OP Malhotra Class 11 Maths Solutions Chapter 21 Measures of Dispersion Exercise 21 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Ex 21(a)

 

Question 1. 15, 17, 19, 25, 30, 35, 48
Answer: To find the mean deviation, we first arrange the given values and calculate the mean. The values are given as:

Here, we first calculate the mean \( \bar{x} \):
\( \bar{x} = \frac{\Sigma x_i}{n} = \frac{189}{7} = 27 \)
Now, we can find the Mean Deviation (M.D.) about the mean:
\( \text{M.D. about mean} = \frac{\Sigma |x_i - \bar{x}|}{n} = \frac{64}{7} \approx 9.143 \)
Next, we calculate the Coefficient of Mean Deviation:
\( \text{Coeff. of M.D} = \frac{\text{M.D}}{\text{Mean}} = \frac{64/7}{27} = \frac{64}{7 \times 27} = \frac{64}{189} \approx 0.339 \)
Mean deviation helps us understand how spread out the data points are from the average value.In simple words: First, we add all the numbers and divide by how many there are to get the average (mean). Then, for each number, we find how far it is from the average. We add up all these differences and divide by the total count again to get the mean deviation. The coefficient then shows this deviation as a fraction of the mean.

🎯 Exam Tip: Remember to calculate the mean accurately as the first step, as all subsequent calculations of deviation depend on it. Ensure absolute differences are used for mean deviation.

 

Question 2. 21, 23, 25, 28,30, 32, 38, 39, 46, 48
Answer: We begin by listing the data points and calculating their mean. The given values are:
Here, the mean \( \bar{x} \) is calculated as:
\( \bar{x} = \frac{21+23+25+28+30+32+38+39+46+48}{10} = \frac{330}{10} = 33 \)
Now, we construct a table to find the deviations from the mean:

Now, we can find the Mean Deviation (M.D.) about the mean:
\( \text{M.D. about Mean} = \frac{\Sigma|d_i|}{n} = \frac{78}{10} = 7.8 \)
Next, we calculate the Coefficient of Mean Deviation:
\( \text{Coeff. of M.D.} = \frac{\text{M.D.}}{\bar{x}} = \frac{7.8}{33} \approx 0.236 \)
These calculations help in understanding how much individual data points vary from the central tendency.In simple words: First, we find the average of all numbers. Then, for each number, we subtract the average to see how far it is, ignoring if it's bigger or smaller. We add up all these differences and divide by the total count to get the mean deviation. Finally, we divide the mean deviation by the average to get the coefficient, which shows the deviation relative to the mean.

🎯 Exam Tip: When calculating the mean, be careful to add all numbers correctly and divide by the exact count of observations. A small error in the mean will affect all subsequent deviation calculations.

 

Question 3. 10, 70, 50, 53, 20, 95, 55, 42, 60, 48, 80 deviation from the mean for the following frequency distributions.
Answer: First, we calculate the mean of the given data values. The values are:
\( \text{Mean } \bar{x} = \frac{10+70+50+53+20+95+55+42+60+48+80}{11} = \frac{583}{11} = 53 \)
Now, we create a table to calculate the deviations from the mean:

The required Mean Deviation (M.D.) about the mean is:
\( \text{M.D. about Mean} = \frac{\Sigma|d_i|}{n} = \frac{190}{11} \approx 17.27 \)
And the coefficient of Mean Deviation is:
\( \text{Coeff. of M.D} = \frac{\text{M.D}}{\text{Mean}} = \frac{190/11}{53} = \frac{190}{11 \times 53} = \frac{190}{583} \approx 0.326 \)
This method effectively quantifies the average spread of data points around the mean.In simple words: First, we add up all the numbers and divide by the count to get the average. Then, we find how far each number is from this average, always using a positive value. We add all these "how far" values and divide by the total count again to get the mean deviation. To find the coefficient, we divide the mean deviation by the average itself.

🎯 Exam Tip: Be mindful of negative signs when calculating deviations. The absolute value \(|d_i|\) is crucial for mean deviation, as it ensures all distances from the mean contribute positively to the total spread.

