GSEB Class 11 Maths Solutions Chapter 15 Statistics Exercise 15.1

GSEB Solutions

 

Find the mean deviation about the mean for the data in questions 1 and 2:

 

Question 1. Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17.
Answer: For the given data: 4, 7, 8, 9, 10, 12, 13, 17
The number of observations \( n = 8 \).
First, we find the arithmetic mean \( \bar{x} \):
\( \bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} \)
\( \bar{x} = 10 \)
Next, we calculate the sum of the absolute deviations from the mean:
\( \Sigma|x_i - \bar{x}| = |4-10| + |7-10| + |8-10| + |9-10| + |10-10| + |12-10| + |13-10| + |17-10| \)
\( = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 \)
\( = 24 \)
Finally, the mean deviation about the mean \( \text{M.D. } (\bar{x}) \) is:
\( \text{M.D. } (\bar{x}) = \frac{\Sigma|x_i - \bar{x}|}{n} = \frac{24}{8} \)
\( = 3 \)
In simple words: We first find the average of all numbers, which is 10. Then, we find how far each number is from this average, ignoring if it's bigger or smaller, and add up all these distances. Finally, we divide that total by how many numbers we have to get the mean deviation.

Exam Tip: Remember to calculate the mean correctly before finding the absolute deviations, and ensure you sum all deviations before dividing by the count of observations.

 

Question 2. Find the mean deviation about the mean for the data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Answer: For the given data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
The number of observations \( n = 10 \).
First, we find the mean \( \bar{x} \):
\( \bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} = \frac{500}{10} \)
\( \bar{x} = 50 \)
Next, we calculate the sum of the absolute deviations from the mean:
\( \Sigma|x_i - \bar{x}| = |38-50| + |70-50| + |48-50| + |40-50| + |42-50| + |55-50| + |63-50| + |46-50| + |54-50| + |44-50| \)
\( = 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 \)
\( = 84 \)
Finally, the mean deviation about the mean \( \text{M.D. } (\bar{x}) \) is:
\( \text{M.D. } (\bar{x}) = \frac{\Sigma|x_i - \bar{x}|}{n} = \frac{84}{10} \)
\( = 8.4 \)
In simple words: We first find the average of these numbers, which is 50. Then, we find the positive difference between each number and 50, and add up all those differences. Finally, we divide that total by the count of numbers (10) to get the mean deviation.

Exam Tip: Pay careful attention to arithmetic, especially when summing large numbers or performing subtractions for deviations. Double-checking your calculations can avoid common errors.

 

Find the mean deviation about the median for the data in questions 3 and 4:

 

Question 3. Find the mean deviation about the median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Answer: For the given data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
First, arrange the data in ascending order:
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
The number of observations \( n = 12 \).
Since \( n \) is even, the median (M) is the average of the \( (\frac{n}{2}) \)th and \( (\frac{n}{2} + 1) \)th items.
\( \frac{12}{2}\text{th item} = 6\text{th item} = 13 \).
\( (\frac{12}{2} + 1)\text{th item} = 7\text{th item} = 14 \).
Median \( M = \frac{13+14}{2} = \frac{27}{2} = 13.5 \).
Next, calculate the sum of the absolute deviations from the median:
\( \Sigma|x_i - M| = |10-13.5| + |11-13.5| + |11-13.5| + |12-13.5| + |13-13.5| + |13-13.5| + |14-13.5| + |16-13.5| + |16-13.5| + |17-13.5| + |17-13.5| + |18-13.5| \)
\( = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 \)
\( = 28 \).
Finally, the mean deviation about the median \( \text{M.D. (M)} \) is:
\( \text{M.D. (M)} = \frac{\Sigma|x_i - M|}{n} = \frac{28}{12} \)
\( = 2.33 \) (approximately).
In simple words: First, we arrange the numbers in order and find the middle value, which is the median (13.5). Then, we find how far each number is from the median, ignoring negative signs, and add up all these distances. Dividing this total by the number of values gives us the mean deviation from the median.

Exam Tip: When \( n \) is even, remember to take the average of the two middle terms for the median. Always arrange the data in order before finding the median.

