GSEB Class 11 Statistics Solutions Chapter 3 Measures of Central Tendency Exercise 3.4

Chapter 3 Measures Of Central Tendency Ex 3.4

Question 1. Find all quartiles for the data given below about marks scored by 15 students in class test.
8, 8, 7, 0, 2, 4, 6, 5, 5, 4, 8, 9, 3, 6, 7
Answer:
First, the given marks are arranged in ascending sequence: 0, 2, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 9. The total number of observations, \(n\), is 15.

First Quartile (Q1):
Q1 = Value of \( \left(\frac{n+1}{4}\right) \)th observation
= Value of \( \left(\frac{15+1}{4}\right) = \left(\frac{16}{4}\right) = 4 \)th observation
The 4th observation in the ordered data is 4. Thus, the first quartile (Q1) is 4 marks.

Second Quartile (Q2):
Q2 = Value of \( 2\left(\frac{n+1}{4}\right) \)th observation
= Value of \( 2 \times 4 \) (since \( \left(\frac{n+1}{4}\right) = 4 \))
= 8th observation
The 8th observation in the sorted data is 6. Therefore, the second quartile (Q2) is 6 marks.

Third Quartile (Q3):
Q3 = Value of \( 3\left(\frac{n+1}{4}\right) \)th observation
= Value of \( 3 \times 4 \) (since \( \left(\frac{n+1}{4}\right) = 4 \))
= 12th observation
The 12th observation in the arranged data is 7. Hence, the third quartile (Q3) is 7 marks.
In simple words: Quartiles divide the data into four equal parts. We first sort the data and then locate the values corresponding to the 25th, 50th, and 75th percentiles (Q1, Q2, Q3 respectively) using specific positional formulas.

🎯 Exam Tip: Always sort the data in ascending order before calculating quartiles. The correct application of the \( \left(\frac{n+1}{k}\right) \) formula for quantiles is crucial for accuracy.

Question 2. The following table shows data about the distance travelled (in km) by a salesman on different days. Find median, Q3, D8, P62 and interpret them

Distance travelled (km)	0-100	100-200	200-300	300-400	400-500	500-600
No. of days	5	18	24	7	5	1

Answer:
To compute the required measures, we first construct the cumulative frequency distribution from the provided data.

Distance travelled (km)	No. of days (f)	Cumulative frequency (cf)
0-100	5	5
100-200	18	23
200-300	24	47
300-400	7	54
400-500	5	59
500-600	1	60
Total	n = 60	-

Median (M):
The median class is identified as the class containing the \( \left(\frac{n}{2}\right) \)th observation.
For \( n=60 \), this is the \( \left(\frac{60}{2}\right) = 30 \)th observation.
Referring to the cumulative frequency (cf) column, the 30th observation falls within the 200-300 km class interval.

Using the formula for median of grouped data:
\( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Where \( L = 200 \) (lower limit of median class), \( \frac{n}{2} = 30 \), \( cf = 23 \) (cumulative frequency of the class preceding the median class), \( f = 24 \) (frequency of the median class), and \( c = 100 \) (class width).

\( M = 200 + \frac{30-23}{24} \times 100 \)
\( = 200 + \frac{7 \times 100}{24} \)
\( = 200 + \frac{700}{24} \)
\( = 200 + 29.17 \)
\( = 229.17 \) km
Hence, the median \(M\) is 229.17 km.

Interpretation:
A median of 229.17 km indicates that on 50% of the days, the salesman traveled a distance less than or equal to 229.17 km.

Third Quartile (Q3):
The Q3 class is the class that includes the \( 3\left(\frac{n}{4}\right) \)th observation.
For \( n=60 \), this is the \( 3\left(\frac{60}{4}\right) = 3 \times 15 = 45 \)th observation.
Referring to the cumulative frequency (cf) column, the 45th observation falls within the 200-300 km class interval.

