CBSE Class 10 Maths HOTs Statistics Set C

Short Answer Type Questions

Question. Find the class marks of the class 15-35 and class 45-60.
Answer: We know that, Class mark = \( \frac{\text{Lower limit + Upper limit}}{2} \)
∴ Class mark of 15-35 is = \( \frac{15 + 35}{2} = \frac{50}{2} = 25 \)
Class mark of 45-60 is = \( \frac{45 + 60}{2} = \frac{105}{2} = 52.5 \)

Question. What is the arithmetic mean of first n natural numbers?
Answer: Arithmetic mean = \( \frac{\text{Sum of all the observations}}{\text{Number of observations}} \)
= \( \frac{1 + 2 + ... + n}{n} \)
= \( \frac{\frac{n}{2} [2 \times 1 + (n - 1)1]}{n} \)
= \( \frac{2 + n - 1}{2} = \frac{n + 1}{2} \)

Question. Calculate the mean of the following data.
Class: 4-7, 8-11, 12-15, 16-19
Frequency: 5, 4, 9, 10
Answer: First, we find the class mark \( x_i \) of each class:
Class 3.5-7.5: Class mark 5.5, Frequency 5, \( f_i x_i \) = 27.5
Class 7.5-11.5: Class mark 9.5, Frequency 4, \( f_i x_i \) = 38
Class 11.5-15.5: Class mark 13.5, Frequency 9, \( f_i x_i \) = 121.5
Class 15.5-19.5: Class mark 17.5, Frequency 10, \( f_i x_i \) = 175
Total \( \Sigma f_i = 28 \), \( \Sigma f_i x_i = 362 \)
Therefore, Mean \( (\bar{x}) = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{362}{28} = 12.93 \)

Question. Find the mean of the distribution.
Class: 1-3, 3-5, 5-7, 7-10
Frequency: 9, 22, 27, 17
Answer: Class mark \( x_i \) and \( f_i x_i \) table:
1-3: \( x_i = 2, f_i = 9, f_i x_i = 18 \)
3-5: \( x_i = 4, f_i = 22, f_i x_i = 88 \)
5-7: \( x_i = 6, f_i = 27, f_i x_i = 162 \)
7-10: \( x_i = 8.5, f_i = 17, f_i x_i = 144.5 \)
Total \( \Sigma f_i = 75, \Sigma f_i x_i = 412.5 \)
Mean \( (\bar{x}) = \frac{412.5}{75} = 5.5 \)

Question. The following table gives the number of pages written by Sarika for completing her own book for 30 days.
Number of pages written per day: 16-18, 19-21, 22-24, 25-27, 28-30
Number of days: 1, 3, 4, 9, 13
Find the mean number of pages written per day.
Answer: Since data is not continuous, adjust limits by 0.5. Continuous intervals and mid-values:
15.5-18.5: Mid-value 17, Frequency 1, \( f_i x_i = 17 \)
18.5-21.5: Mid-value 20, Frequency 3, \( f_i x_i = 60 \)
21.5-24.5: Mid-value 23, Frequency 4, \( f_i x_i = 92 \)
24.5-27.5: Mid-value 26, Frequency 9, \( f_i x_i = 234 \)
27.5-30.5: Mid-value 29, Frequency 13, \( f_i x_i = 377 \)
Total \( \Sigma f_i = 30, \Sigma f_i x_i = 780 \)
Mean \( (\bar{x}) = \frac{780}{30} = 26 \)

Question. The mean of the following data is 14. Find the value of k.
x: 5, 10, 15, 20, 25
f: 7, k, 8, 4, 5
Answer: Data table:
\( x_i f_i \): 35, 10k, 120, 80, 125
Total \( \Sigma f_i = k + 24 \), \( \Sigma f_i x_i = 10k + 360 \)
Mean = 14 \( \Rightarrow \frac{10k + 360}{k + 24} = 14 \)
\( 10k + 360 = 14k + 336 \)
\( 4k = 24 \Rightarrow k = 6 \)

Question. The mean of the following frequency distribution is 18. The frequency f in the class interval 19-21 is missing. Determine f.
Class interval: 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25
Frequency: 3, 6, 9, 13, f, 5, 4
Answer: Mid-values \( x_i \): 12, 14, 16, 18, 20, 22, 24
\( f_i x_i \): 36, 84, 144, 234, 20f, 110, 96
Total \( \Sigma f_i = 40 + f \), \( \Sigma f_i x_i = 704 + 20f \)
Mean = 18 \( \Rightarrow 18 = \frac{704 + 20f}{40 + f} \)
\( 720 + 18f = 704 + 20f \Rightarrow 16 = 2f \Rightarrow f = 8 \)

Question. The mileage (\( km L^{-1} \)) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below.
Mileage (\( km L^{-1} \)): 10-12, 12-14, 14-16, 16-18
Number of cars: 7, 12, 18, 13
Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 \( km L^{-1} \). Do you agree with this claim?
Answer: Mid-values \( x_i \): 11, 13, 15, 17
\( f_i x_i \): 77, 156, 270, 221
Total \( \Sigma f_i = 50, \Sigma f_i x_i = 724 \)
Mean \( \bar{x} = \frac{724}{50} = 14.48 km L^{-1} \).
The claim of 16 \( km L^{-1} \) is incorrect as the actual mean is 14.48 \( km L^{-1} \).

