Maharashtra Board Class 11 Maths Part 2 Chapter 1 Partition Values 1.2 Solutions

Question 1. Calculate D6 and P85 for the following data:

79, 82, 36, 38, 51, 72, 68, 70, 64, 63
Answer:
The given data can be arranged in ascending order as follows:
36, 38, 51, 63, 64, 68, 70, 72, 79, 82
Here, n = 10
\( D_6 = \text{value of } 6 \left( \frac{n+1}{10} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 6 \left( \frac{10+1}{10} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (6 \times 1.1)^\text{th} \text{ observation} \)
\( = \text{value of } (6.6)^\text{th} \text{ observation} \)
\( = \text{value of } 6^\text{th} \text{ observation} + 0.6(\text{value of } 7^\text{th} \text{ observation} - \text{value of } 6^\text{th} \text{ observation}) \)
\( = 68 + 0.6(70 - 68) \)
\( = 68 + 0.6(2) \)
\( = 68 + 1.2 \)
\( \therefore D_6 = 69.2 \)

\( P_{85} = \text{value of } \left( \frac{n+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } \left( \frac{10+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (85 \times 0.11)^\text{th} \text{ observation} \)
\( = \text{value of } (9.35)^\text{th} \text{ observation} \)
\( = \text{value of } 9^\text{th} \text{ observation} + 0.35(\text{value of } 10^\text{th} \text{ observation} - \text{value of } 9^\text{th} \text{ observation}) \)
\( = 79 + 0.35(82 - 79) \)
\( = 79 + 0.35(3) \)
\( = 79 + 1.05 \)
\( \therefore P_{85} = 80.05 \)
In simple words: To find D6 and P85 for ungrouped data, first arrange the data in ascending order. Then, use the respective formulas to find the position of the decile or percentile, and if the position is not a whole number, interpolate between the two nearest observations.

🎯 Exam Tip: Always arrange the data in ascending order before applying any decile or percentile formulas. Incorrect ordering is a common source of errors.

Question 2. The daily wages (in Rs.) of 15 labourers are as follows:
230, 400, 350, 200, 250, 380, 210, 225, 375, 180, 375, 450, 300, 350, 250
Calculate D8 and P90.
Answer:
The given data can be arranged in ascending order as follows:
180, 200, 210, 225, 230, 250, 250, 300, 350, 350, 375, 375, 380, 400, 450
Here, n = 15
\( D_8 = \text{value of } 8 \left( \frac{n+1}{10} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 8 \left( \frac{15+1}{10} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (8 \times 1.6)^\text{th} \text{ observation} \)
\( = \text{value of } (12.8)^\text{th} \text{ observation} \)
\( = \text{value of } 12^\text{th} \text{ observation} - 0.8(\text{value of } 13^\text{th} \text{ observation} - \text{value of } 12^\text{th} \text{ observation}) \)
\( = 375 + 0.8(380 - 375) \)
\( = 375 + 0.8(5) \)
\( = 375 + 4 \)
\( \therefore D_8 = 379 \)

\( P_{90} = \text{value of } 90 \left( \frac{n+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 90 \left( \frac{15+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (90 \times 0.16)^\text{th} \text{ observation} \)
\( = \text{value of } (14.4)^\text{th} \text{ observation} \)
\( = \text{value of } 14^\text{th} \text{ observation} + 0.4 (\text{value of } 15^\text{th} \text{ observation} - \text{value of } 14^\text{th} \text{ observation}) \)
\( = 400 + 0.4(450 - 400) \)
\( = 400 + 0.4(50) \)
\( = 400 + 20 \)
\( \therefore P_{90} = 420 \)
In simple words: For ungrouped data, deciles and percentiles locate specific data points in a sorted list. D8 represents the value below which 80% of the data falls, and P90 represents the value below which 90% of the data falls, often requiring interpolation for non-integer positions.

🎯 Exam Tip: Pay close attention to the decimal part of the calculated observation position; it indicates the fraction to use for interpolation between adjacent data points.

