OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Exercise 28 (B)

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(b)

 

Question 1. Find the median of the following sets of data:
(i) 2, 3, 5, 7, 9
(ii) 4, 8, 12, 16, 20, 23, 28, 32
(iii) 60, 33, 63, 61, 44, 48, 51
(iv) 13, 22, 25, 8, 11, 19, 17, 31, 16, 10
Answer:
(i) First, arrange the numbers in increasing order: 2, 3, 5, 7, 9.
There are \( n = 5 \) observations, which is an odd number.
The median is the size of the \( \left(\frac{n+1}{2}\right) \)th observation.
\( \implies \) Median \( = \) size of \( \left(\frac{5+1}{2}\right) \)th term
\( \implies \) Median \( = \) size of the 3rd term
\( \implies \) Median \( = 5 \).
(ii) The numbers are already in increasing order: 4, 8, 12, 16, 20, 23, 28, 32.
There are \( n = 8 \) observations, which is an even number.
The median is the average of the \( \left(\frac{n}{2}\right) \)th term and the \( \left(\frac{n}{2}+1\right) \)th term.
\( \implies \) Median \( = \frac{\text{4th term} + \text{5th term}}{2} \)
\( \implies \) Median \( = \frac{16+20}{2} \)
\( \implies \) Median \( = \frac{36}{2} \)
\( \implies \) Median \( = 18 \).
(iii) First, arrange the numbers in increasing order: 33, 44, 48, 51, 60, 61, 63.
There are \( n = 7 \) observations, which is an odd number.
The median is the size of the \( \left(\frac{n+1}{2}\right) \)th observation.
\( \implies \) Median \( = \) size of \( \left(\frac{7+1}{2}\right) \)th term
\( \implies \) Median \( = \) size of the 4th term
\( \implies \) Median \( = 51 \).
(iv) First, arrange the numbers in increasing order: 8, 10, 11, 13, 16, 17, 19, 22, 25, 31.
There are \( n = 10 \) observations, which is an even number.
The median is the average of the \( \left(\frac{n}{2}\right) \)th term and the \( \left(\frac{n}{2}+1\right) \)th term.
\( \implies \) Median \( = \frac{\text{5th term} + \text{6th term}}{2} \)
\( \implies \) Median \( = \frac{16+17}{2} \)
\( \implies \) Median \( = \frac{33}{2} \)
\( \implies \) Median \( = 16.5 \).
In simple words: To find the median, first put all the numbers in order from smallest to largest. If there's an odd number of items, the median is the middle one. If there's an even number, take the two middle numbers, add them up, and divide by two. The median shows the central value of the data.

🎯 Exam Tip: Always arrange the data in ascending or descending order before attempting to find the median. A common mistake is to find the middle value without sorting first.

 

Question 2. Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median?
Answer: First, let's arrange the given data in ascending order:
41, 43, 57, 58, 61, 71, 92, 99, 127.
Here, the number of observations is \( n = 9 \) (an odd number).
The median is the size of the \( \left(\frac{n+1}{2}\right) \)th term.
\( \implies \) Median \( = \) size of \( \left(\frac{9+1}{2}\right) \)th term
\( \implies \) Median \( = \) size of the 5th term
\( \implies \) Median \( = 61 \).
Now, if the observation 58 is replaced by 85, the new data in ascending order will be:
41, 43, 57, 61, 71, 85, 92, 99, 127. (Note: The value 58 is replaced by 85, and 85 is placed correctly in order.)
The number of observations is still \( n = 9 \) (odd).
The new median will be the size of the 5th term.
\( \implies \) New median \( = 71 \).
In simple words: First, line up all the numbers from smallest to largest and find the middle number. That's the first median. Then, change one number and put all the numbers in order again. Find the new middle number, and that's the new median. The median is a good way to see the center of your data.

🎯 Exam Tip: Remember to re-sort the entire data set when an observation is changed or added. The median changes based on the position of numbers.

 

Question 3. Calculate the median of the following incomes.
Answer: To calculate the median for this grouped data, we first need to create a cumulative frequency (c.f.) table.

The total number of observations is \( N = 115 \).
The median position is the \( \left(\frac{N+1}{2}\right) \)th item.
\( \implies \) Median position \( = \left(\frac{115+1}{2}\right) \)th item
\( \implies \) Median position \( = \left(\frac{116}{2}\right) \)th item
\( \implies \) Median position \( = 58 \)th item.
Looking at the c.f. column, the 58th item falls within the cumulative frequency of 60, which corresponds to a wage of Rs. 24.
Therefore, the median income of workers is Rs. 24. This represents the point where half the workers earn more and half earn less.
In simple words: First, make a list that shows how many workers earn 'up to' a certain wage. Then, find the middle position in this list. The wage at that middle position is the median wage.

🎯 Exam Tip: For discrete data, find the \( \left(\frac{N+1}{2}\right) \)th item and locate it in the cumulative frequency column to determine the median value.

 

Question 4. Compute the median of the following distributions:
Answer: To compute the median for this distribution, we first need to create a cumulative frequency (c.f.) table.

