Maharashtra Board Class 11 Maths Part 2 Chapter 1 Partition Values 1.1 Solutions

Question 1. Compute all the quartiles for the following series of observations: 16, 14.9, 11.5, 11.8, 11.1, 14.5, 14, 12, 10.9, 10.7, 10.6, 10.5, 13.5, 13, 12.6
Answer: Solution: The given data can be arranged in ascending order as follows: 10.5, 10.6, 10.7, 10.9, 11.1, 11.5, 11.8, 12, 12.6, 13, 13.5, 14, 14.5, 14.9, 16 Here, n = 15 Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of \(\left(\frac{15+1}{4}\right)^{\text{th}}\) observation = value of 4th observation
\(\implies\) Q1 = 10.9 Q2 = value of 2\(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of 2 \(\left(\frac{15+1}{4}\right)^{\text{th}}\) observation = value of (2 × 4)th observation = value of 8th observation
\(\implies\) Q2 = 12 Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of 3 \(\left(\frac{15+1}{4}\right)^{\text{th}}\) observation = value of (3 × 4)th observation = value of 12th observation
\(\implies\) Q3 = 14
In simple words: To find the quartiles, first arrange the data in ascending order. Then use the formulas for Q1, Q2, and Q3, which involve finding the value at specific positions in the ordered data.

🎯 Exam Tip: Always arrange data in ascending order before calculating quartiles for ungrouped data. Ensure correct application of the position formula for each quartile.

Question 2. The heights (in cm.) of 10 students are given below: 148, 171, 158, 151, 154, 159, 152, 163, 171, 145 Calculate Q₁ and Q3 for the above data.
Answer: Solution: The given data can be arranged in ascending order as follows: 145, 148, 151, 152, 154, 158, 159, 163, 171, 171 Here, n = 10 Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of \(\left(\frac{10+1}{4}\right)^{\text{th}}\) observation = value of (2.75)th observation = value of 2nd observation + 0.75 (value of 3rd observation – value of 2nd observation) = 148 + 0.75 (151 – 148) = 148 + 0.75(3) = 148 + 2.25
\(\implies\) Q₁ = 150.25 Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of 3\(\left(\frac{10+1}{4}\right)^{\text{th}}\) 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) = 163 + 0.25(171 – 163) = 163 + 0.25(8) = 163 + 2
\(\implies\) Q3 = 165
In simple words: After arranging the data, calculate Q1 and Q3 using their respective formulas. For fractional positions, interpolate by adding a fraction of the difference between the two surrounding observations to the lower observation.

🎯 Exam Tip: When the quartile position is a decimal, remember to interpolate between the integer positions. A common mistake is to round the position instead of interpolating.

Question 3. The monthly consumption of electricity (in units) of families in a certain locality is given below: 205, 201, 190, 188, 195, 172, 210, 225, 215, 232, 260, 230 Calculate electricity consumption (in units) below which 25% of the families lie.
Answer: Solution: To find the consumption of electricity below which 25% of the families lie, we have to find Q1. Monthly consumption of electricity (in units) can be arranged in ascending order as follows: 172, 188, 190, 195, 201, 205, 210, 215, 225, 230, 232, 260. Here, n = 12 Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of \(\left(\frac{12+1}{4}\right)^{\text{th}}\) observation = value of (3.25)th observation = value of 3rd observation + 0.25 (value of 4th observation – value of 3rd observation) = 190 + 0.25(195 – 190) = 190 + 0.25(5) = 190 + 1.25 = 191.25
\(\implies\) the consumption of electricity below which 25% of the families lie is 191.25.
In simple words: To find the value below which 25% of data falls, calculate the first quartile (Q1) using the ordered data. If the position is a decimal, interpolate between the two nearest data points.

🎯 Exam Tip: "Below which 25% of the families lie" directly translates to finding the first quartile (Q1). Clearly state this interpretation in your answer.