Using the formula for quartiles of grouped data:
\( Q_3 = L + \frac{3\left(\frac{n}{4}\right)-cf}{f} \times c \)
Where \( L = 200 \) (lower limit of Q3 class), \( 3\left(\frac{n}{4}\right) = 45 \), \( cf = 23 \) (cumulative frequency of the class preceding Q3 class), \( f = 24 \) (frequency of Q3 class), and \( c = 100 \) (class width).

\( Q_3 = 200 + \frac{45-23}{24} \times 100 \)
\( = 200 + \frac{22}{24} \times 100 \)
\( = 200 + \frac{2200}{24} \)
\( = 200 + 91.67 \)
\( = 291.67 \) km
Hence, the third quartile Q3 is 291.67 km.

Interpretation:
A third quartile of 291.67 km signifies that on 75% of the days, the salesman's travel distance was less than or equal to 291.67 km. This also means the maximum travel on 75% of days was 291.67 km.

Eighth Decile (D8):
The D8 class is the class that includes the \( 8\left(\frac{n}{10}\right) \)th observation.
For \( n=60 \), this is the \( 8\left(\frac{60}{10}\right) = 8 \times 6 = 48 \)th observation.
Referring to the cumulative frequency (cf) column, the 48th observation falls within the 300-400 km class interval.

Using the formula for deciles of grouped data:
\( D_8 = L + \frac{8\left(\frac{n}{10}\right)-cf}{f} \times c \)
Where \( L = 300 \) (lower limit of D8 class), \( 8\left(\frac{n}{10}\right) = 48 \), \( cf = 47 \) (cumulative frequency of the class preceding D8 class), \( f = 7 \) (frequency of D8 class), and \( c = 100 \) (class width).

\( D_8 = 300 + \frac{48-47}{7} \times 100 \)
\( = 300 + \frac{1 \times 100}{7} \)
\( = 300 + 14.29 \)
\( = 314.29 \) km
Hence, the eighth decile D8 is 314.29 km.

Interpretation:
An eighth decile of 314.29 km indicates that on 80% of the days, the salesman traveled a distance less than or equal to 314.29 km. This implies that the maximum distance traveled on 80% of days was 314.29 km.

Sixty-two th Percentile (P62):
The P62 class is the class that includes the \( 62\left(\frac{n}{100}\right) \)th observation.
For \( n=60 \), this is the \( 62\left(\frac{60}{100}\right) = 62 \times 0.60 = 37.2 \)th observation.
Referring to the cumulative frequency (cf) column, the 37.2th observation falls within the 200-300 km class interval.

Using the formula for percentiles of grouped data:
\( P_{62} = L + \frac{62\left(\frac{n}{100}\right)-cf}{f} \times c \)
Where \( L = 200 \) (lower limit of P62 class), \( 62\left(\frac{n}{100}\right) = 37.2 \), \( cf = 23 \) (cumulative frequency of the class preceding P62 class), \( f = 24 \) (frequency of P62 class), and \( c = 100 \) (class width).

\( P_{62} = 200 + \frac{37.2-23}{24} \times 100 \)
\( = 200 + \frac{14.2 \times 100}{24} \)
\( = 200 + \frac{1420}{24} \)
\( = 200 + 59.17 \)
\( = 259.17 \) km
Hence, the sixty-second percentile P62 is 259.17 km.

Interpretation:
A sixty-second percentile of 259.17 km means that on 62% of the days, the salesman's travel distance was less than or equal to 259.17 km.
In simple words: These calculations help us understand the distribution of travel distances. The median is the middle value, Q3 is the 75th percentile, D8 is the 80th percentile, and P62 is the 62nd percentile, each giving us insight into specific points in the data range.

🎯 Exam Tip: When dealing with grouped data, correctly identifying the class interval for the desired quantile (median, quartile, decile, percentile) is the first critical step. Then, ensure you use the correct values for L, cf, f, and c from that specific class and its preceding one in the formula.

Question 3. The following table gives ages of 80 students selected from a college:

Age (years)	17	18	19	20	21	22	23
No. of students	11	14	22	15	8	6	4

Answer:
We first create a cumulative frequency distribution for the given discrete data.