 

Question 4.
Answer: To find the mean deviation for the given frequency distribution, we first need to calculate the mean. We will set up a table for calculations:

First, calculate the mean \( \bar{x} \):
\( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{660}{44} = 15 \)
Now, we find the Mean Deviation (M.D.) about the mean:
\( \text{M.D about Mean} = \frac{\Sigma f_i |d_i|}{\Sigma f_i} = \frac{312}{44} \approx 7.09 \)
In frequency distributions, each data point's deviation is weighted by its frequency, providing a more accurate measure of spread.In simple words: For data with frequencies, first multiply each value by its frequency and add them up, then divide by the total frequency to get the mean. Next, find how far each value is from the mean. Multiply this distance by the frequency of that value. Add these up and divide by the total frequency again to get the mean deviation.

🎯 Exam Tip: When dealing with frequency distributions, remember to multiply each deviation by its corresponding frequency before summing them up. This ensures the correct weighted average for the mean deviation.

 

Question 5.
Answer: For this frequency distribution, we first find the mean. We will prepare a table to facilitate the calculations:

First, calculate the mean \( \bar{x} \):
\( \bar{x} = \frac{576}{48} = 12 \)
Now, we find the Mean Deviation (M.D.) about the mean:
\( \text{M.D about Mean} = \frac{\Sigma f_i |d_i|}{48} = \frac{36}{48} = \frac{3}{4} = 0.75 \)
This provides a clear numerical measure of how much the values in the distribution vary from their average.In simple words: To find the mean deviation, first calculate the average by summing all values (multiplied by their frequency) and dividing by the total frequency. Then, for each value, find its absolute distance from the average. Multiply this distance by the value's frequency. Sum these products and divide by the total frequency again to get the mean deviation.

🎯 Exam Tip: Always verify that the sum of frequencies (\(\Sigma f_i\)) is correct, as this is used for both mean and mean deviation calculations. Errors here will propagate throughout the solution.

 

Question 6.
Answer: We need to calculate the mean deviation for this grouped frequency distribution. First, we find the mid-points for each class interval, then the mean, and finally the mean deviation. The table of values is constructed as:

Using the direct method, the mean \( \bar{x} \) is:
\( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1350}{50} = 27 \)
Now, the Mean Deviation (M.D.) from the mean is:
\( \text{M.D from mean} = \frac{\Sigma f_i|x_i - \bar{x}|}{\Sigma f_i} = \frac{472}{50} = 9.44 \)
This method is commonly used for analyzing data spread in grouped distributions.In simple words: For grouped data, find the middle point of each group. Then, find the average of these mid-points, considering how many times each group appears. After that, calculate how far each mid-point is from this average. Multiply these distances by their group's count, add them up, and divide by the total count again to get the mean deviation.

🎯 Exam Tip: When dealing with grouped data, always correctly identify the class mark (mid-point) for each interval. Any error in the class mark will lead to incorrect calculations for the mean and mean deviation.

 

Question 7.
Answer: To calculate the mean deviation for this grouped data, we first determine the mid-points of the scores, calculate the mean, and then the deviations. The calculation table is as follows:

First, we calculate the mean \( \bar{x} \):
\( \text{Mean } \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{8560}{50} = 171.2 \)
Then, the Mean Deviation (M.D.) about the mean is:
\( \text{M.D about mean} = \frac{\Sigma f_i|d_i|}{\Sigma f_i} = \frac{528}{50} = 10.56 \)
This provides a measure of the typical deviation of scores from the average score in the distribution.In simple words: For data given in groups, find the middle value of each group. Calculate the overall average using these middle values and how many items are in each group. Then, find how far each middle value is from the overall average. Multiply this distance by the count in its group, add all these products, and finally divide by the total count to get the mean deviation.

🎯 Exam Tip: For continuous class intervals, the mid-point is calculated as (lower limit + upper limit) / 2. Ensure these are accurately determined before proceeding with further calculations.

