Maharashtra Board Class 11 Maths Part 2 Chapter 2 Measures of Dispersion 2.1 Solutions

Question 1. Find range of the following data:
575, 609, 335, 280, 729, 544, 852, 427, 967, 250
Answer:
Here, largest value (L) = 967, smallest value (S) = 250
Therefore, Range = L - S
= 967 - 250
= 717
In simple words: The range is the difference between the largest and smallest values in a dataset, providing a simple measure of data spread.

🎯 Exam Tip: To calculate the range, always identify the maximum and minimum values accurately from the given data. A simple subtraction yields the result.

Question 2. The following data gives number of typing mistakes done by Radha during a week.
Find the range of the data.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका राधा द्वारा एक सप्ताह में की गई टाइपिंग गलतियों की संख्या दर्शाती है, जिसमें प्रत्येक दिन (सोमवार से शनिवार) की गलतियों की संख्या सूचीबद्ध है।

Answer:
Here, largest value (L) = 21, smallest value (S) = 10
Therefore, Range = L - S
= 21 - 10
= 11
In simple words: By identifying the highest and lowest number of mistakes Radha made, we calculate the range to see the variation in her typing errors over the week.

🎯 Exam Tip: When given data in a table, carefully scan all values to find the absolute maximum and minimum before calculating the range. Ensure proper identification of categories like "Day" and "No. of mistakes".

Question 3. Find range for the following data:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका वर्गीकृत डेटा के लिए कक्षा अंतराल और उनकी संबंधित आवृत्तियों को दर्शाती है, जिसका उपयोग सीमा की गणना के लिए किया जाएगा।

Answer:
Here, upper limit of the highest class (L) = 72, lower limit of the lowest class (S) = 62
Therefore, Range = L - S
= 72 - 62
= 10
In simple words: For grouped data, the range is found by subtracting the lower limit of the first class from the upper limit of the last class.

🎯 Exam Tip: For grouped data, the range is typically the difference between the upper limit of the highest class interval and the lower limit of the lowest class interval. Do not use the frequencies for range calculation.

Question 4. Find the Q. D. for the following data.
3, 16, 8, 15, 19, 11, 5, 17, 9, 5, 3.
Answer:
The given data can be arranged in ascending order as follows:
3, 3, 5, 5, 8, 9, 11, 15, 16, 17, 19
Here, n = 11
Q1 = value of \((\frac{n+1}{4})\)^th observation
= value of \((\frac{11+1}{4})\)^th observation
= value of 3rd observation
Therefore, Q1 = 5
Q3 = value of \(3(\frac{n+1}{4})\)^th observation
= value of \(3(\frac{11+1}{4})\)^th observation
= value of (3 × 3)th observation
= value of 9th observation
= 16
Therefore, Q.D. = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{16-5}{2}\)
= \(\frac{11}{2}\)
= 5.5
In simple words: The Quartile Deviation (Q.D.) measures the spread of the middle 50% of the data, calculated as half the difference between the third and first quartiles. First, sort the data, then find the positions and values of Q1 and Q3.

🎯 Exam Tip: Always arrange the data in ascending order before calculating quartiles. If the quartile position is not an integer, use interpolation, but for simple cases like this, directly pick the observation.

Question 5. Given below are the prices of shares of a company for the last 10 days. Find Q.D.:
172, 164, 188, 214, 190, 237, 200, 195, 208, 230.
Answer:
The given data can be arranged in ascending order as follows:
164, 172, 188, 190, 195, 200, 208, 214, 230, 237
Here, n = 10
Q₁ = value of \((\frac{n+1}{4})\)^th observation
= value of \((\frac{10+1}{4})\)^th observation
= value of (2.75)th observation
= value of 2nd observation + 0.75(value of 3rd observation - value of 2nd observation)
= 172 + 0.75(188 - 172)
= 172 + 0.75(16)
= 172 + 12
= 184
Therefore, Q₃ = value of \(3(\frac{n+1}{4})\)^th observation
= value of \(3(\frac{10+1}{4})\)^th observation
= value of (3 × 2.75)th observation
= value of (8.25)th observation
= value of 8th observation + 0.25(value of 9th observation - value of 8th observation)
= 214 + 0.25(230 - 214)
= 214 + 0.25(16)
= 214 + 4
= 218
Therefore, Q.D. = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{218-184}{2}\)
= \(\frac{34}{2}\)
= 17
In simple words: For a given set of share prices, after ordering them, we calculate the first and third quartiles using interpolation for fractional ranks, and then find the Quartile Deviation by taking half of their difference.

🎯 Exam Tip: When the quartile position results in a fractional value (e.g., 2.75th observation), always use interpolation to find the exact quartile value. Meticulous calculation is key to avoiding errors.

Question 6. Calculate Q.D. for the following data.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका असतत आवृत्ति वितरण को दर्शाती है, जिसमें विभिन्न मान (X) और उनकी संबंधित आवृत्तियाँ (F) शामिल हैं, जिसका उपयोग चतुर्थक विचलन की गणना के लिए किया जाएगा।

Answer:
Since the given data is arranged in ascending order, we construct less than cumulative frequency table as follows:

Here, n = 30
Q1 = value of \((\frac{n+1}{4})\)^th observation
= value of \((\frac{30+1}{4})\)^th observation
= value of (7.75)th observation
Cumulative frequency which is just greater than (or equal to) 7.75 is 11.
Therefore, Q1 = 25
Q3 = value of \(3(\frac{n+1}{4})\)^th observation
= value of \(3(\frac{30+1}{4})\)^th observation
= value of (3 × 7.75)th observation
= value of (23.25)th observation
Cumulative frequency which is just greater than (or equal to) 23.25 is 27.
Therefore, Q3 = 29
Therefore, Q.D. = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{29-25}{2}\)
Therefore, Q.D. = 2
In simple words: To find the Quartile Deviation for discrete frequency distribution, first calculate the cumulative frequency, then find the Q1 and Q3 values corresponding to their respective cumulative frequencies. Finally, use the formula Q.D. = (Q3 - Q1) / 2.