Age (in years)	No. of students (f)	Cumulative frequency (cf)
17	11	11
18	14	25
19	22	47
20	15	62
21	8	70
22	6	76
23	4	80
Total	n = 80	-

Median (M):
The median is the value of the \( \left(\frac{n+1}{2}\right) \)th observation.
For \( n=80 \), this is the \( \left(\frac{80+1}{2}\right) = \left(\frac{81}{2}\right) = 40.5 \)th observation.
Referring to the cumulative frequency (cf) column, the 40.5th observation corresponds to an age of 19 years. Hence, the median \(M\) is 19 years.

Interpretation:
A median age of 19 years implies that 50% of the students selected from the college are 19 years old or younger.

First Quartile (Q1):
The first quartile is the value of the \( \left(\frac{n+1}{4}\right) \)th observation.
For \( n=80 \), this is the \( \left(\frac{80+1}{4}\right) = \left(\frac{81}{4}\right) = 20.25 \)th observation.
Referring to the cumulative frequency (cf) column, the 20.25th observation corresponds to an age of 18 years. Thus, Q1 is 18 years.

Interpretation:
A first quartile of 18 years indicates that 25% of the students selected are 18 years old or younger.

Fourth Decile (D4):
The fourth decile is the value of the \( 4\left(\frac{n+1}{10}\right) \)th observation.
For \( n=80 \), this is the \( 4\left(\frac{80+1}{10}\right) = 4(8.1) = 32.4 \)th observation.
Referring to the cumulative frequency (cf) column, the 32.4th observation corresponds to an age of 19 years. Hence, D4 is 19 years.

Interpretation:
A fourth decile of 19 years implies that 40% of the students selected are 19 years old or younger.

Thirty-second Percentile (P32):
The thirty-second percentile is the value of the \( 32\left(\frac{n+1}{100}\right) \)th observation.
For \( n=80 \), this is the \( 32\left(\frac{80+1}{100}\right) = 32 \times 0.81 = 25.92 \)th observation.
Referring to the cumulative frequency (cf) column, the 25.92th observation is between the 25th (age 18) and 47th (age 19) observations.

We use interpolation for P32:
P32 = Value of 25th observation + 0.92 (Value of 26th observation - Value of 25th observation)
= 18 + 0.92 (19 - 18)
= 18 + 0.92
= 18.92 years
Therefore, P32 = 18.92 years.

Interpretation:
A thirty-second percentile of 18.92 years means that 32% of the students are 18.92 years old or younger.
In simple words: For discrete data, we calculate the cumulative frequency and then directly locate the observation value that corresponds to the required percentile position (median, quartile, decile, or percentile). If the position is fractional, interpolation may be used.

🎯 Exam Tip: When calculating quantiles for discrete data, always find the position using \( \left(\frac{n+1}{k}\right) \) (for un-grouped data) or \( \left(\frac{n}{k}\right) \) (for grouped data in exclusive form). If the position is a fraction, interpolate between the two nearest integer positions.

Question 4. Use the following data to find the median salary of employees in a firm. Also find the lower limit for the richest 20% employees.

Salary (thousand Rs.)	5 or more	10 or more	15 or more	20 or more	25 or more	30 or more
No. of employees	120	117	106	76	31	12

Answer:
The provided information is a 'more than' type cumulative frequency distribution. The class length \(c\) is 5, as the difference between successive lower boundary points (e.g., 10-5) is 5.

We convert this into an original frequency distribution and a 'less than' type cumulative frequency distribution.

Wages (thousand Rs.)	'More than' cf	frequency (f)	'Less than' cumulative frequency (cf)
5-10	120	120 - 117 = 3	3
10-15	117	117 - 106 = 11	14
15-20	106	106 - 76 = 30	44
20-25	76	76 - 31 = 45	89
25-30	31	31 - 12 = 19	108
30 and more	12	12	120
Total	-	n = 120	-

Lower limit for the richest 20% employees (P80):
The richest 20% of employees implies that we are looking for the salary level above which 20% of employees fall. Equivalently, 80% of employees earn a wage less than or equal to this point, which is represented by the 80th percentile (P80).