The total number of observations is \( N = 63 \).
The median position is the \( \left(\frac{N+1}{2}\right) \)th item.
\( \implies \) Median position \( = \left(\frac{63+1}{2}\right) \)th item
\( \implies \) Median position \( = \left(\frac{64}{2}\right) \)th item
\( \implies \) Median position \( = 32 \)th item.
Looking at the c.f. column, the 32nd item falls within the cumulative frequency of 33, which corresponds to an 'x' value of 5.
Thus, the median of the distribution is 5.
In simple words: Make a list of how many times each 'x' value appears, and then a running total. Find the middle spot in the running total, and the 'x' value next to it is your median. The median helps us understand the typical value in a dataset.

🎯 Exam Tip: For discrete frequency distributions, calculate the cumulative frequency and then use the formula for the median position \( \left(\frac{N+1}{2}\right) \).

 

Question 5. Marks obtained by 38 students are given below. Calculate the median marks:
Answer: To calculate the median marks, we first need to arrange the data and create a cumulative frequency (c.f.) table.

The total number of observations (students) is \( N = 38 \).
Since \( N \) is an even number, the median is the average of the \( \left(\frac{N}{2}\right) \)th term and the \( \left(\frac{N}{2}+1\right) \)th term.
\( \implies \) Median position \( = \frac{\left(\frac{38}{2}\right)\text{th term} + \left(\frac{38}{2}+1\right)\text{th term}}{2} \)
\( \implies \) Median position \( = \frac{19\text{th term} + 20\text{th term}}{2} \).
Looking at the c.f. column, both the 19th and 20th terms fall within the cumulative frequency of 28, which corresponds to marks of 70.
So, 19th term = 70 and 20th term = 70.
\( \implies \) Median \( = \frac{70+70}{2} = \frac{140}{2} = 70 \).
Therefore, the median marks obtained by the students are 70. This means half the students scored 70 or less, and half scored 70 or more.
In simple words: First, sort the marks and make a running total of students. Since there's an even number of students, find the two middle marks, add them, and divide by two. This gives you the median mark.

🎯 Exam Tip: For even-numbered data points in discrete frequency distributions, remember to average the two middle values. Make sure your data is ordered (like marks here) before creating the c.f. table.

 

Question 6. Find whether the following statements are true or false:
(i) The median of a frequency distribution is the most commonly occurring value.
(ii) The median of a discrete ungrouped frequency distribution containing a number of items is the value of the middle item, the data being arranged in ascending or descending order.
Answer:
(i) False. The most commonly occurring value in a frequency distribution is called the Mode, not the Median. The median is the middle value when data is ordered.
(ii) True. The definition of a median for discrete ungrouped data is indeed the value of the middle item after the data has been arranged in either ascending or descending order. This helps to find the central point of the dataset.
In simple words: For the first part, the median is not the number that shows up most often; that's the mode. For the second part, yes, the median is exactly the middle number after you've put all the numbers in order.

🎯 Exam Tip: Clearly understand the definitions of mean, median, and mode. They are all measures of central tendency but describe different aspects of a dataset.

 

Question 7. In a school examination it is decided that exactly half the pupils will pass. Name the tendency that is used.
Answer: The measure of central tendency used here is the Median.
The median is defined as the middle point of an array of given numbers, where an equal number of items lie above and below it. If exactly half the pupils are to pass, then the passing mark would be set at the median, ensuring 50% pass and 50% fail. This makes the median ideal for dividing a group into two equal halves.
In simple words: When you want to divide a group exactly in half, so that half are above a certain point and half are below, you use the median.

🎯 Exam Tip: The median is particularly useful when you need to divide a dataset into two equal parts or when the data has extreme values that would skew the mean.

 

Question 8. (1, 2, 3, 6, 8) is a set of five positive integers whose mean is 4 and median is 3. Write down two other sets of five positive integers, each having the same mean and median as this set.
Answer: For a set of five numbers \( a, b, c, d, e \) arranged in ascending order, the median is the 3rd term, \( c \). The mean is \( \frac{a+b+c+d+e}{5} \).
Given set: (1, 2, 3, 6, 8)
Mean \( = \frac{1+2+3+6+8}{5} = \frac{20}{5} = 4 \).
Median \( = 3 \) (the 3rd term).

Here are two other sets of five positive integers with a mean of 4 and a median of 3:
1. **Set 1: (1, 2, 3, 5, 9)**
Arranged in ascending order: 1, 2, 3, 5, 9.
Mean \( = \frac{1+2+3+5+9}{5} = \frac{20}{5} = 4 \).
Median \( = 3 \) (the 3rd term).
2. **Set 2: (1, 2, 3, 4, 10)**
Arranged in ascending order: 1, 2, 3, 4, 10.
Mean \( = \frac{1+2+3+4+10}{5} = \frac{20}{5} = 4 \).
Median \( = 3 \) (the 3rd term).
The key is to keep the middle element as 3 and adjust the other numbers so their sum equals 20 (for a mean of 4).
In simple words: We need to find other sets of five numbers where the middle number is 3 and when you add them all up and divide by five, you get 4. The order of numbers matters for finding the median.

🎯 Exam Tip: When creating new sets, fix the median first. Then, ensure the sum of all numbers (5 x mean) is correct. Adjust the remaining numbers carefully to maintain the ascending order.

