RBSE Solutions Class 11 Maths Chapter 13 Measures of Dispersion Exercise 13.2

Find the Mean Deviation From Mean for the Data Given in Q1 and Q2.

Question 1. 4, 7, 8, 9, 10, 12, 13, 17
Answer:

Here, \( N = 8 \)
\( \Sigma x_i = 80 \)
Mean \( (\bar{x}) = \frac { \Sigma x_i }{ N } = \frac { 80 }{ 8 } = 10 \)
\( \Sigma |x_i - \bar{x}| = 24 \)
Mean deviation \( = \frac { \Sigma |x_i - \bar{x}| }{ N } = \frac { 24 }{ 8 } = 3 \)
In simple words: First, we find the average (mean) of all the numbers. Then, for each number, we find how far it is from the mean (ignoring if it's bigger or smaller). Finally, we average these distances to get the mean deviation.

🎯 Exam Tip: Always double-check your mean calculation, as any error there will affect all subsequent deviation values.

Question 2. For the following data, find the mean deviation from the mean: 60, 38, 30, 32, 45, 53, 36, 44, 34.
Answer:

Here, \( N = 10 \)
Mean \( (\bar{x}) = \frac { \Sigma x_i }{ N } = \frac { 400 }{ 10 } = 40 \)
Mean deviation \( = \frac { \Sigma |x_i - \bar{x}| }{ N } = \frac { 84 }{ 10 } = 8.4 \)
In simple words: We calculate the mean first, which is 40. Then, for each number, we find how much it differs from 40, taking only the positive difference. Finally, we sum these differences and divide by the total count of numbers (10) to get the average deviation.

🎯 Exam Tip: Remember that \( N \) represents the total count of data points, even if some raw data points are not explicitly listed in an intermediate step, ensure your calculations align with the given \( \Sigma x_i \) and \( N \) values.

Find the Mean Deviation From Median for the Data Given in Q. 3 and Q.4.

Question 3. 13, 10, 12, 13, 15, 18, 17, 11, 14, 16, 12
Answer:
Arranging the data in ascending order, the values are:
10, 11, 12, 12, 13, 13, 14, 15, 16, 17, 18
Here, \( n = 11 \). Since \( n \) is odd, the median is the \( \frac{n+1}{2} \)th term.
Median \( (M) = \frac{11+1}{2} \)th term \( = 6 \)th term \( = 13 \).

Mean deviation from median \( = \frac { \Sigma |x_i - M| }{ N } = \frac { 22 }{ 11 } = 2 \)
In simple words: We first put the numbers in order and find the middle number, which is the median (13). Then we calculate how far each number is from this median. Finally, we average these distances to find the mean deviation from the median.

🎯 Exam Tip: When finding the median for an odd number of data points, it's simply the middle value after arranging the data in order. For an even number, it's the average of the two middle values.

Question 4. 26, 32, 35, 39, 41, 62, 36, 50, 43.
Answer:
Arranging the data in ascending order, the values are:
26, 32, 35, 36, 39, 41, 43, 50, 62
Here, \( N = 9 \). Since \( N \) is odd, the median is the \( \frac{N+1}{2} \)th term.
Median \( (M) = \frac{9+1}{2} \)th term \( = 5 \)th term \( = 39 \).

Mean deviation from median \( = \frac { \Sigma |x_i - M| }{ N } = \frac { 67 }{ 9 } = 7.44 \)
In simple words: After arranging the numbers from smallest to largest, the middle number is 39, which is our median. We then find how far each number is from 39, add all these distances, and divide by the total count of numbers (9) to get the mean deviation.

🎯 Exam Tip: Always sort the data first when calculating the median; otherwise, you might pick the wrong middle value.

Find the Mean Deviation From Mode for the Data Given in Q. 5. and Q. 6.

Question 5. 2, 4, 6, 4, 8, 6, 4, 10, 4, 8
Answer:
In the given data, the number 4 appears a maximum of 4 times.
So, Mode \( (z) = 4 \).

Thus, mean deviation from mode \( (\delta_z) = \frac { \Sigma |x_i - z| }{ N } = \frac { 20 }{ 10 } = 2 \)
In simple words: The mode is the number that appears most often, which is 4 in this case. We then find how far each number is from the mode, add these differences, and divide by the total number of items (10) to find the mean deviation from the mode.