The P80 class is the class that includes the \( 80\left(\frac{n}{100}\right) \)th observation.
For \( n=120 \), this is the \( 80\left(\frac{120}{100}\right) = 80 \times 1.2 = 96 \)th observation.
Referring to the 'less than' cumulative frequency (cf) column, the 96th observation falls within the 25-30 thousand Rs. class interval.

Using the formula for percentiles of grouped data:
\( P_{80} = L + \frac{80\left(\frac{n}{100}\right)-cf}{f} \times c \)
Where \( L = 25 \) (lower limit of P80 class), \( 80\left(\frac{n}{100}\right) = 96 \), \( cf = 89 \) (cumulative frequency of the class preceding P80 class), \( f = 19 \) (frequency of P80 class), and \( c = 5 \) (class width).

\( P_{80} = 25 + \frac{96-89}{19} \times 5 \)
\( = 25 + \frac{7 \times 5}{19} \)
\( = 25 + \frac{35}{19} \)
\( = 25 + 1.84 \)
\( = \text{Rs. } 26.84 \) thousand
Hence, the lower limit for the richest 20% of employees is Rs. 26.84 thousand.

Median (M):
The median class is the class that includes the \( \left(\frac{n}{2}\right) \)th observation.
For \( n=120 \), this is the \( \left(\frac{120}{2}\right) = 60 \)th observation.
Referring to the 'less than' cumulative frequency (cf) column, the 60th observation falls within the 20-25 thousand Rs. class interval.

Using the formula for median of grouped data:
\( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Where \( L = 20 \) (lower limit of median class), \( \frac{n}{2} = 60 \), \( cf = 44 \) (cumulative frequency of the class preceding the median class), \( f = 45 \) (frequency of the median class), and \( c = 5 \) (class width).

\( M = 20 + \frac{60-44}{45} \times 5 \)
\( = 20 + \frac{16 \times 5}{45} \)
\( = 20 + \frac{80}{45} \)
\( = 20 + 1.78 \)
\( = \text{Rs. } 21.78 \) thousand
Hence, the median salary is Rs. 21.78 thousand.
In simple words: We converted the 'more than' cumulative frequency distribution into a standard frequency distribution and then a 'less than' cumulative frequency distribution to calculate the median and the 80th percentile (P80), which represents the lower limit for the top 20% earners.

🎯 Exam Tip: When data is presented as 'more than' or 'less than' cumulative frequencies, the first step is always to convert it into a simple frequency distribution and then into the desired cumulative frequency type ('less than' is usually preferred for calculating quantiles). This conversion is crucial for accurate quantile calculations.

Question 5. The following table shows the monthly expense for entertainment in a group of 100 students. Find the median of this expense.

Expense (Rs.)	Less than 200	200-400	400-600	600-700	700-800	800 and above
No. of students	8	23	40	17	7	5

Answer:
We construct the cumulative frequency distribution for the given frequency distribution:

Expense (in Rs.)	No. of students (f)	Cumulative frequency (cf)
Less than 200	8	8
200-400	23	31
400-600	40	71
600-700	17	88
700-800	7	95
800 and more	5	100
Total	n = 100	-

Median (M):
The median class is the class that includes the \( \left(\frac{n}{2}\right) \)th observation.
For \( n=100 \), this is the \( \left(\frac{100}{2}\right) = 50 \)th observation.
Referring to the cumulative frequency (cf) column, the 50th observation falls within the 400-600 Rs. class interval.

Using the formula for median of grouped data:
\( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Where \( L = 400 \) (lower limit of median class), \( \frac{n}{2} = 50 \), \( cf = 31 \) (cumulative frequency of the class preceding the median class), \( f = 40 \) (frequency of the median class), and \( c = 200 \) (class width).

\( M = 400 + \frac{50-31}{40} \times 200 \)
\( = 400 + (19 \times 5) \)
\( = 400 + 95 \)
\( = \text{Rs. } 495 \)
Hence, the median monthly entertainment expense \(M\) is Rs. 495.
In simple words: The median expense is the middle value in the sorted dataset. By creating a cumulative frequency distribution, we identify the class containing the median and apply a specific formula to find its exact value within that class.