 

Question 9. Find the median from the following distribution:
Answer: To find the median for this grouped frequency distribution, we first create a cumulative frequency (c.f.) table.

The total frequency is \( N = 100 \).
The median position is \( \frac{N}{2} = \frac{100}{2} = 50 \).
The cumulative frequency just greater than or equal to 50 is 70. This corresponds to the median class 40-55.
For this median class:
Lower limit (\( L \)) = 40
Frequency (\( f \)) = 44
Cumulative frequency of the class preceding the median class (\( C \)) = 26
Class width (\( i \)) = \( 55 - 40 = 15 \)
Now, we use the formula for the median of grouped data:
\( Md = L + \frac{\frac{N}{2} - C}{f} \times i \)
\( \implies Md = 40 + \frac{50 - 26}{44} \times 15 \)
\( \implies Md = 40 + \frac{24}{44} \times 15 \)
\( \implies Md = 40 + \frac{6}{11} \times 15 \)
\( \implies Md = 40 + \frac{90}{11} \)
\( \implies Md = 40 + 8.1818... \)
\( \implies Md \approx 48.18 \)
Thus, the required median marks are approximately 48.18. This value represents the central tendency of marks for these students.
In simple words: First, add up the frequencies to find the total. Then, find the middle spot in this total. Locate which "Marks" group this middle spot falls into. Use a special math rule with the group's numbers to find the exact median.

🎯 Exam Tip: Remember to identify the median class correctly using \( \frac{N}{2} \) and carefully plug in all values into the median formula. Pay close attention to the class width.

 

Question 10. Find the median from the following distribution:
Answer: The given class intervals are discontinuous (e.g., 11-15, 16-20). To calculate the median, we first need to make them continuous and then create a cumulative frequency (c.f.) table.
The adjustment factor is \( \frac{16-15}{2} = \frac{1}{2} = 0.5 \).
We subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class interval to make them continuous.

The total frequency is \( N = 130 \).
The median position is \( \frac{N}{2} = \frac{130}{2} = 65 \).
The cumulative frequency just greater than or equal to 65 is 91. This corresponds to the median class 30.5-35.5.
For this median class:
Lower limit (\( L \)) = 30.5
Frequency (\( f \)) = 35
Cumulative frequency of the class preceding the median class (\( C \)) = 56
Class width (\( i \)) = \( 35.5 - 30.5 = 5 \)
Now, we use the formula for the median of grouped data:
\( Md = L + \frac{\frac{N}{2} - C}{f} \times i \)
\( \implies Md = 30.5 + \frac{65 - 56}{35} \times 5 \)
\( \implies Md = 30.5 + \frac{9}{35} \times 5 \)
\( \implies Md = 30.5 + \frac{9}{7} \)
\( \implies Md = 30.5 + 1.2857... \)
\( \implies Md \approx 31.785 \)
Thus, the median for this distribution is approximately 31.785. Making class intervals continuous is a crucial first step for this type of problem.
In simple words: First, adjust the class intervals so they don't have gaps. Then, make a running total of the frequencies. Find the middle spot in the total, pick the group it falls into, and use the median rule for that group to get the exact answer.

🎯 Exam Tip: Always remember to convert discontinuous class intervals into continuous ones by applying the adjustment factor before calculating the median for grouped data.

 

Question 11. Find the median from the following distribution:
Answer: The given distribution is in "Less than" form. To find the median, we first convert it into a regular class interval frequency distribution and then create a cumulative frequency (c.f.) table.

The total frequency is \( N = 125 \).
The median position is \( \frac{N}{2} = \frac{125}{2} = 62.5 \).
The cumulative frequency just greater than or equal to 62.5 is 76. This corresponds to the median class 30-40.
For this median class:
Lower limit (\( L \)) = 30
Frequency (\( f \)) = 36
Cumulative frequency of the class preceding the median class (\( C \)) = 40
Class width (\( i \)) = \( 40 - 30 = 10 \)
Now, we use the formula for the median of grouped data:
\( Md = L + \frac{\frac{N}{2} - C}{f} \times i \)
\( \implies Md = 30 + \frac{62.5 - 40}{36} \times 10 \)
\( \implies Md = 30 + \frac{22.5}{36} \times 10 \)
\( \implies Md = 30 + \frac{225}{36} \)
\( \implies Md = 30 + 6.25 \)
\( \implies Md = 36.25 \)
Thus, the median for this distribution is 36.25. Converting "Less than" distributions is a standard step.
In simple words: First, change the "less than" numbers into normal groups with a frequency for each. Then, make a running total. Find the middle spot and use the median math rule for that group to get the exact median number.

🎯 Exam Tip: When dealing with "less than" or "more than" type distributions, always convert them into standard class intervals with their respective frequencies before calculating any measure of central tendency.

 

Question 12. Find the median from the following distribution:
Answer: The given distribution is in "More than" form. To find the median, we first convert it into a regular class interval frequency distribution and then create a cumulative frequency (c.f.) table.