Question 4. For the following data of daily expenditure of families (in Rs.), compute the expenditure below which 75% of families include their expenditure.
Answer: Solution: To find the expenditure below which 75% of families have their expenditure, we have to find Q3. We construct the less than cumulative frequency table as given below:

Here, n = 100 Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of 3\(\left(\frac{100+1}{4}\right)^{\text{th}}\) observation = value of (3 × 25.25)th observation = value of (75.75)th observation Cumulative frequency which is just greater than (or equal to) 75.75 is 87.
\(\implies\) Q3 = 650
\(\implies\) the expenditure below which 75% of families include their expenditure is Rs. 650.
In simple words: To find the expenditure below which 75% of families lie, construct a cumulative frequency table. Calculate the position of the third quartile (Q3) and find the corresponding value from the cumulative frequency.

🎯 Exam Tip: For discrete frequency distributions, calculate the (N+1)/4 * k position, then locate the first cumulative frequency greater than or equal to this position to find the quartile value.

Question 5. Calculate all the quartiles for the following frequency distribution:

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

Here, n = 300 Q1 = value of \(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of \(\left(\frac{300+1}{4}\right)^{\text{th}}\) observation = value of (75.25)th observation Cumulative frequency which is just greater than (or equal to) 75.25 is 90.
\(\implies\) Q1 = 2 Q2 = value of 2\(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of 2\(\left(\frac{300+1}{4}\right)^{\text{th}}\) observation = value of (2 × 75.25)th observation = value of (150.50)th observation
\(\implies\) Cumulative frequency which is just greater than (or equal to) 150.50 is 185.
\(\implies\) Q2 = 3 Q3 = value of 3\(\left(\frac{n+1}{4}\right)^{\text{th}}\) observation = value of 3\(\left(\frac{300+1}{4}\right)^{\text{th}}\) observation = value of (3 × 75.25)th observation = value of (225.75)th observation Cumulative frequency which is just greater than (or equal to) 225.75 is 249.
\(\implies\) Q3 = 4
In simple words: For a frequency distribution, construct a cumulative frequency table. Then, calculate the position for each quartile (Q1, Q2, Q3) and find the corresponding value from the cumulative frequency column.

🎯 Exam Tip: When calculating quartiles for frequency distributions, correctly identify the cumulative frequency that just exceeds or equals the calculated position to find the corresponding quartile value.

Question 6. The following is the frequency distribution of heights of 200 male adults in a factory:

Find the central height.
Answer: Solution: To find the central height, we have to find Q2. We construct the less than cumulative frequency table as given below:

Here, N = 200 Q2 class = class containing \(\left(\frac{2N}{4}\right)^{\text{th}}\) observation
\(\implies \frac{2N}{4} = \frac{2 \times 200}{4} = 100\) Cumulative frequency which is just greater than (or equal to) 100 is 156.
\(\implies\) Q2 lies in the class 165 – 170.
\(\implies L = 165, h = 5, f = 64, \text{c.f.} = 92\)
\(\implies Q_2 = L + \frac{h}{f}\left(\frac{2N}{4} - \text{c.f.}\right)\)
\(\implies = 165 + \frac{5}{64}(100-92)\)
\(\implies = 165 + \frac{5}{64} \times 8\)
\(\implies = 165 + \frac{5}{8}\)
\(\implies = 165 + 0.625\)
\(\implies = 165.625\)
\(\implies\) Central height is 165.625 cm.
In simple words: The central height is the median (Q2). For grouped data, calculate the median class by finding the \(N/2\) position, then use the median formula \(L + \frac{h}{f}(\frac{N}{2} - c.f.)\) to find the precise value.

🎯 Exam Tip: For grouped data, correctly identify the quartile class using the cumulative frequency before applying the interpolation formula. Ensure all components (L, h, f, c.f.) are picked from the correct class and preceding class interval.

Question 7. The following is the data of pocket expenditure per week of 50 students in a class. It is known that the median of the distribution is Rs. 120. Find the missing frequencies.