Question 3. Calculate 2nd decile and 65th percentile for the following:

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

Here, n = 200
\( D_2 = \text{value of } 2 \left( \frac{n+1}{10} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 2 \left( \frac{200+1}{10} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (2 \times 20.1)^\text{th} \text{ observation} \)
\( = \text{value of } (40.2)^\text{th} \text{ observation} \)
Cumulative frequency which is just greater than (or equal to) 40.2 is 58.
\( \therefore D_2 = 120 \)

\( P_{65} = \text{value of } 65 \left( \frac{n+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 65 \left( \frac{200+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (65 \times 2.01)^\text{th} \text{ observation} \)
\( = \text{value of } (130.65)^\text{th} \text{ observation} \)
The cumulative frequency which is just greater than (or equal to) 130.65 is 150.
\( \therefore P_{65} = 280 \)
In simple words: For discrete data, we first build a cumulative frequency table. Then, D2 is found by locating the observation whose cumulative frequency is just greater than or equal to the calculated position \( \left( \frac{2(N+1)}{10} \right) \), and P65 is found similarly using its position \( \left( \frac{65(N+1)}{100} \right) \).

🎯 Exam Tip: Always construct a cumulative frequency table for grouped or discrete data before attempting to find deciles or percentiles. This step is crucial for accurate location.

Question 4. From the following data calculate the rent of the 15th, 65th, and 92nd house.

Answer:
Arranging the given data in ascending order.

Here, n = 100
\( P_{15} = \text{value of } 15 \left( \frac{n+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 15 \left( \frac{100+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (15 \times 1.01)^\text{th} \text{ observation} \)
\( = \text{value of } (15.15)^\text{th} \text{ observation} \)
Cumulative frequency which is just greater than (or equal to) 15.15 is 25.
\( \therefore P_{15} = 11000 \)

\( P_{65} = \text{value of } 65 \left( \frac{n+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 65 \left( \frac{100+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (65 \times 1.01)^\text{th} \text{ observation} \)
\( = \text{value of } (65.65)^\text{th} \text{ observation} \)
Cumulative frequency which is just greater than (or equal to) 65.65 is 70.
\( \therefore P_{65} = 14000 \)

\( P_{92} = \text{value of } 92 \left( \frac{n+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } 92 \left( \frac{100+1}{100} \right)^\text{th} \text{ observation} \)
\( = \text{value of } (92 \times 1.01)^\text{th} \text{ observation} \)
\( = \text{value of } (92.92)^\text{th} \text{ observation} \)
Cumulative frequency which is just greater than (or equal to) 92.92 is 98.
\( \therefore P_{92} = 17000 \)
In simple words: For discrete data, after arranging the data and creating a cumulative frequency table, percentiles like P15, P65, and P92 are found by calculating their respective positions and identifying the data value corresponding to the first cumulative frequency that is greater than or equal to that position.

🎯 Exam Tip: When given raw data, always re-arrange it in ascending order first. For grouped discrete data, constructing a cumulative frequency table is a mandatory initial step to accurately locate percentiles.

Question 5. The following frequency distribution shows the weight of students in a class.

(a) Find the percentage of students whose weight is more than 50 kg.
(b) If the weight column provided is of mid values then find the percentage of students whose weight is more than 50 kg.
Answer:
(a) Let the percentage of students weighing less than 50 kg be x.
\( \therefore P_x = 50 \)

From the table, out of 120 students, 84 students have their weight less than 50 kg.
\( \therefore \) Number of students weighing more than 50 kg \( = 120 - 84 = 36 \)
\( \therefore \) Percentage of students having their weight more than 50 kg \( = \frac{36}{120} \times 100 = 30\% \)

