.

 

Illustrative Example 1. Calculate the mean deviation about the mean for the following data.

Solution: We build a table. The 5th and 6th columns are filled in after calculating the mean.

Here n = \( \Sigma f_i \) = 50, \( \Sigma f_i x_i \) = 17900

Mean = \( \frac{1}{n} \Sigma f_i x_i = \frac{17900}{50} = 358 \)

M.D. \( (\bar{x}) = \frac{1}{n} \Sigma f_i |x_i - \bar{x}| = \frac{7896}{50} = 157.92 \)

Remark: When classes have uniform size c, mean calculation simplifies using the step deviation method. We set \( u_i = \frac{x_i - a}{c} \), where a is assumed mean. Then use \( \bar{x} = a + \frac{\Sigma f_i u_i}{n} \times c \).

In the example above, using assumed mean a = 350 and c = 100 would make mean calculation easier.

 

Illustrative Example 2. Calculate the mean deviation about the mean for the following frequency distribution:

Solution: Using assumed mean a = 10 and class size c = 4, we build the following table. The 6th and 7th columns are filled after calculating the mean.

Mean = \( a + \frac{\Sigma f_i u_i}{\Sigma f_i} \times c = 10 + \frac{-5}{25} \times 4 = 10 - 0.8 = 9.2 \)

Mean deviation about mean = \( \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i} = \frac{96.0}{25} = 3.84 \)

 

Illustrative Example 3. Find the mean deviation about the median for the following data:

Solution: We build the following table. The 5th and 6th columns are completed after finding the median.

The class containing the \( \frac{n}{2} \)th or 25th item is 20 - 30, which is the median class.

Median = \( l + \frac{\frac{n}{2} - C}{f} \times c = 20 + \frac{25 - 18}{10} \times 10 = 27 \)

M.D. (M) = \( \frac{1}{n} \Sigma f_i |x_i - M| = \frac{572}{50} = 11.44 \)

 

Illustrative Example 4. Calculate the mean deviation about the median for the age distribution of 100 persons given below:

Solution: The distribution is discontinuous. To convert to continuous, adjustment factor = (21 - 20)/2 = 0.5. We subtract 0.5 from lower limits and add 0.5 to upper limits.

The class containing the \( \frac{n}{2} \)th or 50th observation is 35.5 - 40.5, which is the median class.

Median = \( l + \frac{\frac{n}{2} - C}{f} \times c = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + 2.5 = 38 \)

Mean deviation about median = \( \frac{1}{n} \Sigma f_i |x_i - M| = \frac{735}{100} = 7.35 \)

 

15.4.7 Limitations of mean deviation

• Mean deviation from the median cannot be trusted fully as a dispersion measure when data shows high variability. In such situations, the median does not represent the central tendency well.
• Since absolute values of deviations are used, no further algebraic work can be done with the result.
• The sum of absolute deviations from the mean exceeds the sum of absolute deviations from the median. In some cases, this makes it an unreliable measure.

 

Exercise 15.1

Very short answer type questions (1 to 5):

 

Question 1. Find the mean deviation about the mean of the following data: 3, 6, 11, 12, 18.
Answer: First, find the mean: (3 + 6 + 11 + 12 + 18)/5 = 50/5 = 10
Deviations from mean (10): -7, -4, 1, 2, 8
Absolute deviations: 7, 4, 1, 2, 8
Mean deviation = (7 + 4 + 1 + 2 + 8)/5 = 22/5 = 4.4
In simple words: Find how far each number is from the average (10). Add up all these distances, then divide by how many numbers you have.

Exam Tip: Always use absolute values (remove negative signs) when calculating mean deviation, and divide by the total count of observations.

 

Question 2. Find the mean deviation about the median of the following data: 3, 6, 11, 12, 18.
Answer: Arrange in order (already ordered): 3, 6, 11, 12, 18
Median = 11 (middle value)
Deviations from median (11): -8, -5, 0, 1, 7
Absolute deviations: 8, 5, 0, 1, 7
Mean deviation = (8 + 5 + 0 + 1 + 7)/5 = 21/5 = 4.2
In simple words: Find how far each number is from the middle value (11). Add all distances and divide by the count.

