GSEB Class 11 Statistics Solutions Chapter 4 Measures of Dispersion Solution Exercise 4.2

Question 1. A shooter missed his target in the last 10 trials by the following distance (mm) during the practice session: 20, 32, 24, 41, 18, 27, 15, 36, 35, 25 Find the quartile deviation and coefficient of quartile deviation of such distance missed by the shooter.

Answer:To determine the quartile deviation and its coefficient, we first arrange the given mistarget distances in ascending order: 15, 18, 20, 24, 25, 27, 32, 35, 36, 41 Here, the total number of observations, \( n = 10 \). **First Quartile (Q₁):** \( Q_1 = \text{Value of } \left(\frac{n+1}{4}\right)\text{th observation} \) \( = \text{Value of } \left(\frac{10+1}{4}\right)\text{th observation} \) \( = \text{Value of } \left(\frac{11}{4}\right)\text{th observation} \) \( = \text{Value of } 2.75\text{th observation} \) \( = \text{Value of 2nd observation} + 0.75 \times (\text{Value of 3rd observation} - \text{Value of 2nd observation}) \) \( = 18 + 0.75 \times (20 - 18) \) \( = 18 + 0.75 \times (2) \) \( = 18 + 1.50 \) \( Q_1 = 19.50 \text{ mm} \) **Third Quartile (Q₃):** \( Q_3 = \text{Value of } 3\left(\frac{n+1}{4}\right)\text{th observation} \) \( = \text{Value of } 3 \times (2.75)\text{th observation} \) \( = \text{Value of } 8.25\text{th observation} \) \( = \text{Value of 8th observation} + 0.25 \times (\text{Value of 9th observation} - \text{Value of 8th observation}) \) \( = 35 + 0.25 \times (36 - 35) \) \( = 35 + 0.25 \times (1) \) \( = 35 + 0.25 \) \( Q_3 = 35.25 \text{ mm} \) **Quartile Deviation (Qd) of Mistarget Measures:** \[ Qd = \frac{Q_3 - Q_1}{2} \] Substituting the calculated values, \( Q_3 = 35.25 \) and \( Q_1 = 19.50 \): \[ Qd = \frac{35.25 - 19.50}{2} \] \[ = \frac{15.75}{2} \] \[ = 7.875 \approx 7.88 \] **Coefficient of Quartile Deviation:** \[ \text{Coefficient of Quartile Deviation} = \frac{Q_3 - Q_1}{Q_3 + Q_1} \] Substituting the values: \[ = \frac{35.25 - 19.50}{35.25 + 19.50} \] \[ = \frac{15.75}{54.75} \] \[ = 0.2876... \] \[ \approx 0.29 \]In simple words: To find how spread out the distances are, we first sort them. Then, we calculate the first and third quartiles, which divide the data into four equal parts. Using these quartiles, we can compute the quartile deviation, which tells us the spread of the middle 50% of the data, and its coefficient, which indicates this spread relative to the average of the quartiles.

🎯 Exam Tip: Remember to always arrange the data in ascending order before calculating quartiles for ungrouped data. Pay close attention to the interpolation steps for fractional quartile positions to ensure accuracy.

Question 2. Find the quartile deviation and coefficient of quartile deviation of the marks from the following frequency distribution of marks of 43 students of a school;

Marks	10	20	30	40	50	60
No. of students	4	7	15	8	7	2

Answer:To calculate the quartile deviation and its coefficient, we first need to construct a cumulative frequency table for the given distribution:

Marks	No. of students (f)	Cumulative frequency (cf)
10	4	4
20	7	11
30	15	26
40	8	34
50	7	41
60	2	43

The total number of students \( n = 43 \). **First Quartile (Q₁):** \( Q_1 = \text{Value of } \left(\frac{n+1}{4}\right)\text{th observation} \) \( = \text{Value of } \left(\frac{43+1}{4}\right)\text{th observation} \) \( = \text{Value of } \frac{44}{4}\text{th observation} \) \( = \text{Value of } 11\text{th observation} \) Referring to the cumulative frequency (cf) column, the 11th observation falls into the group with a cf of 11, which corresponds to Marks = 20. Therefore, \( Q_1 = 20 \text{ marks} \) **Third Quartile (Q₃):** \( Q_3 = \text{Value of } 3\left(\frac{n+1}{4}\right)\text{th observation} \) \( = \text{Value of } 3 \times (11)\text{th observation} \) \( = \text{Value of } 33\text{rd observation} \) Referring to the cumulative frequency (cf) column, the 33rd observation falls into the group with a cf of 34, which corresponds to Marks = 40. Therefore, \( Q_3 = 40 \text{ marks} \) **Quartile Deviation (Qd) of Marks:** \[ Qd = \frac{Q_3 - Q_1}{2} \] Substituting \( Q_3 = 40 \) and \( Q_1 = 20 \): \[ Qd = \frac{40 - 20}{2} \] \[ = \frac{20}{2} \] \[ = 10 \text{ marks} \] **Coefficient of Quartile Deviation of Marks:** \[ \text{Coefficient of Quartile Deviation} = \frac{Q_3 - Q_1}{Q_3 + Q_1} \] Substituting the values: \[ = \frac{40 - 20}{40 + 20} \] \[ = \frac{20}{60} \] \[ = 0.33 \]In simple words: For grouped data, we find the cumulative frequency to locate the positions of the first and third quartiles. Once located in the frequency table, these quartile values help us calculate the quartile deviation, which measures the spread of the middle half of the marks, and its coefficient for relative comparison.