 

Question 4. Find the mean deviation about the median for the data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Answer: For the given data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
First, arrange the data in ascending order:
36, 42, 45, 46, 46, 49, 51, 53, 60, 72.
The number of observations \( n = 10 \).
Since \( n \) is even, the median (M) is the average of the \( (\frac{n}{2}) \)th and \( (\frac{n}{2} + 1) \)th items.
\( \frac{10}{2}\text{th item} = 5\text{th observation} = 46 \).
\( (\frac{10}{2} + 1)\text{th item} = 6\text{th observation} = 49 \).
Median \( M = \frac{46+49}{2} = \frac{95}{2} = 47.5 \).
Next, calculate the sum of the absolute deviations from the median:
\( \Sigma|x_i - M| = |36-47.5| + |42-47.5| + |45-47.5| + |46-47.5| + |46-47.5| + |49-47.5| + |51-47.5| + |53-47.5| + |60-47.5| + |72-47.5| \)
\( = 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 \)
\( = 70 \).
Finally, the mean deviation about the median \( \text{M.D. (M)} \) is:
\( \text{M.D. (M)} = \frac{\Sigma|x_i - M|}{n} = \frac{70}{10} \)
\( = 7 \).
In simple words: First, we sort the numbers and find the median (47.5). Then, we calculate how far each number is from this median value, always using a positive distance, and add them all up. Dividing this total by the count of numbers (10) gives us the mean deviation.

Exam Tip: For longer data sets, carefully list out all absolute deviations to avoid missing any terms. A minor calculation error can affect the final result significantly.

 

Find the mean deviation about the median for the data in questions 7 to 8:

 

Question 7. Find the mean deviation about the median for the following data:

Answer: To find the mean deviation about the median, we first create a table to calculate cumulative frequency (c.f.) and the necessary sums:
The total frequency is \( N = \Sigma f_i = 26 \).
Since \( N \) is even, the median (M) is the average of the \( (\frac{N}{2}) \)th and \( (\frac{N}{2} + 1) \)th values.
\( \frac{N}{2}\text{th value} = \frac{26}{2}\text{th value} = 13\text{th value} \). From the c.f. column, the 13th value is 7.
\( (\frac{N}{2} + 1)\text{th value} = 14\text{th value} \). From the c.f. column, the 14th value is also 7.
Median \( M = \frac{7+7}{2} = 7 \).
From the table, the sum of absolute deviations multiplied by frequencies is \( \Sigma f_i|x_i - M| = 84 \).
Mean deviation from the median is:
\( \text{M.D. (M)} = \frac{\Sigma f_i|x_i - M|}{\Sigma f_i} = \frac{84}{26} = \frac{42}{13} \)
\( = 3.23 \) (approximately).
In simple words: First, we find the cumulative frequencies to locate the median, which is 7. Then, for each data point, we calculate how far it is from the median and multiply that by its frequency. We add up all these values and divide by the total frequency to get the mean deviation.

Exam Tip: Constructing the cumulative frequency table correctly is crucial for finding the median in grouped data. Ensure all columns for calculation are accurate.

 

Question 8. Find the mean deviation about the median for the following data:

Answer: To find the mean deviation about the median, we create a table for cumulative frequency (c.f.) and the required sums:
The total frequency is \( N = \Sigma f_i = 29 \). Since \( N \) is odd, the median is the \( (\frac{N+1}{2}) \)th observation.
\( (\frac{29+1}{2})\text{th observation} = 15\text{th observation} \).
From the c.f. column, the 15th observation falls in the class where \( x_i = 30 \) (because c.f. jumps from 14 to 21 at \( x_i = 30 \)). So, the median \( M = 30 \).
From the table, the sum of absolute deviations multiplied by frequencies is \( \Sigma f_i|x_i - M| = 148 \).
Mean deviation from the median is:
\( \text{M.D. (M)} = \frac{\Sigma f_i|x_i - M|}{\Sigma f_i} = \frac{148}{29} \)
\( = 5.1 \) (approximately).
In simple words: We add up the frequencies to get a total of 29. Since this is an odd number, we find the middle observation, which is the 15th, making the median 30. Then, we find the difference of each value from the median, multiply by its frequency, add all these up, and divide by the total frequency.

Exam Tip: Be careful when identifying the median for discrete frequency distributions. If \( N \) is odd, the median is a specific \( x_i \) value. If \( N \) is even, it's the average of two \( x_i \) values.

 

Find the mean deviation about the mean for the data in questions 9 and 10:

 

Question 9. Find the mean deviation about the mean for the income per day data:

Answer: To find the mean deviation about the mean, we construct the following table using the assumed mean method:
Here, the assumed mean \( A = 350 \), and the class size \( h = 100 \).
The mean \( \bar{x} \) is calculated using the formula:
\( \bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} \times h \)
\( = 350 + \frac{4}{50} \times 100 \)
\( = 350 + 8 \)
\( \bar{x} = 358 \)
From the table, the sum of \( f_i|x_i - \bar{x}| \) is \( 7896 \).
The total frequency \( \Sigma f_i = 50 \).
Mean deviation from the mean is:
\( \text{M.D. } (\bar{x}) = \frac{\Sigma f_i|x_i - \bar{x}|}{\Sigma f_i} = \frac{7896}{50} \)
\( = 157.92 \).
In simple words: We set an estimated average, then find how much each mid-point deviates from this estimate, adjusting for the class width. We calculate the true mean using a formula, then find the absolute difference of each mid-point from the true mean, multiply by its frequency, sum these, and divide by the total number of people.