The total frequency is \( N = 165 \).
The median position is \( \frac{N}{2} = \frac{165}{2} = 82.5 \).
The cumulative frequency just greater than or equal to 82.5 is 125. This corresponds to the median class 30-40.
For this median class:
Lower limit (\( L \)) = 30
Frequency (\( f \)) = 58
Cumulative frequency of the class preceding the median class (\( C \)) = 67
Class width (\( i \)) = \( 40 - 30 = 10 \)
Now, we use the formula for the median of grouped data:
\( Md = L + \frac{\frac{N}{2} - C}{f} \times i \)
\( \implies Md = 30 + \frac{82.5 - 67}{58} \times 10 \)
\( \implies Md = 30 + \frac{15.5}{58} \times 10 \)
\( \implies Md = 30 + \frac{155}{58} \)
\( \implies Md = 30 + 2.6724... \)
\( \implies Md \approx 32.67 \)
Thus, the median for this distribution is approximately 32.67. Converting "More than" data correctly is essential.
In simple words: First, change the "more than" numbers into normal groups, then find the frequency for each group. Make a running total. Find the middle spot, see which group it's in, and then use the median rule for that specific group to get the final answer.

🎯 Exam Tip: Remember that "more than" cumulative frequencies are decreasing. To convert to standard frequencies, subtract the next c.f. from the current one, starting from the total frequency at the lowest value.

 

Question 13. Find the first and third quartile for the following data: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Answer: The given data is already in ascending order: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
The number of observations is \( n = 11 \) (odd).
**First Quartile (\( Q_1 \))**:
The position of the first quartile is the \( \left(\frac{n+1}{4}\right) \)th item.
\( \implies Q_1 = \left(\frac{11+1}{4}\right) \)th item
\( \implies Q_1 = \left(\frac{12}{4}\right) \)th item
\( \implies Q_1 = \) 3rd item
From the data, the 3rd item is 6.
So, \( Q_1 = 6 \).
**Third Quartile (\( Q_3 \))**:
The position of the third quartile is the \( 3\left(\frac{n+1}{4}\right) \)th item.
\( \implies Q_3 = 3\left(\frac{11+1}{4}\right) \)th item
\( \implies Q_3 = 3\left(\frac{12}{4}\right) \)th item
\( \implies Q_3 = 3 \times \) 3rd item
\( \implies Q_3 = \) 9th item
From the data, the 9th item is 18.
So, \( Q_3 = 18 \).
The quartiles divide the data into four equal parts. \( Q_1 \) means 25% of the data is below it, and \( Q_3 \) means 75% is below it.
In simple words: First, put all the numbers in order. The first quartile is the number that is one-quarter of the way through the list. The third quartile is the number that is three-quarters of the way through the list.

🎯 Exam Tip: Always sort the data first for quartiles. For ungrouped data, use \( \frac{n+1}{4} \) for \( Q_1 \) position and \( 3\frac{n+1}{4} \) for \( Q_3 \) position. If the position is a fraction, interpolate.

 

Question 14. Compute Q1, Q3, D3, D6 and D8 for the following data: 14, 7, 13, 12, 13, 17, 8, 10, 6, 15, 18, 21, 20
Answer: First, arrange the given data in ascending order:
6, 7, 8, 10, 12, 13, 13, 14, 15, 17, 18, 20, 21.
The number of observations is \( n = 13 \) (odd).
**First Quartile (\( Q_1 \))**:
The position of \( Q_1 \) is the \( \left(\frac{n+1}{4}\right) \)th item.
\( \implies Q_1 = \left(\frac{13+1}{4}\right) \)th item \( = \left(\frac{14}{4}\right) \)th item \( = 3.5 \)th item.
Since it's a decimal, we interpolate: 3rd item \( + 0.5 \times \) (4th item - 3rd item)
\( \implies Q_1 = 8 + 0.5 \times (10 - 8) \)
\( \implies Q_1 = 8 + 0.5 \times 2 \)
\( \implies Q_1 = 8 + 1 = 9 \).
**Third Quartile (\( Q_3 \))**:
The position of \( Q_3 \) is the \( 3\left(\frac{n+1}{4}\right) \)th item.
\( \implies Q_3 = 3\left(\frac{13+1}{4}\right) \)th item \( = 3 \times 3.5 \)th item \( = 10.5 \)th item.
Since it's a decimal, we interpolate: 10th item \( + 0.5 \times \) (11th item - 10th item)
\( \implies Q_3 = 17 + 0.5 \times (18 - 17) \)
\( \implies Q_3 = 17 + 0.5 \times 1 \)
\( \implies Q_3 = 17 + 0.5 = 17.5 \).
**Third Decile (\( D_3 \))**:
The position of \( D_3 \) is the \( 3\left(\frac{n+1}{10}\right) \)th item.
\( \implies D_3 = 3\left(\frac{13+1}{10}\right) \)th item \( = 3 \times \left(\frac{14}{10}\right) \)th item \( = 3 \times 1.4 \)th item \( = 4.2 \)th item.
Since it's a decimal, we interpolate: 4th item \( + 0.2 \times \) (5th item - 4th item)
\( \implies D_3 = 10 + 0.2 \times (12 - 10) \)
\( \implies D_3 = 10 + 0.2 \times 2 \)
\( \implies D_3 = 10 + 0.4 = 10.4 \).
**Sixth Decile (\( D_6 \))**:
The position of \( D_6 \) is the \( 6\left(\frac{n+1}{10}\right) \)th item.
\( \implies D_6 = 6\left(\frac{13+1}{10}\right) \)th item \( = 6 \times \left(\frac{14}{10}\right) \)th item \( = 6 \times 1.4 \)th item \( = 8.4 \)th item.
Since it's a decimal, we interpolate: 8th item \( + 0.4 \times \) (9th item - 8th item)
\( \implies D_6 = 14 + 0.4 \times (15 - 14) \)
\( \implies D_6 = 14 + 0.4 \times 1 \)
\( \implies D_6 = 14 + 0.4 = 14.4 \).
**Eighth Decile (\( D_8 \))**:
The position of \( D_8 \) is the \( 8\left(\frac{n+1}{10}\right) \)th item.
\( \implies D_8 = 8\left(\frac{13+1}{10}\right) \)th item \( = 8 \times \left(\frac{14}{10}\right) \)th item \( = 8 \times 1.4 \)th item \( = 11.2 \)th item.
Since it's a decimal, we interpolate: 11th item \( + 0.2 \times \) (12th item - 11th item)
\( \implies D_8 = 18 + 0.2 \times (20 - 18) \)
\( \implies D_8 = 18 + 0.2 \times 2 \)
\( \implies D_8 = 18 + 0.4 = 18.4 \).
These measures (quartiles and deciles) help to understand the spread and distribution of the data beyond just the median.
In simple words: After putting numbers in order, \( Q_1 \) is the point that cuts off the bottom quarter of the numbers, and \( Q_3 \) cuts off the bottom three-quarters. \( D_3, D_6, D_8 \) are like different quarter or tenth points in the list. To find these exactly, you might need to guess a bit between two numbers if the position isn't whole.