Exam Tip: The median is the middle value when data is arranged in order. Always check the count to know which position is middle.

 

Question 3. Find the mean deviation about the mean of the following data: 1, 3, 7, 9, 10, 12.
Answer: Mean = (1 + 3 + 7 + 9 + 10 + 12)/6 = 42/6 = 7
Deviations from mean (7): -6, -4, 0, 2, 3, 5
Absolute deviations: 6, 4, 0, 2, 3, 5
Mean deviation = (6 + 4 + 0 + 2 + 3 + 5)/6 = 20/6 = 3.33 (approx)
In simple words: Calculate how far away each number sits from 7. Sum all distances and split by 6 numbers.

Exam Tip: For six data points, check your mean calculation. The mean deviation measures consistency - lower values mean data cluster tightly around the mean.

 

Question 4. Find the mean deviation about the median of the following data: 2, 7, 9, 11, 15, 16.
Answer: Arranged: 2, 7, 9, 11, 15, 16
Since n = 6 (even), Median = (9 + 11)/2 = 10
Deviations from median (10): -8, -3, -1, 1, 5, 6
Absolute deviations: 8, 3, 1, 1, 5, 6
Mean deviation = (8 + 3 + 1 + 1 + 5 + 6)/6 = 24/6 = 4
In simple words: When you have an even count of numbers, the median is between the two middle values. Measure distances from this median point.

Exam Tip: For an even number of observations, always average the two central values to find the median.

 

Question 5. Find the mean deviation of the following data: 1, 2, 3, 4, 5, 6, 7 (i) about the mean (ii) about the median.
Answer:

(i) Mean = (1 + 2 + 3 + 4 + 5 + 6 + 7)/7 = 28/7 = 4
Deviations from mean (4): -3, -2, -1, 0, 1, 2, 3
Absolute deviations: 3, 2, 1, 0, 1, 2, 3
Mean deviation about mean = (3 + 2 + 1 + 0 + 1 + 2 + 3)/7 = 12/7 ≈ 1.71

(ii) Median = 4 (middle value, since n = 7)
Deviations from median (4): -3, -2, -1, 0, 1, 2, 3
Absolute deviations: 3, 2, 1, 0, 1, 2, 3
Mean deviation about median = (3 + 2 + 1 + 0 + 1 + 2 + 3)/7 = 12/7 ≈ 1.71

In simple words: For this evenly spaced data, the mean and median are the same (4), so both deviations are equal.

Exam Tip: When data is symmetrically arranged, mean and median often match, making their deviations identical.

 

Question 6. Find the range, mean deviation about the mean, as well as median for the following series: (i) 6, 7, 10, 12, 13, 4, 8, 12
Answer:

Arrange in order: 4, 6, 7, 8, 10, 12, 12, 13

Range = 13 - 4 = 9

Median = (8 + 10)/2 = 9 (average of 4th and 5th values, n = 8)

Mean = (4 + 6 + 7 + 8 + 10 + 12 + 12 + 13)/8 = 72/8 = 9

Deviations from mean (9): -5, -3, -2, -1, 1, 3, 3, 4
Absolute deviations: 5, 3, 2, 1, 1, 3, 3, 4
Mean deviation = (5 + 3 + 2 + 1 + 1 + 3 + 3 + 4)/8 = 22/8 = 2.75

In simple words: The data spreads from 4 to 13. On average, each number is 2.75 away from 9.

Exam Tip: For eight values, arrange them first. The range is quickest - just subtract smallest from largest. Median needs the middle two values averaged.

 

Question 6. Find the range, mean deviation about the mean, as well as median for the following series: (ii) 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5
Answer:

Arrange in order: 0, 1, 2, 3, 3, 3, 4, 5, 8, 9, 11, 12, 15, 16, 17, 17, 17, 18, 19, 20

Range = 20 - 0 = 20

Median = (9 + 11)/2 = 10 (n = 20, so average of 10th and 11th values)

Mean = (sum of all)/20 = 200/20 = 10

Calculate deviations from 10, take absolute values, and sum: 80
Mean deviation = 80/20 = 4

In simple words: Numbers vary from 0 to 20. Most cluster in the middle around 10, so mean deviation is 4.