🎯 Exam Tip: The mode is simply the most frequent value in the data set, so identify it correctly before calculating deviations.

Question 6. 2.2, 2.5, 2.1, 2.5, 2.9, 2.8, 2.5, 2.3
Answer:
In the given data, the number 2.5 appears a maximum of 3 times.
So, Mode \( (z) = 2.5 \).

Thus, Mean deviation \( (\delta_z) \) about mode \( = \frac { \Sigma |x_i - z| }{ N } = \frac { 1.6 }{ 8 } = 0.2 \)
In simple words: Here, the most frequent number (mode) is 2.5. We find the difference between each data point and 2.5, add up these differences, and then divide by the total number of data points (8) to get the mean deviation from the mode.

🎯 Exam Tip: Be careful with decimal calculations, ensuring precision when finding deviations and sums.

Find the Mean Deviation From Mean for the Data Given in Q. 7 to Q. 8.

Question 7. The data is given in the table below.

Answer:

Mean \( (\bar{x}) = \frac { \Sigma f_i x_i }{ N } = \frac { 350 }{ 25 } = 14 \)
Mean deviation \( (M.D.) = \frac { \Sigma f_i |x_i - \bar{x}| }{ N } = \frac { 158 }{ 25 } = 6.32 \)
In simple words: For grouped data, we first find the mean by multiplying each \( x_i \) by its frequency \( f_i \) and summing them up, then dividing by the total frequency \( N \). After finding the mean (14), we calculate how far each \( x_i \) is from the mean, multiply that by its frequency, sum those results, and divide by \( N \) to get the mean deviation.

🎯 Exam Tip: When dealing with frequency distributions, remember to multiply the absolute deviation by the frequency of each class before summing them up.

Question 8. The data is given in the table below.

Answer:

Mean \( (\bar{x}) = \frac { \Sigma f_i x_i }{ N } = \frac { 2400 }{ 40 } = 60 \)
Mean deviation \( (M.D.) = \frac { \Sigma f_i |x_i - \bar{x}| }{ N } = \frac { 640 }{ 40 } = 46 \)
In simple words: First, we find the mean (average) of the data, which is 60. Then, for each data point, we find its absolute difference from the mean, multiply by its frequency, sum all these products, and finally divide by the total number of data points (40) to get the mean deviation.

🎯 Exam Tip: Always construct a detailed table with \( f_i x_i \) and \( f_i |x_i - \bar{x}| \) columns to avoid calculation errors for grouped data.

Question 9. For the following data, find the mean deviation from the median:

Answer:

Here, \( N = 26 \). The median is the value corresponding to the \( \frac{N}{2} \)th term and \( (\frac{N}{2} + 1) \)th term.
\( \frac{N}{2} = \frac{26}{2} = 13 \)
And the next term is 14.
The 13th and 14th terms both fall into the class with \( x_i = 7 \) (since cumulative frequency up to 7 is 14).
So, Median \( (M) = 7 \).
Mean deviation \( = \frac { \Sigma f_i |x_i - M| }{ N } = \frac { 84 }{ 26 } = 3.23 \)
In simple words: For this grouped data, we find the median by looking at the cumulative frequencies to locate the middle value, which is 7. Then, we calculate the absolute difference of each \( x_i \) from this median, multiply by its frequency, sum them all, and finally divide by the total number of items \( N \) to get the mean deviation.

🎯 Exam Tip: For discrete frequency distributions, locate the median by finding the value corresponding to the \( \frac{N}{2} \)th or \( (\frac{N+1}{2}) \)th cumulative frequency.

Question 10. The data is given in the table below.

Answer:

Here, \( N = 29 \). Since \( N \) is odd, the median is the \( \frac{N+1}{2} \)th term.
Median \( (M) = \frac{29+1}{2} \)th term \( = 15 \)th term.
From the cumulative frequency column, the 15th term is 25.
So, Median \( (M) = 25 \).
Mean deviation \( = \frac { \Sigma f_i |x_i - M| }{ N } = \frac { 148 }{ 29 } = 5.1 \)
In simple words: First, we find the median, which is 25. Then we calculate how far each data point \( x_i \) is from the median, multiply this difference by its frequency \( f_i \), sum all these products, and divide by the total frequency \( N \) to get the mean deviation.