🎯 Exam Tip: Pay close attention to the class boundaries, especially for 'less than' type classes or open-ended classes. Ensure the cumulative frequency is correctly calculated, as any error will propagate through the median calculation.

Question 6. The following data indicate records of hospital stays (in days) of 30 patients admitted to a hospital:
1, 10, 2, 6, 3, 4, 15, 1, 5, 9, 2, 4, 3, 1, 10,
7, 3, 5, 4, 2, 4, 8, 5, 3, 1, 9, 6, 2, 3, 7
Find the median of stay. Further convert this information in a continuous frequency distribution (inclusive type) by taking classes of equal length starting from 1-3. Find the median from the frequency distribution and compare it with your earlier answer.
Answer:
Median of unclassified data:
For the unclassified data, we begin by arranging the number of hospital stay days in ascending order: 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 15. The total number of observations, \(n\), is 30.

The median \(M\) for unclassified data is determined by the value at the \( \left(\frac{n+1}{2}\right) \)th position.
With \( n=30 \), this corresponds to the \( \left(\frac{30+1}{2}\right) = 15.5 \)th observation.
Since this is between two values, the median is the average of the 15th and 16th observations. Both the 15th and 16th observations in the sorted list are 4.
Therefore, \( M = \frac{\text{value of 15th observation} + \text{value of 16th observation}}{2} = \frac{4+4}{2} = 4 \).
The median for the unclassified data is 4 days.

Median of classified data:
For calculating the median from classified data, we first note that the minimum observation is 1 day and the maximum is 15 days. The initial class interval is given as 1-3. We need to create an inclusive frequency distribution with equal class lengths. Following this pattern, the next class would be 4-6, and the final class, encompassing the maximum observation of 15, would be 13-15. The complete inclusive continuous frequency distribution is constructed as follows:

No. of days of stay in hospital	Tally Marks	No. of patients (f)	Cumulative frequency (cf)
1-3	\( \cancel{N} \cancel{N} \cancel{N} \text{ III} \)	13	13
4-6	\( \cancel{N} \cancel{N} \cancel{N} \text{ IIII} \)	9	22
7-9	\( \cancel{N} \)	5	27
10-12	II	2	29
13-15	I	1	30
Total	-	n = 30	-

The median class is the class that includes the \( \left(\frac{n}{2}\right) \)th observation.
For \( n=30 \), this is the \( \left(\frac{30}{2}\right) = 15 \)th observation.
Referring to the cumulative frequency (cf) column, the 15th observation falls within the 4-6 days class interval.

For the exclusive form, the median class is 3.5-6.5 (adjusting for inclusive boundaries).
Using the formula for median of grouped data:
\( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Where \( L = 3.5 \) (lower boundary of median class in exclusive form), \( \frac{n}{2} = 15 \), \( cf = 13 \) (cumulative frequency of the class preceding the median class), \( f = 9 \) (frequency of the median class), and \( c = 3 \) (class width).

\( M = 3.5 + \frac{15-13}{9} \times 3 \)
\( = 3.5 + \frac{2}{9} \times 3 \)
\( = 3.5 + \frac{2}{3} \)
\( = 3.5 + 0.67 \)
\( = 4.17 \) days
Hence, the median of the classified data is 4.17 days.

Comparison:
The median for unclassified data was 4 days, and for classified data, it is approximately 4.17 days. Both values are very close, indicating consistency between the two calculation methods.
In simple words: We calculated the median for raw data by sorting and finding the middle value. Then, we grouped the data into classes and used a formula for grouped data median. The results from both methods were similar, showing that grouping data can approximate the central tendency well.

🎯 Exam Tip: When converting inclusive classes to exclusive classes for median calculation, remember to adjust the lower and upper boundaries by subtracting/adding 0.5 (or half the gap between classes) respectively. This ensures continuity for the formula. Also, remember to clearly state the comparison between the medians of classified and unclassified data if asked.