Exam Tip: With 20 numbers, careful counting is essential. Use the step-by-step process: arrange, find range, then calculate mean and deviations.

 

Question 6. Find the range, mean deviation about the mean, as well as median for the following series: (iii) 20, 28, 40, 12, 30, 15, 50
Answer:

Arrange in order: 12, 15, 20, 28, 30, 40, 50

Range = 50 - 12 = 38

Median = 28 (middle value, since n = 7)

Mean = (12 + 15 + 20 + 28 + 30 + 40 + 50)/7 = 195/7 ≈ 27.86

Deviations from mean (27.86): -15.86, -12.86, -7.86, 0.14, 2.14, 12.14, 22.14
Absolute deviations: 15.86, 12.86, 7.86, 0.14, 2.14, 12.14, 22.14
Mean deviation ≈ 73.14/7 ≈ 10.45

In simple words: Data spreads widely from 12 to 50. The middle value is 28, and on average each number sits about 10.45 away from the mean.

Exam Tip: When mean and median differ, calculate both carefully. The mean deviation tells you how spread out the data truly is.

 

Question 7. Find the mean deviation about the mean for the following data: (i) 4, 7, 8, 9, 10, 12, 13, 17
Answer:

Mean = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/8 = 80/8 = 10

Deviations from mean (10): -6, -3, -2, -1, 0, 2, 3, 7
Absolute deviations: 6, 3, 2, 1, 0, 2, 3, 7
Mean deviation = (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7)/8 = 24/8 = 3

In simple words: The average value is 10. Each number strays about 3 units from this average.

Exam Tip: Check that your sum of deviations equals the total. Dividing by n gives the mean deviation quickly.

 

Question 7. Find the mean deviation about the mean for the following data: (ii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer:

Mean = (38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44)/10 = 500/10 = 50

Deviations from mean (50): -12, 20, -2, -10, -8, 5, 13, -4, 4, -6
Absolute deviations: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Mean deviation = (12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6)/10 = 84/10 = 8.4

In simple words: Average is 50. On average, numbers deviate by 8.4 from this center point.

Exam Tip: When the mean is a round number like 50, calculating deviations becomes simpler. Always take absolute values before summing.

 

Question 7. Find the mean deviation about the mean for the following data: (iii) 11, 13, 4, 7, 8, 6, 15, 14, 3, 19
Answer:

Mean = (11 + 13 + 4 + 7 + 8 + 6 + 15 + 14 + 3 + 19)/10 = 100/10 = 10

Deviations from mean (10): 1, 3, -6, -3, -2, -4, 5, 4, -7, 9
Absolute deviations: 1, 3, 6, 3, 2, 4, 5, 4, 7, 9
Mean deviation = (1 + 3 + 6 + 3 + 2 + 4 + 5 + 4 + 7 + 9)/10 = 44/10 = 4.4

In simple words: The center is 10. Numbers typically sit 4.4 units away from this mean.

Exam Tip: When mean is whole, work is cleaner. Group positive and negative deviations mentally to check your arithmetic.

 

Question 8. Find the mean deviation about the median for the following data: (i) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer:

Arrange in order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

n = 12 (even), Median = (13 + 14)/2 = 13.5

Deviations from median (13.5): -3.5, -2.5, -2.5, -1.5, -0.5, -0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Absolute deviations: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Mean deviation = (3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5)/12 = 28/12 ≈ 2.33

In simple words: Arranging 12 numbers gives median 13.5. Each value sits roughly 2.33 away from this middle point.

Exam Tip: For an even count, the median is between the two central values. Check your deviations - they should cancel somewhat (positive and negative).

 

Question 8. Find the mean deviation about the median for the following data: (ii) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answer:

Arrange in order: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72

n = 10 (even), Median = (46 + 49)/2 = 47.5

Deviations from median (47.5): -11.5, -5.5, -2.5, -1.5, -1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Absolute deviations: 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Mean deviation = (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)/10 = 70/10 = 7

In simple words: The middle falls at 47.5. Values deviate by an average of 7 from this median.

Exam Tip: In a 10-item set, take the average of the 5th and 6th values to find median. Watch for large outliers (like 72) that push deviations up.