🎯 Exam Tip: For discrete series, identify Q1 and Q3 by locating the cumulative frequency just greater than or equal to \((\frac{n+1}{4})\) and \(3(\frac{n+1}{4})\) respectively. The corresponding 'X' value will be the quartile.

Question 7. Following data gives the age distribution of 240 employees of a firm. Calculate Q.D. of the distribution.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका एक फर्म के 240 कर्मचारियों का आयु वितरण (वर्ग अंतराल में) और कर्मचारियों की संख्या दर्शाती है, जिसका उपयोग चतुर्थक विचलन की गणना के लिए किया जाएगा।

Answer:
We construct the less than cumulative frequency table as follows:

Here, N = 240
Q₁ class = class containing \((\frac{N}{4})\)^th observation
Therefore, \(\frac{N}{4}\) = \(\frac{240}{4}\) = 60
Cumulative frequency which is just greater than (or equal to) 60 is 70.
Therefore, Q₁ lies in the class 25 - 30.
Therefore, L = 25, c.f. = 30, f = 40, h = 5
Therefore, Q₁ = L + \(\frac{h}{f}\) \((\frac{N}{4}\) - c.f.)
= 25 + \(\frac{5}{40}\) (60 - 30)
= 25 + \(\frac{1}{8}\) × 30
= 25 + 3.75
Q1 = 28.75
Q₃ class = class containing \((\frac{3N}{4})\)^th observation
Therefore, \(\frac{3N}{4}\) = \(\frac{3 \times 240}{4}\) = 180
Cumulative frequency which is just greater than (or equal to) 180 is 180.
Therefore, Q₃ lies in the class 35-40.
Therefore, L = 35, c.f. = 130, f = 50, h = 5
Therefore, Q₃ = L + \(\frac{h}{f}\) \((\frac{3N}{4}\) - c.f.)
= 35 + \(\frac{5}{50}\) (180 - 130)
= 35 + \(\frac{1}{10}\) × 50
= 35 + 5
Therefore, Q₃ = 40
Therefore, Q.D. = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{40-28.75}{2}\)
= \(\frac{11.25}{2}\)
= 5.625
In simple words: For grouped data, we calculate Q1 and Q3 by first finding the respective quartile classes using cumulative frequencies, then applying the interpolation formula to find their precise values. Finally, the Quartile Deviation is half the difference between Q3 and Q1.

🎯 Exam Tip: For continuous frequency distribution, use the formula Q = L + \(\frac{h}{f}\) (\(\frac{QN}{4}\) - c.f.) carefully for both Q1 and Q3, where Q is the quartile number (1 or 3).

Question 8. Following data gives the weight of boxes. Calculate Q.D. for the data.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका बॉक्स के वजन (किलोग्राम में वर्ग अंतराल) और बक्सों की संख्या दर्शाती है, जिसका उपयोग चतुर्थक विचलन की गणना के लिए किया जाएगा।

Answer:
We construct the less than cumulative frequency table as follows:

Here, N = 60
Q₁ class = class containing \((\frac{N}{4})\)^th observation
Therefore, \(\frac{N}{4}\) = \(\frac{60}{4}\) = 15
Cumulative frequency which is just greater than (or equal to) 15 is 26.
Therefore, Q₁ lies in the class 14 - 16.
Therefore, L = 14, c.f. = 10, f = 16, h = 2
Therefore, Q₁ = L + \(\frac{h}{f}\) \((\frac{N}{4}\) - c.f.)
= 14 + \(\frac{2}{16}\) (15 - 10)
= 14 + \(\frac{1}{8}\) × 5
= 14 + 0.625
Q1 = 14.625
Q₃ class = class containing \((\frac{3N}{4})\)^th observation
Therefore, \(\frac{3N}{4}\) = \(\frac{3 \times 60}{4}\) = 45
Cumulative frequency which is just greater than (or equal to) 45 is 58.
Therefore, Q₃ lies in the class 18 - 20.
Therefore, L = 18, c.f. = 40, f = 18, h = 2
Therefore, Q₃ = L + \(\frac{h}{f}\) \((\frac{3N}{4}\) - c.f.)
= 18 + \(\frac{2}{18}\) (45 - 40)
= 18 + \(\frac{1}{9}\) × 5
= 18 + 0.5556
Q3 = 18.5556
Therefore, Q.D. = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{18.5556-14.625}{2}\)
= \(\frac{3.9306}{2}\)
= 1.9653
In simple words: To calculate the Quartile Deviation for grouped frequency data, first compute the cumulative frequency. Then, identify the classes for Q1 and Q3 using their respective positions and apply the interpolation formula to find their exact values. Finally, compute Q.D. as half the difference between Q3 and Q1.

🎯 Exam Tip: Ensure that the class intervals are continuous before applying the grouped data formulas. If they are not, make them continuous by adjusting boundaries. Double-check all values for L, c.f., f, and h during calculations.