(b) The difference between any two consecutive mid values of weight is 5 kg.
The class intervals must of width 5, with 40, 45,..... as their mid values.
\( \therefore \) The class intervals will be 37.5 - 42.5, 42.5 - 47.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 120
Let \( P_x = 50 \)
The value 50 lies in the class 47.5 - 52.5
\( \therefore L = 47.5, h = 5, f = 29, c.f. = 55 \)
\( P_x = L + \frac{h}{f} \left( \frac{xN}{100} - c.f. \right) \)
\( \implies 50 = 47.5 + \frac{5}{29} \left( \frac{x \times 120}{100} - 55 \right) \)
\( \implies 50 - 47.5 = \frac{5}{29} \left( \frac{6x}{5} - 55 \right) \)
\( \implies 2.5 = \frac{5}{29} \left( \frac{6x - 275}{5} \right) \)
\( \implies 2.5 \times 29 = 6x - 275 \)
\( \implies 72.5 = 6x - 275 \)
\( \implies 6x = 72.5 + 275 \)
\( \implies 6x = 347.5 \)
\( \implies x = \frac{347.5}{6} \)
\( \implies x \approx 57.92 \)
\( \therefore 58\% \) of students are having weight below 50 kg (approximately).
\( \therefore \) Percentage of students having weight above 50 kg is \( 100 - 58 = 42 \)
\( \therefore 42\% \) of students are having weight above 50 kg.
In simple words: Part (a) directly calculates the percentage of students above 50 kg from a simple frequency count. Part (b) re-evaluates this by first converting mid-values into continuous class intervals, then uses the percentile formula for grouped data to find the percentage of students whose weight is more than 50 kg, requiring interpolation.

🎯 Exam Tip: When dealing with mid-values, remember to construct continuous class intervals before proceeding with percentile calculations for grouped data. Clearly distinguish between discrete and continuous data approaches.

Question 6. Calculate D4 and P48 from the following data:

Answer:
The difference between any two consecutive mid values is 5, the width of class interval = 5
\( \therefore \) Class interval with mid-value 2.5 is 0 - 5
Class interval with mid value 7.5 is 5 - 10, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 100
\( D_4 \text{ class} = \text{class containing } \left( \frac{4N}{10} \right)^\text{th} \text{ observation} \)
\( \therefore \frac{4N}{10} = \frac{4 \times 100}{10} = 40 \)
Cumulative frequency which is just greater than (or equal to) 40 is 50.
\( \therefore D_4 \text{ lies in the class } 10 - 15. \)
\( \therefore L = 10, h = 5, f = 25, c.f. = 25 \)
\( \therefore D_4 = L + \frac{h}{f} \left( \frac{4N}{10} - c.f. \right) \)
\( = 10 + \frac{5}{25} (40 - 25) \)
\( = 10 + \frac{1}{5} (15) \)
\( = 10 + 3 \)
\( \therefore D_4 = 13 \)

\( P_{48} \text{ class} = \text{class containing } \left( \frac{48N}{100} \right)^\text{th} \text{ observation} \)
\( \therefore \frac{48N}{100} = \frac{48 \times 100}{100} = 48 \)
Cumulative frequency which is just greater than (or equal to) 48 is 50.
\( \therefore P_{48} \text{ lies in the class } 10 - 15. \)
\( \therefore L = 10, h = 5, f = 25, c.f. = 25 \)
\( \therefore P_{48} = L + \frac{h}{f} \left( \frac{48N}{100} - c.f. \right) \)
\( = 10 + \frac{5}{25} (48 - 25) \)
\( = 10 + \frac{1}{5} (23) \)
\( = 10 + 4.6 \)
\( \therefore P_{48} = 14.6 \)
In simple words: To calculate D4 and P48 from data with mid-values, first convert the mid-values into continuous class intervals and create a cumulative frequency table. Then, apply the interpolation formula for deciles and percentiles in grouped data, identifying the correct class and its corresponding lower limit, frequency, and cumulative frequency.

🎯 Exam Tip: When transforming mid-values into class intervals, ensure the width of the interval is correctly determined (difference between consecutive mid-values) and applied to both upper and lower bounds symmetrically.

Question 7. Calculate D9 and P20 of the following distribution.

