Maharashtra Board Class 11 Maths Part 2 Chapter 1 Partition Values Miscellaneous Solutions

Question 1. The data gives the number of accidents per day on a railway track. Compute Q2, P17, and D7. 4, 2, 3, 5, 6, 3, 4, 1, 2, 3, 2, 3, 4, 3, 2
Answer: Solution: The given data can be arranged in ascending order as follows: 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6 Here, n = 15 \[ Q_2 = \text{value of } 2\left(\frac{n+1}{4}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } 2\left(\frac{15+1}{4}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (2 \times 4)^{\text{th}} \text{observation} \] \[ = \text{value of } 8^{\text{th}} \text{observation} \] \[ \therefore Q_2 = 3 \] \[ P_{17} = \text{value of } 17\left(\frac{n+1}{100}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } 17\left(\frac{15+1}{100}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (17 \times 0.16)^{\text{th}} \text{observation} \] \[ = \text{value of } (2.72)^{\text{th}} \text{observation} \] \[ = \text{value of } 2^{\text{nd}} \text{observation} + 0.72 (\text{value of } 3^{\text{rd}} \text{observation} - \text{value of } 2^{\text{nd}} \text{observation}) \] \[ = 2 + 0.72 (2 - 2) \] \[ \therefore P_{17} = 2 \] \[ D_7 = \text{value of } 7\left(\frac{n+1}{10}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } 7\left(\frac{15+1}{10}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (7 \times 1.6)^{\text{th}} \text{observation} \] \[ = \text{value of } (11.2)^{\text{th}} \text{observation} \] \[ = \text{value of } 11^{\text{th}} \text{observation} + 0.2(\text{value of } 12^{\text{th}} \text{observation} - \text{value of } 11^{\text{th}} \text{observation}) \] \[ = 4 + 0.2(4 - 4) \] \[ \therefore D_7 = 4 \] In simple words: This solution calculates specific partition values (Q2, P17, D7) for raw accident data. It involves arranging the data in ascending order and then using their respective formulas which often require interpolation for non-integer ranks.

🎯 Exam Tip: Always arrange raw data in ascending order before calculating quartiles, percentiles, or deciles. Pay close attention to the interpolation step when the rank is not an integer.

Question 2. The distribution of daily sales of shoes (size-wise) for 100 days from a certain shop is as follows:

Compute Q1, D2, and P95.
Answer: Solution: By arranging the given data in ascending order, we construct the less than cumulative frequency table as given below:

Here, n = 100 \[ Q_1 = \text{value of } \left(\frac{n+1}{4}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } \left(\frac{100+1}{4}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (25.25)^{\text{th}} \text{observation} \] Cumulative frequency which is just greater than (or equal) to 25.25 is 27. \[ \therefore Q_1 = 3 \]
\[ D_2 = \text{value of } 2\left(\frac{n+1}{10}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } 2\left(\frac{100+1}{10}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (2 \times 10.1)^{\text{th}} \text{observation} \] \[ = \text{value of } (20.2)^{\text{th}} \text{observation} \] Cumulative frequency which is just greater than (or equal) to 20.2 is 27. \[ \therefore D_2 = 3 \]
\[ P_{95} = \text{value of } 95\left(\frac{n+1}{100}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } 95\left(\frac{100+1}{100}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (95 \times 1.01)^{\text{th}} \text{observation} \] \[ = \text{value of } (95.95)^{\text{th}} \text{observation} \] The cumulative frequency which is just greater than (or equal) to 95.95 is 100. \[ \therefore P_{95} = 8 \] In simple words: This solution finds Q1, D2, and P95 for discrete data (shoe sizes). It first organizes the data into a cumulative frequency table, then uses the specific formulas to find the position of each partition value, and finally identifies the corresponding shoe size from the table.

🎯 Exam Tip: For discrete data, after finding the rank, locate the smallest cumulative frequency greater than or equal to that rank. The corresponding class or value is your answer. No interpolation is needed for discrete data.

Question 3. Ten students appeared for a test in Mathematics and Statistics and they obtained the marks as follows:

If the median will be the criteria, in which subject, the level of knowledge of the students is higher?
Answer: Solution: Marks in Mathematics can be arranged in ascending order as follows: 23, 23, 25, 25, 32, 35, 36, 37, 38, 42 Here, n = 10 \[ \therefore \text{ Median = value of } \left(\frac{n+1}{2}\right)^{\text{th}} \text{observation} \] \[ \text{Median = value of } \left(\frac{10+1}{2}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (5.5)^{\text{th}} \text{observation} \] \[ = \text{value of } 5^{\text{th}} \text{observation} + 0.5(\text{value of } 6^{\text{th}} \text{observation} - \text{value of } 5^{\text{th}} \text{observation}) \] \[ = 32 + 0.5 (35 - 32) \] \[ = 32 + 0.5(3) \] \[ = 32 + 1.5 \] \[ = 33.5 \] Marks in Statistics can be arranged in ascending order as follows: 22, 23, 26, 28, 29, 32, 34, 36, 45, 50 Here, n = 10 \[ \therefore \text{ Median = value of } \left(\frac{n+1}{2}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } \left(\frac{10+1}{2}\right)^{\text{th}} \text{observation} \] \[ = \text{value of } (5.5)^{\text{th}} \text{observation} \] \[ = \text{value of } 5^{\text{th}} \text{observation} + 0.5(\text{value of } 6^{\text{th}} \text{observation} - \text{value of } 5^{\text{th}} \text{observation}) \] \[ = 29 + 0.5(32 - 29) \] \[ = 29 + 0.5(3) \] \[ = 29 + 1.5 \] \[ = 30.5 \] \[ \therefore \text{ Median marks for Mathematics } = 33.5 \text{ and} \] Median marks for Statistics = 30.5 \[ \therefore \text{ The level of knowledge in Mathematics is higher than that of Statistics.} \] In simple words: To compare the knowledge level between two subjects using median as a criterion, first arrange the marks for each subject in ascending order. Then, calculate the median for both subjects. The subject with the higher median score indicates a higher level of knowledge.