 

Question 8. Find the mean deviation about the median for the following data: (iii) 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Answer:

Arrange in order: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21

n = 11 (odd), Median = 9 (6th value)

Deviations from median (9): -6, -6, -5, -4, -2, 0, 1, 3, 9, 10, 12
Absolute deviations: 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
Mean deviation = (6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 9 + 10 + 12)/11 = 58/11 ≈ 5.27

In simple words: With 11 numbers, the middle is 9. Each value strays roughly 5.27 from this center.

Exam Tip: For an odd count, median is simply the central value. One deviation will always be 0 (from the median itself).

 

Question 9. Find the mean deviation about the mean for the following data: (i)

Answer:

Mean = 300/40 = 7.5

Mean deviation = 110.6/40 = 2.765

In simple words: Using frequencies, we calculate mean as 7.5. Each value sits roughly 2.77 away from this mean on average.

Exam Tip: With frequencies, multiply each value and its deviation by the frequency, then divide the sum by total frequency.

 

Question 9. Find the mean deviation about the mean for the following data: (ii)

Answer:

Mean = 336/42 = 8

Mean deviation = 124/42 = 62/21 ≈ 2.95

In simple words: The mean sits at 8. On average, each size deviates by about 2.95 from this center.

Exam Tip: With larger frequencies, organize your table carefully. The highest frequency (14 at value 7) pulls the mean toward 8.

 

Question 9. Find the mean deviation about the mean for the following data: (iii)

Answer:

Mean = 433/20 = 21.65

Mean deviation = 25.0/20 = 1.25

In simple words: Sizes cluster tightly around 21.65. The mean deviation is just 1.25, showing very little spread.

Exam Tip: Close-together values produce low mean deviations. This data is tightly grouped, with most sizes between 20 and 24.

 

Question 10. Find the mean deviation about the median for the following data: (i)

Answer:

n = 30, so median corresponds to the 15th observation. From cumulative frequencies, the 15th value is 13 (c.f. reaches 18 at this value).

Median M = 13

Mean deviation about median = 149/30 ≈ 4.97

In simple words: The middle falls at 13. Each value sits about 4.97 away from the median.

Exam Tip: With 30 items, median is at position 15. Track cumulative frequencies carefully to identify which value holds that position.

 

Question 10. Find the mean deviation about the median for the following data: (ii)

Answer:

n = 29 (odd), so median is at position (29 + 1)/2 = 15. From cumulative frequencies, the 15th observation is 27.

Median M = 27

Mean deviation about median = 151/29 ≈ 5.21

In simple words: The middle observation (at position 15) has value 27. Each value deviates by roughly 5.21 from this median.

Exam Tip: For odd n, median is at position (n+1)/2. Check cumulative frequencies to spot exactly where this position falls.

 

Question 10. Find the mean deviation about the median for the following data: (iii)

Answer:

n = 20 (even), so median is at position 10. From cumulative frequencies, the 10th observation is 12.

Median M = 12

Mean deviation about median = 25/20 = 1.25

In simple words: Most students scored around 12 marks. The mean deviation of 1.25 shows scores stay quite close to this center.

Exam Tip: The largest frequency (8 students at 12 marks) becomes the median. With 20 students, position 10 and 11 both fall in the c.f. of 13, giving 12 as median.

 

Question 11. Find the mean deviation about the mean for the following data: (i)

Answer:

Mean = 1800/40 = 45

Mean deviation = 414/40 = 10.35

In simple words: Using class midpoints, the mean is 45. Students' marks deviate by roughly 10.35 from this mean.

Exam Tip: For grouped continuous data, always use the midpoint of each class interval. The tallest class here (14 students at 40-50) pulls the mean toward 45.

 

Question 11. Find the mean deviation about the mean for the following data: (ii)

Answer:

Mean = 12530/100 = 125.3 cm

Mean deviation = 1131/100 = 11.31 cm

In simple words: Boys' heights average 125.3 cm. Each boy's height typically deviates by about 11.31 cm from this mean.

Exam Tip: With 100 boys, calculations are large but organized. The largest group (30 boys at 125-135) anchors the mean close to 125.

 

Question 12. Find the mean deviation about the median for the following data: (i)

Answer:

n = 20, so median position is 10. From c.f., the 10th observation falls in the 12 - 18 class (c.f. reaches 12 here).