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

Here, N = 240
\( D_9 \text{ class} = \text{class containing } \left( \frac{9N}{10} \right)^\text{th} \text{ observation} \)
\( \therefore \frac{9N}{10} = \frac{9 \times 240}{10} = 216 \)
Cumulative frequency which is just greater than (or equal to) 216 is 225.
\( \therefore D_9 \text{ lies in the class } 80 - 100. \)
\( \therefore L = 80, h = 20, f = 90, c.f. = 135 \)
\( \therefore D_9 = L + \frac{h}{f} \left( \frac{9N}{10} - c.f. \right) \)
\( = 80 + \frac{20}{90} (216 - 135) \)
\( = 80 + \frac{2}{9} (81) \)
\( = 80 + 18 \)
\( \therefore D_9 = 98 \)

\( P_{20} \text{ class} = \text{class containing } \left( \frac{20N}{100} \right)^\text{th} \text{ observation} \)
\( \therefore \frac{20N}{100} = \frac{20 \times 240}{100} = 48 \)
Cumulative frequency which is just greater than (or equal to) 48 is 50.
\( \therefore P_{20} \text{ lies in the class } 40 - 60. \)
\( \therefore L = 40, h = 20, f = 35, c.f. = 15 \)
\( \therefore P_{20} = L + \frac{h}{f} \left( \frac{20N}{100} - c.f. \right) \)
\( = 40 + \frac{20}{35} (48 - 15) \)
\( = 40 + \frac{4}{7} (33) \)
\( = 40 + \frac{132}{7} \)
\( = 40 + 18.86 \)
\( \therefore P_{20} = 58.86 \)
In simple words: For grouped frequency distributions, calculating D9 and P20 involves first determining the cumulative frequencies. Then, locate the respective class intervals where these values fall using their positional formulas (\( \frac{9N}{10} \) and \( \frac{20N}{100} \)), and finally apply the interpolation formula for grouped data using the lower limit, class width, frequency, and preceding cumulative frequency of that class.

🎯 Exam Tip: Be careful with formula substitution for grouped data. Ensure you use the cumulative frequency *of the class immediately preceding* the decile/percentile class (c.f.) and the frequency *of the decile/percentile class itself* (f).

Question 8. Weekly wages for a group of 100 persons are given below:

D3 for this group is Rs. 1100. Calculate the missing frequencies.
Answer:
Let a and b be the missing frequencies of class 500 - 1000 and class 2000 - 2500 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, D3 = 1100
\( \therefore D_3 \text{ lies in the class } 1000 - 1500. \)
\( \therefore L = 1000, h = 500, f = 25, c.f. = 7 + a \)
\( \therefore \frac{3N}{10} = \frac{3 \times 100}{10} = 30 \)
\( \therefore D_3 = L + \frac{h}{f} \left( \frac{3N}{10} - c.f. \right) \)
\( \therefore 1100 = 1000 + \frac{500}{25} [30 - (7 + a)] \)
\( \therefore 1100 - 1000 = 20(30 - 7 - a) \)
\( \therefore 100 = 20(23 - a) \)
\( \therefore 100 = 460 - 20a \)
\( \therefore 20a = 460 - 100 \)
\( \therefore 20a = 360 \)
\( \therefore a = 18 \)
Substituting the value of a in equation (i), we get
\( 18 + b = 38 \)
\( \therefore b = 38 - 18 = 20 \)
\( \therefore 18 \text{ and } 20 \text{ are the missing frequencies of the class } 500 - 1000 \text{ and class } 2000 - 2500 \text{ respectively.} \)
In simple words: To find missing frequencies in a grouped frequency distribution when a decile value is given, first set up a cumulative frequency table with algebraic expressions for missing frequencies. Use the total frequency to form one equation, and then apply the decile formula, substituting the given decile value and known class interval parameters to form a second equation. Solve these simultaneous equations to find the missing frequencies.

🎯 Exam Tip: When solving for missing frequencies, remember that the sum of all frequencies must equal N. This provides a crucial first equation, which combined with the decile/percentile formula, forms a system of equations to solve.