 

Question 8.
Answer: To find the mean deviation for this grouped frequency distribution, we first need to calculate the mid-points for each class interval, then the mean, and finally the mean deviation. The table is set up as follows:

First, calculate the mean \( \bar{x} \):
\( \text{Mean } \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{9200}{182} \approx 50.55 \)
Then, the Mean Deviation (M.D.) about the mean is:
\( \text{M.D about mean} = \frac{\Sigma f_i|d_i|}{\Sigma f_i} = \frac{2390.6}{182} \approx 13.135 \)
This method is particularly helpful for summarizing the variability in large datasets presented as grouped frequency distributions.In simple words: For grouped data, find the middle value of each class. Calculate the average by multiplying each mid-value by its frequency, adding them up, and dividing by the total frequency. Then, for each mid-value, find its positive difference from the average. Multiply this difference by its frequency, add all these products, and divide by the total frequency to get the mean deviation.

🎯 Exam Tip: Rounding errors can accumulate in calculations involving decimals. It's often best to keep fractions until the final step or use sufficient decimal places to maintain accuracy, especially when dealing with mean deviation of grouped data.

 

Question 9. 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Answer: To find the mean deviation about the median, we first arrange the data in ascending order and find the median. The given data points are:
Arranging the data in ascending order, we get: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21.
The number of observations \(n = 11\) (which is an odd number).
The median \( M_d \) is the \( \left(\frac{n+1}{2}\right)\mathrm{th} \) observation.
So, \( M_d = \left(\frac{11+1}{2}\right)\mathrm{th} \text{ observation} = 6\mathrm{th} \text{ observation} = 9 \)
Now, we calculate the absolute deviations from the median:

The Mean Deviation (M.D.) about the median is:
\( \text{M.D about median} = \frac{\Sigma|x_i - M_d|}{n} = \frac{58}{11} \approx 5.273 \)
And the Coefficient of Mean Deviation is:
\( \text{Coeff. of M.D} = \frac{\text{M.D}}{\text{Median}} = \frac{58/11}{9} = \frac{58}{11 \times 9} = \frac{58}{99} \approx 0.586 \)
The median is less affected by extreme values than the mean, making mean deviation about the median a robust measure of dispersion.In simple words: First, sort the numbers from smallest to largest and find the middle number, which is the median. Then, for each number, find how far it is from the median. Add up all these differences and divide by the total count to get the mean deviation about the median. To find the coefficient, divide this mean deviation by the median itself.

🎯 Exam Tip: Always arrange data in ascending order to find the median correctly. For an odd number of observations, the median is the middle value; for an even number, it's the average of the two middle values.

 

Question 10. 100, 150, 200, 250, 360, 490, 500, 600, 671
Answer: To find the mean deviation about the median, we first arrange the data and determine the median. The data is already given in ascending order:
100, 150, 200, 250, 360, 490, 500, 600, 671
The number of observations \(n = 9\) (which is an odd number).
The median \( M_d \) is the \( \left(\frac{n+1}{2}\right)\mathrm{th} \) observation.
So, \( M_d = \left(\frac{9+1}{2}\right)\mathrm{th} \text{ observation} = 5\mathrm{th} \text{ observation} = 360 \)
Now, we calculate the absolute deviations from the median:

The Mean Deviation (M.D.) about the median is:
\( \text{M.D about median} = \frac{\Sigma|x_i - M_d|}{n} = \frac{1561}{9} \approx 173.44 \)
And the Coefficient of Mean Deviation is:
\( \text{Coeff. of M.D} = \frac{\text{M.D}}{\text{Median}} = \frac{1561/9}{360} = \frac{1561}{9 \times 360} = \frac{1561}{3240} \approx 0.482 \)
Using the median for deviation calculations is especially useful when the data contains outliers that might skew the mean.In simple words: First, find the middle number (median) of the given data. Then, for each data point, calculate how far it is from this median, ignoring the sign. Add up all these distances and divide by the total number of data points to get the mean deviation. To find the coefficient, divide this mean deviation by the median itself.