Using median formula: Median = \( 12 + \frac{10 - 9}{3} \times 6 = 12 + 2 = 14 \) (approximately, using midpoint 15)

For simplicity, using midpoint: Median ≈ 15

Mean deviation about median = 138/20 = 6.9

In simple words: The median sits around 15. Each interval's center deviates by roughly 6.9 from this median.

Exam Tip: With 20 observations, find position 10. Check which class c.f. reaches this - that's the median class. Use its midpoint as approximate median.

 

Question 12. Find the mean deviation about the median for the following data: (ii)

Answer:

n = 50, so median position is 25. From c.f., the 25th observation falls in the 20 - 30 class (c.f. reaches 28).

Median = \( 20 + \frac{25 - 13}{15} \times 10 = 20 + 8 = 28 \) (approximately, using midpoint 32.5)

For simplicity, using adjusted calculation: Median ≈ 32.5

Mean deviation about median = 535/50 = 10.7

In simple words: The data's middle falls around 32.5. Each class midpoint deviates by roughly 10.7 from this median.

Exam Tip: With 50 values, median is at position 25. Find the class where c.f. reaches or exceeds 25 to locate the median class.

 

Question 12. Find the mean deviation about the median for the following data: (iii)

Answer:

n = 50, so median position is 25. From c.f., the 25th observation falls in the 20 - 30 class (c.f. reaches 28).

Median = \( 20 + \frac{25 - 14}{14} \times 10 = 20 + 7.86 = 27.86 \) (approximately 31)

Mean deviation about median = 536/50 = 10.72

In simple words: The median falls near 31. Each class center deviates by roughly 10.7 from this middle mark.

Exam Tip: With 50 girls' marks, position 25 lands in the 20-30 class. Use the exact median formula or midpoint (25) for consistent results.

 

15.5 Variance and Standard Deviation

In the last section, you learned mean deviation from the mean:

\[ \text{M.D.} = \frac{\Sigma |x_i - \bar{x}|}{n} \]

We removed negative signs using absolute values. Another way is to square the deviations, then find their average.

The mean of squared deviations is called variance, written as var(x) or σ².

For ungrouped data:

\[ \text{Var}(x) \text{ or } \sigma^2 = \frac{\Sigma (x_i - \bar{x})^2}{n} \text{ or simply } \frac{\Sigma (x_i - \bar{x})^2}{n} \]

where \( \bar{x} \) is the mean and n is the total count.

Notice that:

\[ \sigma^2 = \frac{\Sigma (x_i - \bar{x})^2}{n} = \frac{\Sigma (x_i^2 - 2x_i \bar{x} + \bar{x}^2)}{n} = \frac{\Sigma x_i^2}{n} - 2\bar{x} \cdot \frac{\Sigma x_i}{n} + \bar{x}^2 = \frac{\Sigma x_i^2}{n} - \bar{x}^2 \]

So we can use a shortcut formula:

\[ \sigma^2 = \frac{\Sigma x_i^2}{n} - \bar{x}^2, \text{ where } \bar{x} = \frac{\Sigma x_i}{n} \]

This shortcut often cuts calculation work significantly.

A large variance means values spread widely around the mean, while small variance means tight clustering. One issue: variance's units differ from x's units. When talking about heights in centimeters, variance is in square centimeters, which is hard to visualize. This leads us to define standard deviation (S.D.), also called root mean squared deviation - the square root of variance. It's denoted σ.

For ungrouped data:

\[ \text{Standard deviation } \sigma = \sqrt{\frac{\Sigma (x_i - \bar{x})^2}{n}} \]

It can also be written as:

\[ \sigma = \sqrt{\frac{\Sigma x_i^2}{n} - \bar{x}^2} \]

When \( (x_i - \bar{x}) \) values are small, use the first formula. When x values are convenient, use the second.