🎯 Exam Tip: For calculating the median in a frequency distribution, identify the position of the median term using the cumulative frequency.

Find the Mean Deviation From Mode for Data Given in Q. 11 to Q. 12.

Question 11. The data is given in the table below.

Answer:

Mode \( (z) = 5 \) (since it has the highest frequency of 6).
Mean deviation \( = \frac { \Sigma f_i |x_i - z| }{ \Sigma f_i } = \frac { 18 }{ 18 } = 1 \)
In simple words: The mode (most frequent value) is 5. We calculate how far each \( x_i \) is from 5, multiply by its frequency, sum these products, and then divide by the total number of items \( \Sigma f_i \) to get the mean deviation from the mode.

🎯 Exam Tip: For frequency distributions, the mode is simply the \( x_i \) value with the highest frequency \( f_i \).

Question 12. The data is given in the table below.

Answer:

Mode \( (z) = 40 \) (since it has the highest frequency of 26).
Mean deviation \( = \frac { \Sigma f_i |x_i - z| }{ \Sigma f_i } = \frac { 1310 }{ 100 } = 13.1 \)
In simple words: The number 40 appears most often, making it the mode. We calculate how much each \( x_i \) is different from 40, multiply by its frequency, add all these values, and then divide by the total number of items (100) to find the mean deviation from the mode.

🎯 Exam Tip: Always clearly identify the mode (the class or value with the highest frequency) before starting deviation calculations.

Find the Mean Deviation From Mean for Data Given in Q. 13. and Q. 14.

Question 13. The data is given in the table below.

Answer:

Mean \( (\bar{x}) = \frac { \Sigma f_i x_i }{ N } = \frac { 1790 }{ 50 } = 35.8 \)
Mean deviation \( = \frac { \Sigma f_i |x_i - \bar{x}| }{ N } = \frac { 789.6 }{ 50 } = 15.792 \)
In simple words: For data in class intervals, we first find the midpoint \( x_i \) for each interval. Then we calculate the mean by summing \( f_i x_i \) and dividing by total frequency. After finding the mean (35.8), we determine each midpoint's absolute distance from the mean, multiply by its frequency, sum these values, and divide by \( N \) to find the mean deviation.

🎯 Exam Tip: When dealing with class intervals, always use the midpoint of each interval to represent \( x_i \) for calculations.

Question 14. The data is given in the table below.

Answer:

Mean \( (\bar{x}) = \frac { \Sigma f_i x_i }{ N } = \frac { 12530 }{ 100 } = 125.3 \)
Mean deviation \( = \frac { \Sigma f_i |x_i - \bar{x}| }{ N } = \frac { 1128.8 }{ 100 } = 11.28 \)
In simple words: For data given in ranges, we first find the middle value of each range. Then we calculate the mean by summing (midpoint x frequency) and dividing by total frequency. After finding the mean (125.3), we compute the absolute difference of each midpoint from the mean, multiply by its frequency, sum these products, and finally divide by the total number of items \( N \) to get the mean deviation.

🎯 Exam Tip: Always make sure to use the correct midpoint for each class interval in your calculations to ensure accuracy.

Find the Mean Deviation From Median for Data Given in Q. 15. and Q. 16.

Question 15. The data is given in the table below.

Answer:

Calculation of Median:
\( N = 25 \). Since \( N \) is odd, the median position is the \( \frac{N+1}{2} \)th term.
\( = \frac{25+1}{2} \)th term \( = 13 \)th term.
From the cumulative frequency, the 13th term falls in the class interval 30-40. So, the median class is 30-40.
Here, \( l = 30 \) (lower limit of median class)
\( h = 40 - 30 = 10 \) (class width)
\( F = 7 \) (cumulative frequency of class preceding median class)
\( f = 7 \) (frequency of median class)
Median \( (M) = l + \frac { (\frac{N}{2} - F) }{ f } \times h \)
\( = 30 + \frac { (\frac{25}{2} - 7) }{ 7 } \times 10 \)
\( = 30 + \frac { (12.5 - 7) }{ 7 } \times 10 \)
\( = 30 + \frac { 5.5 }{ 7 } \times 10 \)
\( = 30 + \frac { 55 }{ 7 } \)
\( = 30 + 7.857 \approx 37.86 \)
Mean deviation \( (M.D.) = \frac { \Sigma f_i |x_i - M| }{ N } = \frac { 258.58 }{ 25 } = 10.34 \)
In simple words: First, we find the median class by looking at cumulative frequencies. Then we use a special formula to calculate the exact median (M = 37.86). After that, we find how far each class midpoint is from this median, multiply by its frequency, sum them, and divide by the total number of items \( N \) to get the mean deviation.