🎯 Exam Tip: Always arrange the data in ascending order first. For fractional positions (like 3.5th term), use interpolation: Value of lower term + fractional part * (Value of higher term - Value of lower term).

 

Question 15. 18, 20, 9, 15, 21, 26, 14, 13, 27, 22, 16, 28 Find D7 and P33.
Answer: First, arrange the given data in ascending order:
9, 13, 14, 15, 16, 18, 20, 21, 22, 26, 27, 28.
The number of observations is \( n = 12 \) (even).
**Seventh Decile (\( D_7 \))**:
The position of \( D_7 \) is the \( 7\left(\frac{n+1}{10}\right) \)th item.
\( \implies D_7 = 7\left(\frac{12+1}{10}\right) \)th item \( = 7 \times \left(\frac{13}{10}\right) \)th item \( = 7 \times 1.3 \)th item \( = 9.1 \)th item.
Since it's a decimal, we interpolate: 9th item \( + 0.1 \times \) (10th item - 9th item)
\( \implies D_7 = 22 + 0.1 \times (26 - 22) \)
\( \implies D_7 = 22 + 0.1 \times 4 \)
\( \implies D_7 = 22 + 0.4 = 22.4 \).
**Thirty-third Percentile (\( P_{33} \))**:
The position of \( P_{33} \) is the \( 33\left(\frac{n+1}{100}\right) \)th item.
\( \implies P_{33} = 33\left(\frac{12+1}{100}\right) \)th item \( = 33 \times \left(\frac{13}{100}\right) \)th item \( = 33 \times 0.13 \)th item \( = 4.29 \)th item.
Since it's a decimal, we interpolate: 4th item \( + 0.29 \times \) (5th item - 4th item)
\( \implies P_{33} = 15 + 0.29 \times (16 - 15) \)
\( \implies P_{33} = 15 + 0.29 \times 1 \)
\( \implies P_{33} = 15 + 0.29 = 15.29 \).
Deciles divide data into 10 parts, and percentiles divide it into 100 parts, offering a detailed view of data distribution.
In simple words: First, line up all the numbers in order. To find \( D_7 \), you look for the spot 7-tenths of the way through the list. For \( P_{33} \), you look for the spot 33-hundredths of the way through. If the spot is between two numbers, you guess the value in the middle.

🎯 Exam Tip: Ensure data is sorted. The formulas for deciles and percentiles are similar to quartiles, just with denominators of 10 and 100 respectively. Practice interpolation for non-integer positions.

 

Question 16. Compute Q1, Q3, D6 and P45 for the following data:
Answer: To compute the quartiles, deciles, and percentiles for this grouped data, we first need to create a cumulative frequency (c.f.) table.