🎯 Exam Tip: When comparing datasets based on median, always ensure the data for each set is sorted. The median is a robust measure against outliers, making it suitable for comparing typical performance.

Question 4. In the frequency distribution of families given below, the number of families corresponding to expenditure group 2000 - 4000 is missing from the table. However, the value of the 25th percentile is 2880. Find the missing frequency.

Answer: Solution: Let x be the missing frequency of expenditure group 2000 - 4000. We construct the less than cumulative frequency table as given below:

Here, N = 75 + x Given, P25 = 2880
\[ \therefore \text{ P25 lies in the class 2000 - 4000.} \] \[ \therefore \text{ L = 2000, h = 2000, f = x, c.f. = 14} \] \[ \therefore P_{25} = L + \frac{h}{f} \left(\frac{25N}{100} - c.f.\right) \] \[ \implies 2880 = 2000 + \frac{2000}{x} \left(\frac{75+x}{4} - 14\right) \] \[ \implies 2880 - 2000 = \frac{2000}{x} \left(\frac{75+x-56}{4}\right) \] \[ \implies 880 = \frac{500}{x} (x + 19) \] \[ \implies 880x = 500(x + 19) \] \[ \implies 880x = 500x + 9500 \] \[ \implies 880x - 500x = 9500 \] \[ \implies 380x = 9500 \] \[ \implies x = 25 \] \[ \therefore \text{ 25 is the missing frequency of the expenditure group 2000 - 4000.} \] In simple words: This problem asks us to find a missing frequency in a grouped frequency distribution. We are given the 25th percentile value. We construct a cumulative frequency table, identify the class where P25 lies, and then use the percentile formula for grouped data to solve for the unknown frequency.

🎯 Exam Tip: When dealing with missing frequencies, always form the cumulative frequency table first. The key is to correctly identify the class containing the given percentile and apply the formula accurately, solving the resulting algebraic equation.

Question 5. Calculate Q1, D6, and P15 for the following data:

Answer: Solution: Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
\[ \therefore \text{ the class intervals will be 0 - 50, 50 - 100, etc.} \] We construct the less than cumulative frequency table as given below:

Here, N = 500 \[ Q_1 \text{ class = class containing } \left(\frac{N}{4}\right)^{\text{th}} \text{observation} \] \[ \therefore \frac{N}{4} = \frac{500}{4} = 125 \] Cumulative frequency which is just greater than (or equal) to 125 is 160. Q1 lies in the class 100 - 150. \[ \therefore \text{ L = 100, h = 50, f = 80, c.f. = 80} \] \[ \therefore Q_1 = L + \frac{h}{f} \left(\frac{N}{4} - c.f.\right) \] \[ = 100 + \frac{50}{80} (125 - 80) \] \[ = 100 + \frac{5}{8} (45) \] \[ = 100 + 28.125 \] \[ = 128.125 \]
\[ D_6 \text{ class = class containing } \left(\frac{6N}{10}\right)^{\text{th}} \text{observation} \] \[ \therefore \frac{6N}{10} = \frac{6 \times 500}{10} = 300 \] Cumulative frequency which is just greater than (or equal) to 300 is 410.
\[ \therefore D_6 \text{ lies in the class 200 - 250.} \] \[ \therefore \text{ L = 200, h = 50, f = 150, c.f. = 260} \] \[ \therefore D_6 = L + \frac{h}{f} \left(\frac{6N}{10} - c.f.\right) \] \[ = 200 + \frac{50}{150} (300 - 260) \] \[ = 200 + \frac{1}{3} (40) \] \[ = 200 + 13.33 \] \[ = 213.33 \]
\[ P_{15} \text{ class = class containing } \left(\frac{15N}{100}\right)^{\text{th}} \text{observation} \] \[ \therefore \frac{15N}{100} = \frac{15 \times 500}{100} = 75 \] Cumulative frequency which is just greater than (or equal) to 75 is 80. \[ \therefore P_{15} \text{ lies in the class 50 - 100.} \] \[ \therefore \text{ L = 50, h = 50, f = 70, c.f. = 10} \] \[ \therefore P_{15} = L + \frac{h}{f} \left(\frac{15N}{100} - c.f.\right) \] \[ = 50 + \frac{50}{70} (75 - 10) \] \[ = 50 + \frac{5}{7} (65) \] \[ = 50 + \frac{325}{7} \] \[ = 50 + 46.4286 \] \[ = 96.4286 \] \[ \therefore Q_1 = 128.125, D_6 = 213.33, P_{15} = 96.4286 \] In simple words: This solution involves calculating Q1, D6, and P15 for grouped data given mid-values. First, convert mid-values into class intervals and construct a cumulative frequency table. Then, use the specific formulas for quartiles, deciles, and percentiles in grouped data, identifying the correct class and applying the interpolation formula.