🎯 Exam Tip: Always double-check the order of operations when calculating the coefficient of mean deviation. Remember to divide the mean deviation by the median value.

 

Question 11.
Answer: To find the mean deviation about the median for this frequency distribution, we first calculate the cumulative frequencies to find the median. Then we calculate deviations from the median. The table is constructed as follows:

Here, the total number of observations \(n = \Sigma f_i = 48\) (which is an even number).
The median \( M_d \) is the average of the \( \left(\frac{n}{2}\right)\mathrm{th} \) and \( \left(\frac{n}{2}+1\right)\mathrm{th} \) observations.
\( M_d = \frac{ \left(\frac{48}{2}\right)\mathrm{th} \text{ obs} + \left(\frac{48}{2}+1\right)\mathrm{th} \text{ obs} }{2} = \frac{24\mathrm{th} \text{ obs} + 25\mathrm{th} \text{ obs}}{2} \)
From the cumulative frequency table, both the 24th and 25th observations fall in the class corresponding to \(x_i = 12\). So, the value is 12.
\( M_d = \frac{12+12}{2} = 12 \)
Now, the Mean Deviation (M.D.) about the median is:
\( \text{M.D about median} = \frac{\Sigma f_i|x_i - M_d|}{\Sigma f_i} = \frac{36}{48} = 0.75 \)
And the Coefficient of Mean Deviation is:
\( \text{Coeff. of M.D} = \frac{\text{M.D}}{\text{Median}} = \frac{0.75}{12} = 0.0625 \)
Calculating the median and mean deviation for discrete frequency distributions provides insights into data concentration and spread.In simple words: First, calculate the cumulative frequencies to find the median. Since the total frequency is even, the median is the average of the two middle values. Then, for each value, find its positive difference from the median. Multiply this difference by its frequency, add all these products, and divide by the total frequency to get the mean deviation. The coefficient is this mean deviation divided by the median.

🎯 Exam Tip: For even numbers of observations in frequency distributions, the median is the average of the \( (n/2)\mathrm{th} \) and \( (n/2+1)\mathrm{th} \) observation, found by looking at the cumulative frequency column.

 

Question 12.
Answer: To find the mean deviation about the median for this frequency distribution, we first compute the cumulative frequencies to locate the median. Then we calculate the deviations from this median. The table is presented as follows:

Here, the total number of observations \(n = \Sigma f = 30\) (which is an even number).
The median \( M_d \) is the average of the \( \left(\frac{n}{2}\right)\mathrm{th} \) and \( \left(\frac{n}{2}+1\right)\mathrm{th} \) observations.
\( M_d = \frac{ \left(\frac{30}{2}\right)\mathrm{th} \text{ obs} + \left(\frac{30}{2}+1\right)\mathrm{th} \text{ obs} }{2} = \frac{15\mathrm{th} \text{ obs} + 16\mathrm{th} \text{ obs}}{2} \)
From the cumulative frequency table, both the 15th and 16th observations fall in the class corresponding to \(x = 13\). So, the value is 13.
\( M_d = \frac{13+13}{2} = 13 \)
Now, the Mean Deviation (M.D.) about the median is:
\( \text{M.D} = \frac{\Sigma f|x - M_d|}{\Sigma f} = \frac{149}{30} \approx 4.967 \)
And the Coefficient of Mean Deviation is:
\( \text{Coeff. of M.D} = \frac{\text{M.D}}{\text{Median}} = \frac{4.967}{13} \approx 0.382 \)
This provides a measure of how spread out the data points are around the central median value.In simple words: First, find the median by looking at the cumulative frequencies. If the total frequency is even, the median is the average of the two middle values. Then, find the positive difference of each data point from this median. Multiply each difference by its frequency, add them all up, and divide by the total frequency to get the mean deviation. The coefficient is this mean deviation divided by the median.

🎯 Exam Tip: When using cumulative frequency to find the median for even 'n', ensure you pick the correct values for both \( (n/2)\mathrm{th} \) and \( (n/2+1)\mathrm{th} \) observations from the \(x\) column corresponding to their cumulative frequencies.