The total frequency is \( N = 200 \).
**First Quartile (\( Q_1 \))**:
Position of \( Q_1 \) is the \( \left(\frac{N+1}{4}\right) \)th value \( = \left(\frac{200+1}{4}\right) \)th value \( = \left(\frac{201}{4}\right) \)th value \( = 50.25 \)th value.
This means the 50th value \( + 0.25 \times \) (51st value - 50th value).
From the c.f. table, the 50th and 51st values both correspond to \( x = 20 \) (since c.f. 33 to 58 are all 20).
\( \implies Q_1 = 20 + 0.25 \times (20 - 20) = 20 + 0 = 20 \).
**Third Quartile (\( Q_3 \))**:
Position of \( Q_3 \) is the \( 3\left(\frac{N+1}{4}\right) \)th value \( = 3 \times 50.25 \)th value \( = 150.75 \)th value.
This means the 150th value \( + 0.75 \times \) (151st value - 150th value).
From the c.f. table, the 150th value is 23 (since c.f. 125 to 150 are 23). The 151st value is 24 (since c.f. 150 to 169 are 24).
\( \implies Q_3 = 23 + 0.75 \times (24 - 23) = 23 + 0.75 \times 1 = 23.75 \).
**Sixth Decile (\( D_6 \))**:
Position of \( D_6 \) is the \( 6\left(\frac{N+1}{10}\right) \)th value \( = 6\left(\frac{200+1}{10}\right) \)th value \( = 6 \times 20.1 \)th value \( = 120.6 \)th value.
This means the 120th value \( + 0.6 \times \) (121st value - 120th value).
From the c.f. table, the 120th and 121st values both correspond to \( x = 22 \) (since c.f. 85 to 125 are all 22).
\( \implies D_6 = 22 + 0.6 \times (22 - 22) = 22 + 0 = 22 \).
**Forty-fifth Percentile (\( P_{45} \))**:
Position of \( P_{45} \) is the \( 45\left(\frac{N+1}{100}\right) \)th value \( = 45\left(\frac{200+1}{100}\right) \)th value \( = 45 \times 2.01 \)th value \( = 90.45 \)th value.
This means the 90th value \( + 0.45 \times \) (91st value - 90th value).
From the c.f. table, the 90th and 91st values both correspond to \( x = 22 \) (since c.f. 85 to 125 are all 22).
\( \implies P_{45} = 22 + 0.45 \times (22 - 22) = 22 + 0 = 22 \).
These calculations show how to pinpoint specific percentages of the data for a better understanding of its spread.
In simple words: For each item (Q1, Q3, D6, P45), first find its spot in the total count of numbers. Then, use the table to find the 'x' value at that spot. If the spot is between two numbers, you take a smart guess (interpolate) to find the exact value.

🎯 Exam Tip: For discrete grouped data, the values in the 'x' column are directly used. The interpolation logic applies if the calculated rank falls between two distinct 'x' values, but if it falls within the range of one 'x' value due to its frequency, that 'x' value is the answer.

 

Question 17. Following are the different sizes and number of shoes in a shoe shop. Calculate median, first quartile, third quartile, 6th decile and 80th percentile.
Answer: To calculate the median, quartiles, decile, and percentile for this data, we first need to create a cumulative frequency (c.f.) table by arranging the shoe sizes in ascending order.

The total number of shoes is \( N = 253 \).
**Median (\( Md \))**:
Position of \( Md \) is the \( \left(\frac{N+1}{2}\right) \)th item \( = \left(\frac{253+1}{2}\right) \)th item \( = \left(\frac{254}{2}\right) \)th item \( = 127 \)th item.
From the c.f. table, the 127th item corresponds to a shoe size of 7.5.
\( \implies Md = 7.5 \).
**First Quartile (\( Q_1 \))**:
Position of \( Q_1 \) is the \( \left(\frac{N+1}{4}\right) \)th item \( = \left(\frac{253+1}{4}\right) \)th item \( = \left(\frac{254}{4}\right) \)th item \( = 63.5 \)th item.
This means the 63rd value \( + 0.5 \times \) (64th value - 63rd value).
From the c.f. table, both the 63rd and 64th values correspond to a shoe size of 7.0 (since c.f. 59 to 84 are all 7.0).
\( \implies Q_1 = 7.0 + 0.5 \times (7.0 - 7.0) = 7.0 + 0 = 7.0 \).
**Third Quartile (\( Q_3 \))**:
Position of \( Q_3 \) is the \( 3\left(\frac{N+1}{4}\right) \)th item \( = 3 \times 63.5 \)th item \( = 190.5 \)th item.
This means the 190th value \( + 0.5 \times \) (191st value - 190th value).
From the c.f. table, the 190th value is 8.5 (since c.f. 174 to 194 are 8.5). The 191st value is also 8.5.
\( \implies Q_3 = 8.5 + 0.5 \times (8.5 - 8.5) = 8.5 + 0 = 8.5 \).
**Sixth Decile (\( D_6 \))**:
Position of \( D_6 \) is the \( 6\left(\frac{N+1}{10}\right) \)th item \( = 6\left(\frac{253+1}{10}\right) \)th item \( = 6 \times 25.4 \)th item \( = 152.4 \)th item.
This means the 152nd value \( + 0.4 \times \) (153rd value - 152nd value).
From the c.f. table, the 152nd and 153rd values both correspond to a shoe size of 8 (since c.f. 134 to 174 are all 8).
\( \implies D_6 = 8 + 0.4 \times (8 - 8) = 8 + 0 = 8 \).
**Eightieth Percentile (\( P_{80} \))**:
Position of \( P_{80} \) is the \( 80\left(\frac{N+1}{100}\right) \)th item \( = 80\left(\frac{253+1}{100}\right) \)th item \( = 80 \times 2.54 \)th item \( = 203.2 \)th item.
This means the 203rd value \( + 0.2 \times \) (204th value - 203rd value).
From the c.f. table, the 203rd and 204th values both correspond to a shoe size of 9.0 (since c.f. 194 to 209 are all 9.0).
\( \implies P_{80} = 9.0 + 0.2 \times (9.0 - 9.0) = 9.0 + 0 = 9.0 \).
This comprehensive set of measures provides a detailed understanding of shoe size distribution.
In simple words: First, list all the shoe sizes in order and count how many of each there are. Then, make a running total. Using special rules, find the shoe size that is exactly in the middle (median), one-quarter way (Q1), three-quarters way (Q3), six-tenths way (D6), and eighty-hundredths way (P80) through the list.