🎯 Exam Tip: For continuous frequency distributions, remember to use the formula for median \( M = l + \frac { (\frac{N}{2} - F) }{ f } \times h \) and identify all variables correctly.

Question 16. The data is given in the table below.

Answer:

Calculation of Median:
\( N = 100 \). Since \( N \) is even, the median position is between the \( \frac{N}{2} \)th term and \( (\frac{N}{2} + 1) \)th term.
\( \frac{N}{2} = \frac{100}{2} = 50 \). So, it's the 50th term.
From the cumulative frequency, the 50th term falls in the class interval 36-40. So, the median class is 36-40.
Here, \( l = 36 \) (lower limit of median class)
\( h = 4 \) (class width)
\( F = 37 \) (cumulative frequency of class preceding median class)
\( f = 26 \) (frequency of median class)
Median \( (M) = l + \frac { (\frac{N}{2} - F) }{ f } \times h \)
\( = 36 + \frac { (\frac{100}{2} - 37) }{ 26 } \times 4 \)
\( = 36 + \frac { (50 - 37) }{ 26 } \times 4 \)
\( = 36 + \frac { 13 }{ 26 } \times 4 \)
\( = 36 + \frac { 1 }{ 2 } \times 4 \)
\( = 36 + 2 = 38 \)
Mean deviation \( = \frac { \Sigma f_i |x_i - M| }{ N } = \frac { 735 }{ 100 } = 7.35 \)
In simple words: We identify the median class using cumulative frequency and calculate the median using the formula. After finding the median (38), we find the absolute difference of each class midpoint \( x_i \) from the median, multiply by its frequency \( f_i \), sum these products, and then divide by the total items \( N \) to get the mean deviation.

🎯 Exam Tip: Always verify that the sum of frequencies matches \( N \) for accurate median and mean deviation calculations.

Question 18. Calculate the mean deviation from the mode for the given data.
Answer: We need to calculate the mean deviation from the mode. From the provided solution steps, the following parameters are used:
Lower limit of the modal class \( (l) = 58 \)
Frequency of the modal class \( (f_1) = 35 \)
Frequency of the class preceding the modal class \( (f_0) = 20 \)
Frequency of the class succeeding the modal class \( (f_2) = 10 \)
Class interval width \( (h) = 3 \)
Total number of observations \( (N) = 75 \)
Sum of \( f|x-z| \) for mean deviation calculation \( (\Sigma f|x-z|) = 155.625 \)

First, we calculate the Mode \( (z) \):
\[ \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \]
\[ = 58 + \frac{35 - 20}{2(35) - 20 - 10} \times 3 \]
\[ = 58 + \frac{15}{70 - 20 - 10} \times 3 \]
\[ = 58 + \frac{15}{40} \times 3 \]
\[ = 58 + \frac{3}{8} \times 3 \]
\[ = 58 + \frac{9}{8} \]
\[ = 58 + 1.125 \]
\[ = 59.125 \]

Next, we calculate the Mean Deviation from the Mode \( (\delta_z) \):
\[ \text{Mean deviation} (\delta_z) = \frac{\Sigma f|x-z|}{N} \]
\[ = \frac{155.625}{75} \]
\[ = 2.075 \]
The mode helps to find the most frequent observation or class within a dataset, which is important for understanding the central tendency, especially for data that is not symmetrical.
In simple words: We first found the mode, which is the value that appears most often, using a special formula for grouped data. Then, we calculated how much each data point was different from this mode, multiplied it by how many times it appeared, and finally divided by the total number of data points. This gives us the average difference from the most common value.

🎯 Exam Tip: When calculating mode and mean deviation for grouped data, always ensure you correctly identify the modal class and its surrounding frequencies (\(f_0, f_1, f_2\)) to use in the formula. Double-check your arithmetic, especially when dealing with decimals.