Answer: Solution: Let a and b be the missing frequencies of class 50 – 100 and class 150 – 200 respectively. We construct the less than cumulative frequency table as given below:

Here, N = 25 + a + b Since, N = 50
\(\implies\) 25 + a + b = 50
\(\implies\) a + b = 25 .....(i) Given, Median = Q2 = 120
\(\implies\) Q2 lies in the class 100 – 150.
\(\implies L = 100, h = 50, f = 15, \frac{2N}{4} = \frac{2 \times 50}{4} = 25\)
\(\implies Q_2 = L + \frac{h}{f}\left(\frac{2N}{4} - \text{c.f.}\right)\)
\(\implies 120 = 100 + \frac{50}{15}[25 - (7 + a)]\)
\(\implies 120 - 100 = \frac{10}{3}(25-7-a)\)
\(\implies 20 = \frac{10}{3}(18-a)\)
\(\implies 60 = 18-a\)
\(\implies 10 = 18-a\)
\(\implies 6 = 18-a\)
\(\implies a = 18-6 = 12\) Substituting the value of a in equation (i), we get 12 + b = 25
\(\implies\) b = 25 - 12 = 13
\(\implies\) 12 and 13 are the missing frequencies of the class 50 – 100 and class 150 – 200 respectively.
In simple words: To find missing frequencies when the median is given for grouped data, set up the cumulative frequency table with variables for missing frequencies. Use the median formula, substitute the given median value, and solve the resulting equation(s) to find the unknown frequencies.

🎯 Exam Tip: When given the median for grouped data, first determine the median class. Then, use the median formula and set up an algebraic equation to solve for the missing frequencies, often requiring simultaneous equations.

Question 8. The following is the distribution of 160 workers according to the wages in a certain factory:

Determine the values of all quartiles and interpret the results.
Answer: Solution: The given table is a more than cumulative frequency. We transform the given table into less than cumulative frequency. We construct the less than cumulative frequency table as given below:

Here, N = 160
\(\implies\) Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text{th}}\) observation
\(\implies \frac{N}{4} = \frac{160}{4} = 40\) Cumulative frequency which is just greater than (or equal to) 40 is 69.
\(\implies\) Q₁ lies in the class 10000 – 11000
\(\implies L = 10000, h = 1000, f = 46, \text{c.f.} = 23\)
\(\implies Q₁ = L + \frac{h}{f}\left(\frac{N}{4} - \text{c.f.}\right)\)
\(\implies = 10000 + \frac{1000}{46}(40-23)\)
\(\implies = 10000 + \frac{1000}{46}(17)\)
\(\implies = 10000 + \frac{17000}{46}\)
\(\implies = 10000 + 369.57\)
\(\implies = 10369.57\) Q2 class = class containing \(\left(\frac{2N}{4}\right)^{\text{th}}\) observation
\(\implies \frac{2N}{4} = \frac{2 \times 160}{4} = 80\) Cumulative frequency which is just greater than (or equal to) 80 is 103.
\(\implies\) Q2 lies in the class 11000 – 12000.
\(\implies L = 11000, h = 1000, f = 34, \text{c.f.} = 69\)
\(\implies Q_2 = L + \frac{h}{f}\left(\frac{2N}{4} - \text{c.f.}\right)\)
\(\implies = 11000 + \frac{1000}{34}(80-69)\)
\(\implies = 11000 + \frac{1000}{34}(11)\)
\(\implies = 11000 + \frac{11000}{34}\)
\(\implies = 11000 + 323.529\)
\(\implies = 11323.529\) Q3 class = class containing \(\left(\frac{3N}{4}\right)^{\text{th}}\) observation
\(\implies \frac{3N}{4} = \frac{3 \times 160}{4} = 120\) Cumulative frequency which is just greater than (or equal to) 120 is 137.
\(\implies\) Q3 lies in the class 12000 – 13000.
\(\implies L = 12000, h = 1000, f = 34, \text{c.f.} = 103\)
\(\implies Q_3 = L + \frac{h}{f}\left(\frac{3N}{4} - \text{c.f.}\right)\)
\(\implies = 12000 + \frac{1000}{34}(120-103)\)
\(\implies = 12000 + \frac{1000}{34}(17)\)
\(\implies = 12000 + \frac{17000}{34}\)
\(\implies = 12000 + 500\)
\(\implies = 12500\) Interpretation: Q1 < Q2 < Q3
In simple words: First, convert the "more than" cumulative frequency distribution into a "less than" cumulative frequency distribution. Then, for each quartile (Q1, Q2, Q3), identify its corresponding class and apply the quartile formula for grouped data to find its value.