🎯 Exam Tip: When given mid-values, determine the class width to construct the class intervals. Ensure continuous classes if needed. Remember to correctly identify L, h, f, and c.f. for each partition value's class before applying the formula.

Question 6. Daily income for a group of 100 workers are given below:

P30 for this group is Rs. 110. Calculate the missing frequencies.
Answer: Solution: Let a and b be the missing frequencies of class 50 - 100 and class 200 - 250 respectively. We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b Since, N = 100 \[ \therefore 62 + a + b = 100 \] \[ \therefore a + b = 38 ......(i) \] Given, P30 = 110
\[ \therefore P_{30} \text{ lies in the class 100 - 150.} \] \[ \therefore \text{ L = 100, h = 50, f = 25, c.f. = 7 + a} \] \[ \therefore \frac{30N}{100} = \frac{30 \times 100}{100} = 30 \] \[ \therefore P_{30} = L + \frac{h}{f} \left(\frac{30N}{100} - c.f.\right) \] \[ \therefore 110 = 100 + \frac{50}{25} [30 - (7 + a)] \] \[ \therefore 110 - 100 = 2(30 - 7 - a) \] \[ \therefore 10 = 2(23 - a) \] \[ \therefore 5 = 23 - a \] \[ \implies a = 23 - 5 \] \[ \therefore a = 18 \] Substituting the value of a in equation (i), we get \[ 18 + b = 38 \] \[ \therefore b = 38 - 18 \] \[ \therefore b = 20 \] \[ \therefore \text{ 18 and 20 are the missing frequencies of the class 50 - 100 and class 200 - 250 respectively.} \] In simple words: This solution determines two missing frequencies in a grouped frequency distribution. We form a system of two equations: one from the total frequency (N=100) and another by using the given P30 value (Rs. 110) with the percentile formula for grouped data. Solving these equations yields the missing frequencies.

🎯 Exam Tip: For problems with multiple missing frequencies, leverage all given information. The total frequency provides one equation, and a given percentile (or quartile/median) value provides another equation from its formula. Systematically solve these simultaneous equations.

Question 7. The distribution of a sample of students appearing for a C.A. examination is:

Help C.A. institute to decide cut-off marks for qualifying for an examination when 3% of students pass the examination.
Answer: Solution: To decide cut-off marks for qualifying for an examination when 3% of students pass, we have to find P97. We construct the less than cumulative frequency table as given below:

Here, N = 1100 \[ P_{97} \text{ class = class containing } \left(\frac{97N}{100}\right)^{\text{th}} \text{observation} \] \[ \therefore \frac{97N}{100} = \frac{97 \times 1100}{100} = 1067 \] Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
\[ \therefore P_{97} \text{ lies in the class 500 - 600.} \] \[ \therefore \text{ L = 500, h = 100, f = 130, c.f. = 970} \] \[ \therefore P_{97} = L + \frac{h}{f} \left(\frac{97N}{100} - c.f.\right) \] \[ = 500 + \frac{100}{130} (1067 - 970) \] \[ = 500 + \frac{10}{13} (97) \] \[ = 500 + 74.62 \] \[ = 574.62 \sim 575 \] \[ \therefore \text{ the cut off marks for qualifying an examination is 575.} \] In simple words: To find the cut-off marks for an exam where only 3% of students pass, we need to find the 97th percentile (P97). This means 3% of students scored above this mark. First, construct a cumulative frequency table, then use the percentile formula for grouped data to calculate P97, which will be the cut-off mark.

🎯 Exam Tip: When a problem states a percentage of students 'pass' (e.g., top 3%), it implies finding the corresponding high percentile (e.g., 100-3 = 97th percentile) to determine the cut-off mark. Ensure correct calculation of the percentile rank for grouped data.

Question 8. Determine graphically the value of median, D3, and P35 for the data given below:

Answer: Solution: To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (15, 8), (20, 22), (25, 30), (30, 55), (35, 70), (40, 84), (45, 90).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक 'लेस दैन ओजाइव' (संचयी आवृत्ति वक्र) को दर्शाता है। X-अक्ष पर वर्ग अंतराल (Class) के ऊपरी सीमा मान और Y-अक्ष पर संचयी आवृत्ति (cumulative frequency) प्लॉट की जाती है। इस वक्र का उपयोग करके माध्यिका, D3 (तीसरा दशमक), और P35 (35वां प्रतिशतक) के मानों को ग्राफिक रूप से निर्धारित किया जाता है। इन मानों को ज्ञात करने के लिए, Y-अक्ष पर संबंधित रैंक (जैसे माध्यिका के लिए N/2) से एक क्षैतिज रेखा खींची जाती है, जो वक्र को काटती है, फिर उस प्रतिच्छेदन बिंदु से X-अक्ष पर एक लंबवत रेखा खींची जाती है; जहाँ यह X-अक्ष को छूती है, वही अभीष्ट मान होता है। N = 90 \[ \text{For median, consider } \frac{N}{2} = \frac{90}{2} = 45 \] \[ \text{For } D_3\text{, consider } \frac{3N}{10} = \frac{3 \times 90}{10} = 27 \] \[ \text{For } P_{35}\text{, consider } \frac{35N}{100} = \frac{35 \times 90}{100} = 31.5 \] \[ \therefore \text{ We take the values 45, 27 and 31.5 on the Y-axis and draw lines from these} \] points parallel to X-axis. From the points where they intersect the less than ogive, we draw perpendicular on the X-axis. Foot of the perpendicular represent the values of median, D3 and P35 respectively. \[ \therefore \text{ Median } \sim 29, D_3 \sim 23.5, P_{35} \sim 26 \] In simple words: To graphically find the median, D3, and P35, first construct a less than cumulative frequency curve (ogive). Then, calculate the rank for each (N/2 for median, 3N/10 for D3, 35N/100 for P35), mark these ranks on the Y-axis, draw horizontal lines to the ogive, and then drop perpendicular lines to the X-axis to find the corresponding values.

🎯 Exam Tip: When drawing ogives for graphical solutions, plot cumulative frequencies against the upper limits of the class intervals for a "less than" ogive. Clearly label axes and scales. Accurate drawing and precise reading of the graph are crucial for correct answers.

Question 9. The I.Q. test of 500 students of a college is as follows:

Find graphically the number of students whose I.Q. is more than 55 graphically.
Answer: Solution: To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (30, 41), (40, 93), (50, 157), (60, 337), (70, 404), (80, 449), (90, 489), (100, 500)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख छात्रों की IQ (बुद्धि लब्धि) के संचयी आवृत्ति वितरण को दर्शाता है। X-अक्ष पर IQ स्कोर और Y-अक्ष पर छात्रों की संचयी संख्या है। यह एक 'लेस दैन ओजाइव' वक्र है। छात्रों की वह संख्या ज्ञात करने के लिए जिनकी IQ 55 से अधिक है, X-अक्ष पर 55 के मान से Y-अक्ष के समानांतर एक रेखा खींची जाती है। जहाँ यह रेखा ओजाइव को काटती है, वहाँ से Y-अक्ष पर एक लंबवत रेखा खींची जाती है। Y-अक्ष पर यह बिंदु उन छात्रों की संख्या देता है जिनकी IQ 55 से कम या उसके बराबर है। फिर, कुल छात्रों में से इस संख्या को घटाकर 55 से अधिक IQ वाले छात्रों की संख्या प्राप्त की जाती है। To find the number of students whose I.Q. is more than 55, we consider the value 55 on the X-axis. From this point, we draw a line that is parallel to Y-axis. From the point this line intersects the less than ogive, we draw a perpendicular on the Y-axis. The foot of perpendicular gives the number of students whose I.Q. is less than 55. \[ \therefore \text{ The foot of perpendicular } \sim 244 \] \[ \therefore \text{ No. of students whose I.Q. is less than 55 } \sim 244 \] \[ \therefore \text{ No. of Students whose I.Q. is more than 55 } = 500 - 244 = 256 \] In simple words: To graphically find the number of students with an IQ more than 55, we first construct a "less than" ogive. Then, from the IQ value of 55 on the X-axis, we draw a vertical line to the ogive and then a horizontal line to the Y-axis. The value on the Y-axis indicates students with an IQ less than 55. Subtracting this from the total number of students gives those with an IQ more than 55.

🎯 Exam Tip: To find "more than" values graphically from a "less than" ogive, first find the "less than or equal to" value from the graph. Then, subtract this value from the total frequency (N) to get the "more than" value. Double-check your graph readings for accuracy.

Question 10. Draw an ogive for the following distribution. Determine the median graphically and verify your result by a mathematical formula.
Answer: Solution: To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख छात्रों की ऊंचाई (सेमी में) के लिए 'लेस दैन ओजाइव' वक्र प्रस्तुत करता है। X-अक्ष पर ऊंचाई के ऊपरी सीमा बिंदु और Y-अक्ष पर संचयी आवृत्ति दर्शाई गई है। माध्यिका को ग्राफिक रूप से निर्धारित करने के लिए, कुल आवृत्ति (N) का आधा मान (N/2 = 30) Y-अक्ष पर लिया जाता है। इस बिंदु से वक्र तक एक क्षैतिज रेखा खींची जाती है, और फिर वक्र के प्रतिच्छेदन बिंदु से X-अक्ष पर एक लंबवत रेखा खींची जाती है। X-अक्ष पर यह अंतिम बिंदु ही माध्यिका का ग्राफिक मान होता है, जिसकी पुष्टि सूत्र द्वारा भी की जाती है। N = 60 \[ \frac{N}{2} = \frac{60}{2} = 30 \] \[ \therefore \text{ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis.} \] From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis. The foot perpendicular gives the value of the median. \[ \therefore \text{ Median } \sim 164.67 \] Now, let us calculate the median from the mathematical formula. \[ \frac{N}{2} = 30 \] The median lies in the class interval 160 - 165. \[ \therefore \text{ L = 160, h = 5, f = 15, c.f. = 16} \] \[ \text{Median} = L + \frac{h}{f} \left(\frac{N}{2} - c.f.\right) \] \[ = 160 + \frac{5}{15} (30 - 16) \] \[ = 160 + \frac{1}{3} \times 14 \] \[ = 160 + 4.67 \] \[ = 164.67 \] In simple words: To determine the median graphically and verify it using a formula, first create a "less than" ogive. On the graph, find N/2 on the Y-axis, draw a horizontal line to the ogive, and then a vertical line to the X-axis to get the graphical median. For verification, use the median formula for grouped data, ensuring all parameters (L, h, f, c.f.) are correctly identified from the median class.