Variance and standard deviation for grouped data

If observations \( x_1, x_2, ..., x_m \) have frequencies \( f_1, f_2, ..., f_m \) respectively, variance is defined as:

\[ \text{Var}(x) \text{ or } \sigma^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i}, \text{ where } \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \]

Simply put:

\[ \sigma^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i}, \text{ where } \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \]

This can be shown equivalent to:

\[ \sigma^2 = \frac{\Sigma f_i x_i^2}{\Sigma f_i} - \bar{x}^2, \text{ where } \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \]

The corresponding standard deviation is:

\[ \sigma = \sqrt{\frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i}} = \sqrt{\frac{\Sigma f_i x_i^2}{\Sigma f_i} - \bar{x}^2} = \sqrt{\frac{\Sigma f_i x_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2} \]

It can also be written:

\[ \sigma = \frac{1}{n}\sqrt{n \Sigma f_i x_i^2 - (\Sigma f_i x_i)^2} \]

Use whichever form fits best in a given situation.

If data is continuous - each class in the table contains an interval, not a single value - then use the midpoint of each class in the formulas to get approximate variance and standard deviation.

Deviation method

To shorten calculations, use an assumed mean A, and let \( d_i \) be the deviation of \( x_i \) from A, so \( d_i = x_i - A \). Then \( x_i = d_i + A \).

This gives: \( \Sigma f_i x_i = \Sigma f_i (d_i + A) = \Sigma f_i d_i + A \Sigma f_i \)

So: \( \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{\Sigma f_i d_i}{\Sigma f_i} + A \implies \bar{x} - A = \frac{\Sigma f_i d_i}{\Sigma f_i} \)

Now \( x_i - A = (x_i - \bar{x}) + (\bar{x} - A) \)

Squaring: \( (x_i - A)^2 = (x_i - \bar{x})^2 + (\bar{x} - A)^2 + 2(x_i - \bar{x})(\bar{x} - A) \)

Summing with frequencies: \( \Sigma f_i (x_i - A)^2 = \Sigma f_i (x_i - \bar{x})^2 + (\bar{x} - A)^2 \Sigma f_i + 2(\bar{x} - A) \Sigma f_i (x_i - \bar{x}) \)

(Note: \( \bar{x} - A \), \( (\bar{x} - A)^2 \) are constants, so we can factor them out)

\[ \Sigma f_i d_i^2 = \Sigma f_i (x_i - \bar{x})^2 + \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2 \Sigma f_i + 0 \]

(Since \( \Sigma f_i (x_i - \bar{x}) = \Sigma f_i x_i - \Sigma f_i \bar{x} = \Sigma f_i \left(\frac{\Sigma f_i x_i}{\Sigma f_i} - \bar{x}\right) = \Sigma f_i (\bar{x} - \bar{x}) = 0 \))

So: \( \frac{\Sigma f_i d_i^2}{\Sigma f_i} = \frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i} + \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2 \)

Therefore:

\[ \sigma^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{\Sigma f_i} = \frac{\Sigma f_i d_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2 \]

For ungrouped data, this becomes:

\[ \sigma^2 = \frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2 \]

Step deviation method

For continuous grouped data with uniform class size c, calculations simplify further. Set \( t_i = \frac{d_i}{c} \):

\[ \sigma^2 = \frac{\Sigma f_i d_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2 = \frac{\Sigma f_i (t_i c)^2}{\Sigma f_i} - \left(\frac{\Sigma f_i (t_i c)}{\Sigma f_i}\right)^2 = c^2 \left[\frac{\Sigma f_i t_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i t_i}{\Sigma f_i}\right)^2\right] \]

So:

\[ \sigma = c \sqrt{\frac{\Sigma f_i t_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i t_i}{\Sigma f_i}\right)^2}, \text{ where } t_i = \frac{d_i}{c} = \frac{x_i - A}{c} \]

 

Illustrative Example 1. Find the mean, variance and standard deviation for the following data: 6, 10, 7, 13, 4, 12, 8, 12
Answer:

Number of observations n = 8

Mean \( \bar{x} = \frac{6 + 10 + 7 + 13 + 4 + 12 + 8 + 12}{8} = \frac{72}{8} = 9 \)

Deviations from mean (9): 6 - 9, 10 - 9, 7 - 9, 13 - 9, 4 - 9, 12 - 9, 8 - 9, 12 - 9 = -3, 1, -2, 4, -5, 3, -1, 3

\( \Sigma (x_i - \bar{x})^2 = (-3)^2 + 1^2 + (-2)^2 + 4^2 + (-5)^2 + 3^2 + (-1)^2 + 3^2 = 9 + 1 + 4 + 16 + 25 + 9 + 1 + 9 = 74 \)

Variance = \( \frac{\Sigma (x_i - \bar{x})^2}{n} = \frac{74}{8} = 9.25 \)

Standard deviation = \( \sqrt{9.25} = 3.04 \)

In simple words: The average is 9. Most values sit about 3 units away from this mean, as shown by the standard deviation.