🎯 Exam Tip: Always start by converting "more than" C.F. to "less than" C.F. for calculating quartiles in grouped data. The interpretation of Q1, Q2, Q3 is also a key scoring element.

Question 9. Following is grouped data for the duration of fixed deposits of 100 senior citizens from a certain bank:

Calculate the limits of fixed deposits of central 50% senior citizens.
Answer: Solution: We construct the less than cumulative frequency table as given below:

To find the limits of fixed deposits of central 50% senior citizens, we have to find Q1 and Q3. Here, N = 100 Q1 class = class containing \(\left(\frac{N}{4}\right)^{\text{th}}\) observation
\(\implies \frac{N}{4} = \frac{100}{4} = 25\) Cumulative frequency which is just greater than (or equal to) 25 is 35.
\(\implies\) Q1 lies in the class 180 – 360.
\(\implies L = 180, h = 180, f = 20, \text{c.f.} = 15\)
\(\implies Q₁ = L + \frac{h}{f}\left(\frac{N}{4} - \text{c.f.}\right)\)
\(\implies = 180 + \frac{180}{20}(25-15)\)
\(\implies = 180 + 9(10)\)
\(\implies = 180 + 90\)
\(\implies = 270\) Q3 class = class containing \(\left(\frac{3N}{4}\right)^{\text{th}}\) observation
\(\implies \frac{3N}{4} = \frac{3 \times 100}{4} = 75\) Cumulative frequency which is just greater than (or equal to) 75 is 90.
\(\implies\) Q3 lies in the class 540 – 720.
\(\implies L = 540, h = 180, f = 30, \text{c.f.} = 60\)
\(\implies Q_3 = L + \frac{h}{f}\left(\frac{3N}{4} - \text{c.f.}\right)\)
\(\implies = 540 + \frac{180}{30}(75-60)\)
\(\implies = 540 + 6(15)\)
\(\implies = 540 + 90\)
\(\implies = 630\)
\(\implies\) Limits of duration of fixed deposits of central 50% senior citizens is from 270 to 630.
In simple words: The central 50% of data lies between the first quartile (Q1) and the third quartile (Q3). Calculate Q1 and Q3 for the given grouped frequency distribution using their respective formulas for grouped data.

🎯 Exam Tip: "Central 50%" always refers to the range between Q1 and Q3. Ensure clear calculation of both quartiles and state the range as your final answer.

Question 10. Find the missing frequency given that the median of the distribution is 1504.

Answer: Solution: Let x be the missing frequency of the class 1550 – 1750. We construct the less than frequency table as given below:

Here, N = 199 + x Given, Median (Q2) = 1504
\(\implies\) Q2 lies in the class 1350 – 1550.
\(\implies L = 1350, h = 200, f = 100, \text{c.f.} = 63, \frac{2N}{4} = \frac{199+x}{2}\)
\(\implies Q_2 = L + \frac{h}{f}\left(\frac{2N}{4} - \text{c.f.}\right)\)
\(\implies 1504 = 1350 + \frac{200}{100}\left(\frac{199+x}{2} - 63\right)\)
\(\implies 1504 - 1350 = 2\left(\frac{199+x-126}{2}\right)\)
\(\implies 154 = 199 + x - 126\)
\(\implies 154 = x + 73\)
\(\implies x = 81\)
In simple words: When the median is known for data with a missing frequency, first create a cumulative frequency table, including a variable for the missing frequency. Identify the median class and use the median formula for grouped data. Set up an equation with the given median and solve for the unknown frequency.

🎯 Exam Tip: Carefully set up the cumulative frequency table and the median formula equation. Algebraic manipulation must be precise to correctly determine the missing frequency value.