Exam Tip: For grouped data, calculate mid-values and use the step-deviation method to simplify the mean calculation. Carefully compute deviations from the mean and their product with frequencies.

 

Question 10. Find the mean deviation about the mean for the height (in cm) data:

Answer: To find the mean deviation about the mean, we make the following table using the assumed mean method:
Here, the assumed mean \( A = 130 \), and the class size \( h = 10 \).
The mean \( \bar{x} \) is calculated using the formula:
\( \bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} \times h \)
\( = 130 + \frac{-47}{100} \times 10 \)
\( = 130 - 4.7 \)
\( \bar{x} = 125.3 \)
From the table, the sum of \( f_i|x_i - \bar{x}| \) is \( 1128.8 \).
The total frequency \( \Sigma f_i = 100 \).
Mean deviation from the mean is:
\( \text{M.D. } (\bar{x}) = \frac{\Sigma f_i|x_i - \bar{x}|}{\Sigma f_i} = \frac{1128.8}{100} \)
\( = 11.288 \).
In simple words: We choose an assumed average, then find how much each mid-point varies from it, adjusting for class size. We then calculate the actual mean. Finally, we find the absolute difference of each mid-point from the actual mean, multiply by its frequency, sum these, and divide by the total number of boys.

Exam Tip: Pay close attention to negative signs when calculating \( f_id_i \). A small error in summing \( f_id_i \) or in the mean calculation will lead to incorrect deviations.

 

Question 11. Find the mean deviation about median for the following data:

Answer: To find the mean deviation about the median, we first prepare a cumulative frequency table and then calculate the median:
The total number of observations \( N = \Sigma f_i = 50 \).
We find \( \frac{N}{2} = \frac{50}{2} = 25 \).
The cumulative frequency just greater than or equal to 25 is 28, which corresponds to the class interval 20-30. So, the median class is 20-30.
For the median class:
Lower limit \( l = 20 \).
Frequency of the median class \( f = 14 \).
Cumulative frequency of the class preceding the median class \( C = 14 \).
Class size \( h = 10 \).
Median \( M = l + \frac{\frac{N}{2}-C}{f} \times h \)
\( M = 20 + \frac{25-14}{14} \times 10 \)
\( M = 20 + \frac{11}{14} \times 10 \)
\( M = 20 + 7.857... \)
\( M \approx 27.86 \).
From the table, \( \Sigma f_i|x_i - M| = 517.16 \).
Mean deviation about the median is:
\( \text{M.D. (M)} = \frac{\Sigma f_i|x_i - M|}{N} = \frac{517.16}{50} \)
\( = 10.34 \) (approximately).
In simple words: We create a table with cumulative frequencies to find the median class and then calculate the exact median using its formula. After finding the median (27.86), we calculate how far each mid-point is from the median, multiply by its frequency, sum these values, and divide by the total number of girls.

Exam Tip: When dealing with continuous frequency distributions, correctly identify the median class, and then carefully apply the median formula using its lower limit, frequency, and preceding cumulative frequency.

 

Question 12. Calculate the mean deviation about median for the age distribution of 100 persons given below:

Answer: To compute the mean deviation about the median, we first convert the given discontinuous class intervals into continuous ones and create a cumulative frequency table:
The total number of observations \( N = \Sigma f_i = 100 \).
We find \( \frac{N}{2} = \frac{100}{2} = 50 \).
The cumulative frequency just greater than or equal to 50 is 63, which corresponds to the class interval 35.5-40.5. So, the median class is 35.5-40.5.
For the median class:
Lower limit \( l = 35.5 \).
Frequency of the median class \( f = 26 \).
Cumulative frequency of the class preceding the median class \( C = 37 \).
Class size \( h = 5 \).
Median \( M = l + \frac{\frac{N}{2}-C}{f} \times h \)
\( M = 35.5 + \frac{50-37}{26} \times 5 \)
\( M = 35.5 + \frac{13}{26} \times 5 \)
\( M = 35.5 + 2.5 \)
\( M = 38 \).
From the table, \( \Sigma f_i|x_i - M| = 735 \).
Mean deviation about the median is:
\( \text{M.D. (M)} = \frac{\Sigma f_i|x_i - M|}{N} = \frac{735}{100} \)
\( = 7.35 \).
In simple words: First, we adjust the age groups to be continuous and find the cumulative frequencies. We use these to find the median class and then calculate the exact median (38). Next, we find how far each mid-point is from this median, multiply by its frequency, add all these up, and divide by the total number of people.

Exam Tip: Remember to convert discontinuous class intervals into continuous ones by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit before calculating the median.