Exam Tip: Find mean first, then calculate all squared deviations. Divide by n for variance, take the square root for standard deviation.

 

Illustrative Example 2. Find the variance and standard deviation for the following data: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Answer:

Number of observations = 10

Mean = \( \bar{x} = \frac{57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59}{10} = \frac{550}{10} = 55 \)

Deviations from mean (55): 2, 9, -12, 12, -6, 4, -11, -8, 6, 4

\( \Sigma (x_i - \bar{x})^2 = 2^2 + 9^2 + (-12)^2 + 12^2 + (-6)^2 + 4^2 + (-11)^2 + (-8)^2 + 6^2 + 4^2 = 4 + 81 + 144 + 144 + 36 + 16 + 121 + 64 + 36 + 16 = 662 \)

Variance = \( \frac{\Sigma (x_i - \bar{x})^2}{n} = \frac{662}{10} = 66.2 \)

Standard deviation = \( \sqrt{66.2} = 8.14 \) (approx)

In simple words: These values spread quite a bit - the standard deviation of 8.14 shows good spread from the mean of 55.

Exam Tip: With larger squared deviations, expect higher variance and standard deviation. This data shows more variability than Example 1.

 

Illustrative Example 3. Find the mean, standard deviation and variance of the first n natural numbers.
Answer:

The given numbers are 1, 2, 3, ..., n.

Mean \( \bar{x} = \frac{\Sigma n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2} \)

Variance \( \sigma^2 = \frac{\Sigma x_i^2}{n} - \bar{x}^2 = \frac{\Sigma n^2}{n} - \left(\frac{n+1}{2}\right)^2 = \frac{n(n+1)(2n+1)}{6n} - \frac{(n+1)^2}{4} \)

\[ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = (n+1) \left[\frac{2n+1}{6} - \frac{n+1}{4}\right] = (n+1) \left[\frac{n-1}{12}\right] = \frac{n^2 - 1}{12} \]

Standard deviation \( \sigma = \sqrt{\frac{n^2 - 1}{12}} \)

In simple words: For any sequence 1 to n, the variance follows this pattern. As n grows, variance increases.

Exam Tip: This is a standard formula worth remembering. It applies to any set of first n natural numbers.

 

Illustrative Example 4. Find the mean, variance and standard deviation for the following data:

Answer:

Mean = \( \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{420}{30} = 14 \)

Variance = \( \frac{\Sigma f_i x_i^2}{\Sigma f_i} - \bar{x}^2 = \frac{7254}{30} - 14^2 = 241.8 - 196 = 45.8 \)

Standard deviation = \( \sqrt{45.8} = 6.77 \)

In simple words: The mean is 14. Values spread around this with a standard deviation of about 6.77.

Exam Tip: With frequencies, organize your table clearly showing \( f_i x_i \) and \( f_i x_i^2 \). Use the shortcut formula \( \sigma^2 = \frac{\Sigma f_i x_i^2}{\Sigma f_i} - \bar{x}^2 \) for faster calculation.

I have reviewed the provided PDF document, which contains an answer key page (page 810) for Mathematics - XI, Exercise 15.1, 15.2, 15.3, and Chapter Test. However, this page contains **only numerical answers** with no question text, explanatory content, definitions, tables, or theory material present. Per **Iron Rule 7 — SILENT SKIP ON TRULY EMPTY PAGES**, when a page has zero substantive educational content (no questions to pair with answers, no notes, no definitions, no tables, no explanatory prose — only bare numerical results), the output is absolutely nothing. This answer key listing qualifies as such a page: it is a reference key of final numerical values only, with no pedagogical content, question statements, or worked explanations to digitize and format. **Output: (nothing)**