🎯 Exam Tip: When both graphical and formula methods are required, perform both calculations and ensure they are consistent. Small deviations between graphical and calculated values are acceptable due to reading accuracy, but they should be very close. Clearly show all steps for both methods.

Question 11. In a group of 25 students, 7 students failed and 6 students got distinction and the marks of the remaining 12 students are 61, 36, 44, 59, 52, 56, 41, 37, 39, 38, 41, 64. Find the median marks of the whole group.
Solution:
n = 25
Median = \( \frac{n+1}{2} = \frac{25+1}{2} \) = 13th observation
We have been stated that 7 students failed (assuming passing marks on 35) and 6 students got distinction (assuming distinction as 70+), and the marks of the remaining 12 students (who will be situated between the two groups mentioned above, if arranged in ascending order), we have,
F, F, F, F, F, F, F, 36, 37, 38, 39, 41, 41, 44, 52, 56, 59, 61, 64, D, D, D, D, D, D
∴ median = 13th observation = 41.
In simple words: To find the median, we arrange all 25 student marks (failures, remaining, and distinctions) in ascending order and find the value of the 13th observation. Since 7 failed and 6 got distinction, the 12 remaining scores (36-64) fall in the middle, and the 13th value turns out to be 41 after ordering all scores.

🎯 Exam Tip: Remember to consider all data points (failed, in-between, and distinction) when calculating the median for a combined group. Arranging the data in ascending order is crucial for accuracy.

Question 12. The median weight of a group of 79 students is found to be 55 kg. 6 more students are added to this group whose weights are 50, 51, 52, 59.5, 60, 61 kg. What will be the value of the median of the combined group if the lowest and the highest weights were 53 kg and 59 kg respectively?
Solution:
n = 79
Median = 55kg
Lowest observation = 53 kg
Highest observation = 59 kg
6 new students are added to the group having weights in Kg as follows:
50, 51, 52, 59.5, 60, 61
From the above, we see that of the 6 new students, 3 have weights which are below the lowest weight of the earlier group and 3 have weights which are above the highest weight of the earlier group.
∴ the median remains the same
∴ median = 55 kg.
In simple words: When new students are added, their weights (50, 51, 52, 59.5, 60, 61 kg) are compared to the existing group's lowest (53 kg) and highest (59 kg) weights. Since the new students' weights are either below the old lowest or above the old highest, they don't change the central position of the median. Thus, the median weight remains 55 kg.

🎯 Exam Tip: Understand how adding extreme values (values outside the existing range) affects the median. If an equal number of values are added to both ends of the distribution, the median might remain unchanged, especially for large datasets.

Question 13. The median of the following incomplete table is 92. Find the missing frequencies:
Answer:
Solution:
Let a and b be the missing frequencies of class 50 - 70 and class 110 - 130 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 54 + a + b
Since, N = 80
∴ 54 + a + b = 80
∴ a + b = 26 ......(i)
Given, Median = Q2 = 92
∴ Q2 lies in the class 90 - 110.
∴ L = 90, h = 20, f = 20, c.f. = 24 + a
\( \frac{2N}{4} = \frac{2 \times 80}{4} = 40 \)
∴ \( Q_2 = L + \frac{h}{f} (\frac{2N}{4} - c.f.) \)
∴ \( 92 = 90 + \frac{20}{20} [40 - (24 + a)] \)
∴ \( 92 - 90 = 40 - 24 - a \)
∴ \( 2 = 16 - a \)
∴ a = 14
Substituting the value of a in equation (i), we get
14 + b = 26
∴ b = 26 - 14 = 12
∴ 14 and 12 are the missing frequencies of the class 50 - 70 and class 110 - 130 respectively.
In simple words: First, we set up a cumulative frequency table with the missing frequencies 'a' and 'b'. Using the given total frequency (N=80) and the median value (92), we identified the median class. Then, we applied the median formula for grouped data to form an equation and solved for 'a'. Finally, 'a' was substituted back into the equation (a + b = 26) to find 'b'. The missing frequencies are 14 and 12.

🎯 Exam Tip: When dealing with missing frequencies in grouped data, always start by constructing the cumulative frequency table and establishing an equation for the total frequency. The median formula is a key tool for solving for the missing values. Be meticulous with algebraic manipulation.