Question 9. The weekly profit (in rupees) of 100 shops are distributed as follows:

Find the limits of the profit of middle 60% of the shops.
Answer:
To find the limits of the profit of the middle 60% of the shops, we have to find P20 and P80.
We construct the less than cumulative frequency table as given below:

Here, N = 100
\( P_{20} \text{ class} = \text{class containing } \left( \frac{20N}{100} \right)^\text{th} \text{ observation} \)
\( \therefore \frac{20N}{100} = \frac{20 \times 100}{100} = 20 \)
Cumulative frequency which is just greater than (or equal to) 20 is 26.
\( \therefore P_{20} \text{ lies in the class } 1000 - 2000. \)
\( \therefore L = 1000, h = 1000, f = 16, c.f. = 10 \)
\( \therefore P_{20} = L + \frac{h}{f} \left( \frac{20N}{100} - c.f. \right) \)
\( = 1000 + \frac{1000}{16} (20 - 10) \)
\( = 1000 + 62.5 (10) \)
\( = 1000 + 625 \)
\( \therefore P_{20} = 1625 \)

\( P_{80} \text{ class} = \text{class containing } \left( \frac{80N}{100} \right)^\text{th} \text{ observation} \)
\( \therefore \frac{80N}{100} = \frac{80 \times 100}{100} = 80 \)
Cumulative frequency which is just greater than (or equal to) 80 is 92.
\( \therefore P_{80} \text{ lies in the class } 4000 - 5000. \)
\( \therefore L = 4000, h = 1000, f = 20, c.f. = 72 \)
\( \therefore P_{80} = L + \frac{h}{f} \left( \frac{80N}{100} - c.f. \right) \)
\( = 4000 + \frac{1000}{20} (80 - 72) \)
\( = 4000 + 50(8) \)
\( = 4000 + 400 \)
\( \therefore P_{80} = 4400 \)
\( \therefore \text{the profit of middle 60\% of the shops lie between the limits Rs. } 1,625 \text{ to Rs. } 4,400. \)
In simple words: To find the middle 60% of data, we calculate the 20th percentile (P20) and the 80th percentile (P80). First, create a cumulative frequency table, then use the grouped data percentile formula to find the values corresponding to P20 and P80. The range between these two values represents the middle 60%.

🎯 Exam Tip: "Middle 60%" implies cutting off 20% from the lower end and 20% from the upper end of the distribution, hence requiring calculation of P20 and P80.

Question 10. In a particular factory, workers produce various types of output units. The following distribution was obtained:

Find the percentage of workers who have produced less than 82 output units.
Answer:
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.
\( \therefore \text{the class intervals will be } 69.5 - 74.5, 74.5 - 79.5, \text{ etc.} \)
We construct the less than cumulative frequency table as given below:

Here, N = 445
Let \( P_x = 82 \)
The value 82 lies in the class 79.5 - 84.5
\( \therefore L = 79.5, h = 5, f = 50, c.f. = 85 \)
\( \therefore P_x = L + \frac{h}{f} \left( \frac{xN}{100} - c.f. \right) \)
\( \implies 82 = 79.5 + \frac{5}{50} \left( \frac{x \times 445}{100} - 85 \right) \)
\( \implies 82 - 79.5 = \frac{1}{10} \left( \frac{89x}{20} - 85 \right) \)
\( \implies 2.5 \times 10 = \frac{89x}{20} - 85 \)
\( \implies 25 = \frac{89x}{20} - 85 \)
\( \implies 25 + 85 = \frac{89x}{20} \)
\( \implies 110 = \frac{89x}{20} \)
\( \implies x = \frac{110 \times 20}{89} \)
\( \implies x \approx 24.72 \)
\( \therefore 24.72\% \) of workers produced less than 82 output units.
In simple words: To find the percentage of workers producing less than a specific output in a discrete distribution, first convert the data into continuous class intervals. Then, use the grouped data percentile formula, where the given output (82) is treated as the Px value, to solve for 'x', which represents the desired percentage.

🎯 Exam Tip: Always check if the given data is continuous. If not, apply the correction factor (subtract 0.5 from lower limit, add 0.5 to upper limit) to create continuous class intervals, which is crucial for accurate grouped data calculations.