🎯 Exam Tip: When dealing with discrete frequency distributions, locate the quartile values directly by referring to the cumulative frequency column. Ensure correct identification of the corresponding observation value for \( \frac{n+1}{4} \) and \( 3\frac{n+1}{4} \).

Question 3. The distribution of amount paid by 200 customers coming for snacks at a restaurant on a particular day is given below: Find the quartile deviation and coefficient of quartile deviation of the amount paid by customers on the day.

Amount (Rs.)	0-50	50-100	100-150	150-200	200-250
No. of customers	25	40	80	30	25

Answer:To calculate the quartile deviation and its coefficient, we first prepare a cumulative frequency distribution for the given amounts paid by customers:

Amount paid (Rs.)	No. of customers (f)	Cumulative frequency (cf)
0-50	25	25
50-100	40	65
100-150	80	145
150-200	30	175
200-250	25	200
Total	\( n = 200 \)	-

**First Quartile (Q₁):** The \( Q_1 \) class is the class that includes the \( \left(\frac{n}{4}\right)\text{th observation} \). \( = \left(\frac{200}{4}\right)\text{th observation} \) \( = 50\text{th observation} \) Referring to the cumulative frequency column, the 50th observation falls in the 50-100 class (since cf = 25 for 0-50, and cf = 65 for 50-100). So, the \( Q_1 \) class is 50-100. For this class: Lower limit \( L = 50 \) Cumulative frequency of preceding class \( cf = 25 \) Frequency of the \( Q_1 \) class \( f = 40 \) Class width \( c = 50 \) Now, using the formula for \( Q_1 \) for grouped data: \[ Q_1 = L + \frac{\left(\frac{n}{4}\right) - cf}{f} \times c \] \[ Q_1 = 50 + \frac{50 - 25}{40} \times 50 \] \[ = 50 + \frac{25}{40} \times 50 \] \[ = 50 + \frac{1250}{40} \] \[ = 50 + 31.25 \] \[ Q_1 = 81.25 \text{ Rs.} \] **Third Quartile (Q₃):** The \( Q_3 \) class is the class that includes the \( 3\left(\frac{n}{4}\right)\text{th observation} \). \( = 3 \times \left(\frac{200}{4}\right)\text{th observation} \) \( = 3 \times 50\text{th observation} \) \( = 150\text{th observation} \) Referring to the cumulative frequency column, the 150th observation falls in the 150-200 class (since cf = 145 for 100-150, and cf = 175 for 150-200). So, the \( Q_3 \) class is 150-200. For this class: Lower limit \( L = 150 \) Cumulative frequency of preceding class \( cf = 145 \) Frequency of the \( Q_3 \) class \( f = 30 \) Class width \( c = 50 \) Now, using the formula for \( Q_3 \) for grouped data: \[ Q_3 = L + \frac{3\left(\frac{n}{4}\right) - cf}{f} \times c \] \[ Q_3 = 150 + \frac{150 - 145}{30} \times 50 \] \[ = 150 + \frac{5}{30} \times 50 \] \[ = 150 + \frac{250}{30} \] \[ = 150 + 8.333... \] \[ Q_3 \approx 158.33 \text{ Rs.} \] **Quartile Deviation (Qd) of Amount Paid:** \[ Qd = \frac{Q_3 - Q_1}{2} \] Substituting the calculated values, \( Q_3 = 158.33 \) and \( Q_1 = 81.25 \): \[ Qd = \frac{158.33 - 81.25}{2} \] \[ = \frac{77.08}{2} \] \[ Qd = 38.54 \text{ Rs.} \] **Coefficient of Quartile Deviation of Amount Paid:** \[ \text{Coefficient of Quartile Deviation} = \frac{Q_3 - Q_1}{Q_3 + Q_1} \] Substituting the values: \[ = \frac{158.33 - 81.25}{158.33 + 81.25} \] \[ = \frac{77.08}{239.58} \] \[ \approx 0.3217... \] \[ \approx 0.32 \]In simple words: For data organized into classes, we first find the classes where the first and third quartiles fall using their positions based on cumulative frequency. Then, we apply a specific formula to interpolate the exact quartile values within those classes. Finally, we use these quartile values to calculate the quartile deviation, which shows the spread of the middle half of customer payments, and its coefficient for relative comparison.

🎯 Exam Tip: When dealing with continuous frequency distributions, correctly identify the quartile class using cumulative frequency, then apply the interpolation formula accurately. Do not forget to calculate the class width 'c'.