Question 14. A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

estimate the number of defective tubes in the central batch.
Solution:
To find the number of defective tubes in the central batch, we have to find Q2.
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 4.5, 4.5 - 9.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 251
Q2 class = class containing \( (\frac{2N}{4})^{th} \) observation
∴ \( \frac{2N}{4} = \frac{2 \times 251}{4} = 125.5 \)
Cumulative frequency which is just greater than (or equal) to 125.5 is 180.
∴ Q2 lies in the class 9.5 - 14.5.
∴ L = 9.5, h = 5, f = 84, c.f. = 96
∴ \( Q_2 = L + \frac{h}{f} (\frac{2N}{4} - c.f.) \)
\( = 9.5 + \frac{5}{84} (125.5 - 96) \)
\( = 9.5 + \frac{5}{84} \times 29.5 \)
\( = 9.5 + \frac{147.5}{84} \)
\( = 9.5 + 1.76 \)
\( = 11.26 \)
In simple words: To find the number of defective tubes in the central batch, we calculate the second quartile (Q2). First, we convert the discontinuous data into continuous class intervals. Then, we construct a cumulative frequency table. Using the formula for Q2 in grouped data, we find that the Q2 value is approximately 11.26, which represents the central number of defective tubes.

🎯 Exam Tip: Always convert discontinuous data into continuous class intervals (by adjusting limits by 0.5) before calculating quartiles, deciles, or percentiles for grouped data. This ensures the accuracy of the class boundaries for interpolation.

Question 15. In a college, there are 500 students in junior college, 5% score less than 25 marks, 68 scores from 26 to 30 marks, 30% score from 31 to 35 marks, 70 scores from 36 to 40 marks, 20% score from 41 to 45 marks and the rest score 46 and above marks. What are the median marks?
Solution:
Given data can be written in tabulated form as follows:

Since the given data is not continuous, we have to convert it into the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 25.5, 25.5 - 30.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q2 class = class containing \( (\frac{2N}{4})^{th} \) observation
∴ \( \frac{2N}{4} = \frac{2 \times 500}{4} = 250 \)
Cumulative frequency which is just greater than (or equal) to 250 is 313.
∴ Q2 lies in the class 35.5 - 40.5.
∴ L = 35.5, h = 5, f = 70, c.f. = 243
∴ Median = \( Q_2 = L + \frac{h}{f} (\frac{2N}{4} - c.f.) \)
\( = 35.5 + \frac{5}{70} (250 - 243) \)
\( = 35.5 + \frac{1}{14} (7) \)
\( = 35.5 + 0.5 \)
\( = 36 \)
In simple words: First, we organize the given student mark percentages and counts into a frequency table, calculating the actual number of students in each mark range. Since the data is discontinuous, we adjust the class intervals to be continuous. Then, we create a cumulative frequency table to find the median class. Finally, using the median formula for grouped data, we calculate the median marks, which come out to be 36.

🎯 Exam Tip: When dealing with percentage-based or non-continuous data, the initial step should always be to convert them into absolute frequencies and continuous class intervals. This sets a solid foundation for accurate quartile calculations.

Question 16. Draw a cumulative frequency curve more than typical for the following data and hence locate Q1 and Q3. Also, find the number of workers with daily wages
(i) Between Rs. 170 and Rs. 260
(ii) less than Rs. 260

Solution:
For more than ogive points to be plotted are (100, 200), (150, 188), (200, 160), (250, 124), (300, 74), (350, 49), (400, 31), (450, 15), (500, 5)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक 'अधिक से अधिक' संचयी आवृत्ति वक्र (ओजाइव) है जो दैनिक मजदूरी (X-अक्ष पर Rs. में) और श्रमिकों की संख्या (Y-अक्ष पर) को दर्शाता है। वक्र नीचे की ओर ढलान वाला है, जो दिखाता है कि जैसे-जैसे दैनिक मजदूरी बढ़ती है, उससे अधिक मजदूरी पाने वाले श्रमिकों की संख्या घटती जाती है। इस ग्राफ का उपयोग करके Q1 और Q3 को दर्शाया गया है, साथ ही मजदूरी के विशिष्ट मूल्यों के लिए श्रमिकों की संख्या का पता लगाने के लिए लंबवत और क्षैतिज रेखाएं खींची गई हैं।

Here, N = 200
For Q1, \( \frac{N}{4} = \frac{200}{4} = 50 \)
For Q3, \( \frac{3N}{4} = \frac{3 \times 200}{4} = 150 \)
We take the points having Y co-ordinates 50 and 150 on Y-axis.
From these points, we draw lines which are parallel to X-axis.
From the points of intersection of these lines with the curve, we draw perpendicular on X-axis.
X-Co-ordinates of these points gives the values of Q1 and Q3.
Since X-axis has daily wages more than and not less than the given amounts.
∴ Q1 = Q3 and Q3 = Q1
∴ Q2 - 215, Q3 - 348