🎯 Exam Tip: When dealing with discrete series (even if values are decimals), if the calculated position falls within a range where the 'x' value is constant in the c.f. table, use that 'x' value directly without further interpolation if the values are identical.

 

Question 18. Find out first quartile, third quartile and first decile.
Answer: The table of values is given as under:

 

Question 19. Calculate the median, 3rd decile and 20th percentile for the following data:

Answer: First, we need to create a cumulative frequency (c.f.) table. This helps us find the position of the median, deciles, and percentiles.**1. Median (\( M_d \)):**
We find the value \( \frac{N}{2} \).
\( \frac{N}{2} = \frac{100}{2} = 50 \) This value (50) falls exactly into the cumulative frequency of the class interval 10-15.
In this median class (10-15):
\( l \) (lower limit) = 10
\( f \) (frequency) = 25
\( c \) (cumulative frequency of class before median class) = 25
\( i \) (class size) = 5
The formula for median is: \( M_d = l + \frac{\frac{N}{2} - c}{f} \times i \)
\( \implies M_d = 10 + \frac{50 - 25}{25} \times 5 \)
\( \implies M_d = 10 + \frac{25}{25} \times 5 \)
\( \implies M_d = 10 + 1 \times 5 \)
\( \implies M_d = 10 + 5 = 15 \)
**2. 3rd Decile (\( D_3 \)):**
Deciles divide the data into 10 equal parts. For the 3rd decile, we find \( \frac{3N}{10} \).
\( \frac{3N}{10} = \frac{3 \times 100}{10} = 30 \) This value (30) falls into the cumulative frequency of the class interval 10-15.
In this decile class (10-15):
\( l \) (lower limit) = 10
\( f \) (frequency) = 25
\( c \) (cumulative frequency of class before decile class) = 25
\( i \) (class size) = 5
The formula for decile is: \( D_k = l + \frac{\frac{kN}{10} - c}{f} \times i \)
\( \implies D_3 = 10 + \frac{30 - 25}{25} \times 5 \)
\( \implies D_3 = 10 + \frac{5}{25} \times 5 \)
\( \implies D_3 = 10 + \frac{1}{5} \times 5 \)
\( \implies D_3 = 10 + 1 = 11 \)
**3. 20th Percentile (\( P_{20} \)):**
Percentiles divide the data into 100 equal parts. For the 20th percentile, we find \( \frac{20N}{100} \).
\( \frac{20N}{100} = \frac{20 \times 100}{100} = 20 \) This value (20) falls into the cumulative frequency of the class interval 5-10.
In this percentile class (5-10):
\( l \) (lower limit) = 5
\( f \) (frequency) = 18
\( c \) (cumulative frequency of class before percentile class) = 7
\( i \) (class size) = 5
The formula for percentile is: \( P_k = l + \frac{\frac{kN}{100} - c}{f} \times i \)
\( \implies P_{20} = 5 + \frac{20 - 7}{18} \times 5 \)
\( \implies P_{20} = 5 + \frac{13}{18} \times 5 \)
\( \implies P_{20} = 5 + \frac{65}{18} \)
\( \implies P_{20} = 5 + 3.611 \)
\( \implies P_{20} = 8.611 \)
In simple words: For grouped data, the median is the middle value, the 3rd decile is the value below which 30% of the data falls, and the 20th percentile is the value below which 20% of the data falls. We find these by first calculating where they sit in the cumulative frequency, then using a formula that adjusts for the class interval.

🎯 Exam Tip: Always construct the cumulative frequency table carefully as the first step. Errors here will carry through all subsequent calculations. Remember the correct formulas for median, deciles, and percentiles, and correctly identify \( l, f, c, \) and \( i \) for the respective class interval.