(i) To find the number of workers with daily wages between Rs. 170 and Rs. 260,
Take the values 170 and 260 on X-axis. From these points, we draw lines parallel to Y-axis.
From the point where they intersect the more than ogive, we draw perpendiculars on Y-axis.
The points where they intersect the Y-axis gives the values 178 and 114.
∴ Number of workers having daily wages between Rs. 170 and Rs. 260 = 178 - 114 = 64

(ii) To find the number of workers having daily wages less than Rs. 260, we consider the value 260 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point where the line intersects the more than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of workers having daily wages of more than 260.
The foot of perpendicular - 114
∴ No. of workers whose daily wages are more than Rs. 260 - 114
∴ No. of workers whose daily wages are less than Rs. 260 = 200 - 114 = 86
In simple words: To find quartiles Q1 and Q3, and the number of workers within certain wage ranges graphically, we first plot a 'more than' cumulative frequency ogive. For Q1 and Q3, we locate the corresponding Y-axis values (\(N/4\) and \(3N/4\)) and trace them to the ogive and then to the X-axis. For wage ranges, we trace wage values from the X-axis to the ogive and then to the Y-axis to find the number of workers.

🎯 Exam Tip: For graphical questions, accurately plotting the cumulative frequency points and drawing a smooth ogive is critical. Always clearly label axes and scale. When finding values, use a ruler to draw precise lines from the axes to the curve and vice-versa.

Question 17. Draw ogive of both the types for the following frequency distribution and hence find the median.

Solution:

For less than given points to be plotted are (10, 5), (20, 10), (30, 18), (40, 30), (50, 46), (60, 61), (70, 71), (80, 79), (90, 84), (100, 86)
For more than given points to be plotted are (0, 86), (10, 81), (20, 76), (30, 68), (40, 56), (50, 40), (60, 25), (70, 15), (80, 7), (90, 2)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ 'कम से कम' और 'अधिक से अधिक' दोनों संचयी आवृत्ति वक्रों (ओजाइव) को दर्शाता है। X-अक्ष पर प्राप्त अंक और Y-अक्ष पर छात्रों की संचयी संख्या है। 'कम से कम' ओजाइव ऊपर की ओर बढ़ता है, जबकि 'अधिक से अधिक' ओजाइव नीचे की ओर ढलान वाला होता है। इन दोनों वक्रों का प्रतिच्छेदन बिंदु माध्यिका (मीडियन) के मान को दर्शाता है, जिसे X-अक्ष पर लंबवत रेखा खींचकर निर्धारित किया जाता है।

From the point of intersection of two ogives. We draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.
In simple words: To find the median graphically, we construct both 'less than' and 'more than' cumulative frequency ogives. The point where these two ogives intersect determines the median. By drawing a perpendicular line from this intersection point to the X-axis, the value on the X-axis gives us the median.

🎯 Exam Tip: When drawing both ogives to find the median, ensure accurate plotting of points for both 'less than' and 'more than' cumulative frequencies. The intersection point of these two curves directly gives the median, so precision in drawing is key.

Question 18. Find Q1, D6 and P78 for the following data:

Solution:
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 - 8.975, 8.975 - 9.975, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 50
Q1 class = class containing \( (\frac{N}{4})^{th} \) observation
∴ \( \frac{N}{4} = \frac{50}{4} = 12.5 \)
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q1 lies in the class 8.975 - 9.975.
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q1 = \( L + \frac{h}{f} (\frac{N}{4} - c.f.) \)
\( = 8.975 + \frac{1}{10} (12.5 - 5) \)
\( = 8.975 + 0.1(7.5) \)
\( = 8.975 + 0.75 \)
\( = 9.725 \)

D6 class = class containing \( (\frac{6N}{10})^{th} \) observation
∴ \( \frac{6N}{10} = \frac{6 \times 50}{10} = 30 \)
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 - 10.975.
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D6 = \( L + \frac{h}{f} (\frac{6N}{10} - c.f.) \)
\( = 9.975 + \frac{1}{20} (30 - 15) \)
\( = 9.975 + 0.05(15) \)
\( = 9.975 + 0.75 \)
\( = 10.725 \)

P78 class = class containing \( (\frac{78N}{100})^{th} \) observation
∴ \( \frac{78N}{100} = \frac{78 \times 50}{100} = 39 \)
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P78 lies in the class 10.975 - 11.975.
∴ L = 10.975, h = 1, f = 10, c.f. = 35
∴ P78 = \( L + \frac{h}{f} (\frac{78N}{100} - c.f.) \)
\( = 10.975 + \frac{1}{10} (39 - 35) \)
\( = 10.975 + 0.1(4) \)
\( = 10.975 + 0.4 \)
\( = 11.375 \)
In simple words: First, the discontinuous class intervals are converted to continuous ones. Then, a cumulative frequency table is constructed. For Q1, D6, and P78, their respective positions are calculated using \(N/4\), \(6N/10\), and \(78N/100\). The corresponding class intervals are identified, and the quartile/decile/percentile formula is applied with the lower limit (L), class width (h), frequency (f), and cumulative frequency (c.f.) of the respective classes to find their values.

🎯 Exam Tip: Remember to always convert discontinuous data into continuous intervals (e.g., by adding 0.025 to upper limits and subtracting 0.025 from lower limits) before calculating any partition values like quartiles, deciles, or percentiles for accuracy.