 

Question 20. Find the interquartile range, quartile deviation for the following data:

Answer: First, we need to create a cumulative frequency (c.f.) table from the given data. This helps us locate the quartiles.**1. First Quartile (\( Q_1 \)):**
\( Q_1 \) is the value of the \( \left(\frac{N+1}{4}\right) \)th item.
\( \implies Q_1 \) = size of \( \left(\frac{543+1}{4}\right) \)th item
\( \implies Q_1 \) = size of \( \frac{544}{4} \)th item
\( \implies Q_1 \) = size of 136th item Looking at the cumulative frequency table, the 136th item falls in the "Age in years" category of 40 (since 64 items are up to age 30, and 196 items are up to age 40). Therefore, \( Q_1 = 40 \).
**2. Third Quartile (\( Q_3 \)):**
\( Q_3 \) is the value of the \( 3\left(\frac{N+1}{4}\right) \)th item.
\( \implies Q_3 \) = size of \( 3\left(\frac{543+1}{4}\right) \)th item
\( \implies Q_3 \) = size of \( 3 \times 136 \)th item
\( \implies Q_3 \) = size of 408th item Looking at the cumulative frequency table, the 408th item falls in the "Age in years" category of 60 (since 349 items are up to age 50, and 489 items are up to age 60). Therefore, \( Q_3 = 60 \).
**3. Interquartile Range (IQR):**
The interquartile range is the difference between the third and first quartiles.
\( \implies \text{Interquartile range} = Q_3 - Q_1 = 60 - 40 = 20 \)
**4. Quartile Deviation (Q.D.):**
The quartile deviation, also known as the semi-interquartile range, is half of the interquartile range.
\( \implies \text{Q.D.} = \frac{Q_3 - Q_1}{2} = \frac{60 - 40}{2} = \frac{20}{2} = 10 \)
In simple words: We calculated the first quartile (\( Q_1 \)) and the third quartile (\( Q_3 \)), which are points that divide the data into four equal parts. The interquartile range (IQR) shows how spread out the middle 50% of the data is, and the quartile deviation (Q.D.) is half of this spread.

🎯 Exam Tip: For ungrouped data or discrete series like this, ensure you correctly identify the item number and its corresponding value from the frequency distribution or cumulative frequency table. The IQR is a key measure of data spread that is less affected by outliers.

 

Question 21. Find the interquartile range, semi-interquartile range, and coefficient of quartile deviation from the following frequency distribution.

Answer: First, we construct the cumulative frequency (c.f.) table. This helps us determine the class intervals where our quartiles lie.**1. First Quartile (\( Q_1 \)):**
We find the value \( \frac{N}{4} \).
\( \frac{N}{4} = \frac{600}{4} = 150 \) This value (150) falls into the cumulative frequency of the class interval 30-40.
In this first quartile class (30-40):
\( l \) (lower limit) = 30
\( f \) (frequency) = 120
\( c \) (cumulative frequency of class before \( Q_1 \) class) = 105
\( i \) (class size) = 10
The formula for \( Q_1 \) is: \( Q_1 = l + \frac{\frac{N}{4} - c}{f} \times i \)
\( \implies Q_1 = 30 + \frac{150 - 105}{120} \times 10 \)
\( \implies Q_1 = 30 + \frac{45}{120} \times 10 \)
\( \implies Q_1 = 30 + \frac{450}{120} \)
\( \implies Q_1 = 30 + 3.75 \)
\( \implies Q_1 = 33.75 \)
**2. Third Quartile (\( Q_3 \)):**
We find the value \( \frac{3N}{4} \).
\( \frac{3N}{4} = \frac{3 \times 600}{4} = 450 \) This value (450) falls into the cumulative frequency of the class interval 70-80.
In this third quartile class (70-80):
\( l \) (lower limit) = 70
\( f \) (frequency) = 120
\( c \) (cumulative frequency of class before \( Q_3 \) class) = 420
\( i \) (class size) = 10
The formula for \( Q_3 \) is: \( Q_3 = l + \frac{\frac{3N}{4} - c}{f} \times i \)
\( \implies Q_3 = 70 + \frac{450 - 420}{120} \times 10 \)
\( \implies Q_3 = 70 + \frac{30}{120} \times 10 \)
\( \implies Q_3 = 70 + \frac{300}{120} \)
\( \implies Q_3 = 70 + 2.5 \)
\( \implies Q_3 = 72.5 \)
**3. Interquartile Range (IQR):**
The interquartile range measures the spread of the middle 50% of the data.
\( \implies \text{Interquartile range} = Q_3 - Q_1 = 72.5 - 33.75 = 38.75 \)
**4. Semi-Interquartile Range (Q.D.):**
The semi-interquartile range is half of the interquartile range. It tells us the average distance of the first and third quartiles from the median.
\( \implies \text{Semi-interquartile range} = \frac{Q_3 - Q_1}{2} = \frac{38.75}{2} = 19.375 \)
**5. Coefficient of Quartile Deviation:**
This is a relative measure of dispersion, useful for comparing variability between different datasets.
\( \implies \text{Coefficient of Q.D.} = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{72.5 - 33.75}{72.5 + 33.75} \)
\( \implies \text{Coefficient of Q.D.} = \frac{38.75}{106.25} \)
\( \implies \text{Coefficient of Q.D.} = 0.365 \)
In simple words: We calculated the first and third quartiles for this data, which helps us understand where the lower and upper 25% marks are. The interquartile range shows the spread of the middle half of the marks. The semi-interquartile range is half of this spread, and the coefficient of quartile deviation helps compare this spread to the average value.

🎯 Exam Tip: Always remember to correctly identify the class interval for \( Q_1 \) and \( Q_3 \) by looking at the cumulative frequency. Pay close attention to the formula components \( l, f, c, \) and \( i \) for each quartile to avoid calculation errors. The coefficient of quartile deviation is a unitless measure, making it good for comparing different datasets.