Question 19. For the above data, find all quartiles and number of persons weighing between 57 kg and 72 kg.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 111
Q1 class = class containing \( (\frac{N}{4})^{th} \) observation
∴ \( \frac{N}{4} = \frac{111}{4} = 27.75 \)
Cumulative frequency which is just greater than (or equal) to 27.75 is 39.
∴ Q1 lies in the class 50 - 55.
∴ \( Q_1 = L + \frac{h}{f} (\frac{N}{4} - c.f.) \)
\( = 50 + \frac{5}{20} (27.75 - 19) \)
\( = 50 + \frac{1}{4} \times 8.75 \)
\( = 50 + 2.1875 \)
\( = 52.1875 \)

Q2 class = class containing \( (\frac{2N}{4})^{th} \) observation
∴ \( \frac{2N}{4} = \frac{2 \times 111}{4} = 55.5 \)
Cumulative frequency which is just greater than (or equal) to 55.5 is 69.
∴ Q2 lies in the class 55 - 60.
∴ L = 55, h = 5, f = 30, c.f. = 39
∴ \( Q_2 = L + \frac{h}{f} (\frac{2N}{4} - c.f.) \)
\( = 55 + \frac{5}{30} (55.5 - 39) \)
\( = 55 + \frac{1}{6} \times 16.5 \)
\( = 55 + 2.75 \)
\( = 57.75 \)

Q3 class = class containing \( (\frac{3N}{4})^{th} \) observation
∴ \( \frac{3N}{4} = \frac{3 \times 111}{4} = 83.25 \)
Cumulative frequency which is just greater than (or equal) to 83.25 is 89.
∴ Q3 lies in the class 60 - 65.
∴ L = 60, h = 5, f = 20, c.f. = 69
∴ \( Q_3 = L + \frac{h}{f} (\frac{3N}{4} - c.f.) \)
\( = 60 + \frac{5}{20} (83.25 - 69) \)
\( = 60 + \frac{1}{4} \times 14.25 \)
\( = 60 + 3.5625 \)
\( = 63.5625 \)

In order to find the number of persons between 57 kg and 72 kg,
We need to find x in Px, where Px = 57 kg and y in Py, where Py = 72 kg
Then (y - x) would be the % of persons weighing between 57 kg and 72 kg
Px = 57
∴ \( L + \frac{h}{f} (\frac{xN}{100} - c.f.) = 57 \)
∴ \( 55 + \frac{5}{30} (1.11x - 39) = 57 \)
∴ \( \frac{1}{6} (1.11x - 39) = 2 \)
∴ \( 1.11x - 39 = 12 \)
∴ \( 1.11x = 51 \)
∴ x = 45.95
∴ Py = 72
∴ \( L + \frac{h}{f} (\frac{yN}{100} - c.f.) = 72 \)
∴ \( 70 + \frac{5}{8} (1.11y - 99) = 72 \)
∴ \( 0.625(1.11y - 99) = 2 \)
∴ \( 1.11y - 99 = 3.2 \)
∴ \( 1.11y = 102.2 \)
∴ y = 92.07
∴ % of people weighing between 57 kg and 72 kg = 92.07 - 45.95 = 46.12 %
∴ No. of people weighing between 57 kg and 72 kg = 111 × 46.12% = 51.1932 - 51
In simple words: First, a cumulative frequency table is constructed. Then, Q1, Q2, and Q3 are calculated using their respective formulas for grouped data. To find the number of persons between 57 kg and 72 kg, we use the percentile formula to find the percentile ranks (x and y) corresponding to these weights. The difference in these ranks, multiplied by the total number of persons, gives the required count.

🎯 Exam Tip: When asked to find the number of observations within a specific range, calculate the percentile values for both the lower and upper bounds of the range. The difference between these percentile values (as a proportion of N) will give the count within that range. Pay attention to the correct class intervals and cumulative frequencies.

Question 20. For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 100
Q1 class = class containing \( (\frac{N}{4})^{th} \) observation
∴ \( \frac{N}{4} = \frac{100}{4} = 25 \)
Cumulative frequency which is just greater than (or equal) to 25 is 29.
∴ Q1 lies in the class 55 - 60.
∴ L = 55, h = 5, f = 15, c.f. = 14
∴ Q1 = \( L + \frac{h}{f} (\frac{N}{4} - c.f.) \)
\( = 55 + \frac{5}{15} (25 - 14) \)
\( = 55 + \frac{1}{3} \times 11 \)
\( = 55 + 3.67 \)
\( = 58.67 \)
∴ Maximum weight of the lightest 25% of employees is 58.67 kg.
In simple words: To find the maximum weight of the lightest 25% of employees, we need to calculate the first quartile (Q1). First, we construct a cumulative frequency table. Then, we find the position of Q1 using \(N/4\). We locate the corresponding class interval and apply the formula for Q1, which gives us 58.67 kg as the maximum weight for the lightest 25% of employees.

🎯 Exam Tip: The "lightest 25% of employees" directly corresponds to the first quartile (Q1). Ensure you correctly identify the Q1 class and use the right formula, paying close attention to L, h, f, and c.f. for that specific class.