Question. An NGO working for welfare of cancer patients, maintained its records as follows:
Age of patients (in years): 0-20, 20-40, 40-60, 60-80
Number of patients: 35, 315, 120, 50
Find mode. [CBSE 2016]
Answer: Maximum frequency is 315, so modal class is 20-40.
\( l = 20, f_1 = 315, f_0 = 35, f_2 = 120, h = 20 \)
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 20 + \left( \frac{315 - 35}{2(315) - 35 - 120} \right) \times 20 \)
= \( 20 + \left( \frac{280}{630 - 155} \right) \times 20 = 20 + \frac{5600}{475} = 20 + 11.79 = 31.79 \)

Question. Find the mode of the following distribution.
Class: 10-15, 15-20, 20-25, 25-30, 30-35, 35-40
Frequency: 45, 30, 75, 20, 35, 15
Answer: Maximum frequency is 75, modal class is 20-25.
\( l = 20, f_1 = 75, f_0 = 30, f_2 = 20, h = 5 \)
Mode = \( 20 + \left( \frac{75 - 30}{2(75) - 30 - 20} \right) \times 5 \)
= \( 20 + \frac{45 \times 5}{150 - 50} = 20 + \frac{225}{100} = 20 + 2.25 = 22.25 \)

Question. Find the mode of the following distribution.
Class: 25-30, 30-35, 35-40, 40-45, 45-50, 50-55
Frequency: 20, 36, 53, 40, 28, 14
Answer: Maximum frequency is 53, modal class is 35-40.
\( l = 35, f_1 = 53, f_0 = 36, f_2 = 40, h = 5 \)
Mode = \( 35 + \left( \frac{53 - 36}{2(53) - 36 - 40} \right) \times 5 \)
= \( 35 + \frac{17 \times 5}{106 - 76} = 35 + \frac{85}{30} \approx 35 + 2.83 = 37.83 \)

Question. Compute the mode for the following frequency distribution.
Size of items (in cm): 0-4, 4-8, 8-12, 12-16, 16-20, 20-24, 24-28
Frequency: 5, 7, 9, 17, 12, 10, 6
Answer: Maximum frequency is 17, modal class is 12-16.
\( l = 12, f_1 = 17, f_0 = 9, f_2 = 12, h = 4 \)
Mode = \( 12 + \left( \frac{17 - 9}{2(17) - 9 - 12} \right) \times 4 \)
= \( 12 + \frac{8 \times 4}{34 - 21} = 12 + \frac{32}{13} \approx 12 + 2.46 = 14.46 \)

Question. Find the mode of the following frequency distribution.
Class: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45
Frequency: 3, 8, 9, 10, 3, 2
Answer: Maximum frequency is 10, modal class is 30-35.
\( l = 30, f_1 = 10, f_0 = 9, f_2 = 3, h = 5 \)
Mode = \( 30 + \left( \frac{10 - 9}{2(10) - 9 - 3} \right) \times 5 \)
= \( 30 + \frac{1 \times 5}{20 - 12} = 30 + \frac{5}{8} = 30 + 0.625 = 30.625 \)

Question. The set of data given below shows the ages of participants in a certain summer camp. Draw a cumulative frequency table for the data.
Age (in years): 10, 11, 12, 13, 14, 15
Frequency: 3, 18, 13, 12, 7, 27
Answer: Age (years) | Frequency | Cumulative Frequency
10 | 3 | 3
11 | 18 | 21
12 | 13 | 34
13 | 12 | 46
14 | 7 | 53
15 | 27 | 80

Question. Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class:
Marks: Below 20, Below 40, Below 60, Below 80, Below 100
Number of students: 17, 22, 29, 37, 50
Form the frequency distribution table for the data.
Answer: Marks | Number of Students
0-20 | 17
20-40 | 22 - 17 = 5
40-60 | 29 - 22 = 7
60-80 | 37 - 29 = 8
80-100 | 50 - 37 = 13

Question. The following table shows the cumulative frequency distribution of marks of 800 students in an examination.
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60, Below 70, Below 80, Below 90, Below 100
Number of students: 10, 50, 130, 270, 440, 570, 670, 740, 780, 800
Construct a frequency distribution table for the data above.
Answer: Marks | Number of Students
0-10 | 10
10-20 | 40
20-30 | 80
30-40 | 140
40-50 | 170
50-60 | 130
60-70 | 100
70-80 | 70
80-90 | 40
90-100 | 20

Question. The following distribution of weights (in kg) of 40 persons.
Weight (in kg): 40-45, 45-50, 50-55, 55-60, 60-65, 65-70, 70-75, 75-80
Number of persons: 4, 4, 13, 5, 6, 5, 2, 1
Construct a cumulative frequency distribution (of the less than type) table for the data above.
Answer: Weight (in kg) | Cumulative Frequency (cf)
Less than 45 | 4
Less than 50 | 8
Less than 55 | 21
Less than 60 | 26
Less than 65 | 32
Less than 70 | 37
Less than 75 | 39
Less than 80 | 40

Question. Form the frequency distribution table from the following data:
Marks (Out of 90): More than or equal to 80, 70, 60, 50, 40, 30, 20, 10, 0
Number of candidates: 4, 6, 11, 17, 23, 27, 30, 32, 34
Answer: Class interval | Number of candidates
0-10 | 34 - 32 = 2
10-20 | 32 - 30 = 2
20-30 | 30 - 27 = 3
30-40 | 27 - 23 = 4
40-50 | 23 - 17 = 6
50-60 | 17 - 11 = 6
60-70 | 11 - 6 = 5
70-80 | 6 - 4 = 2
80-90 | 4

Question. From the following distribution, find the median:
Class: 500-600, 600-700, 700-800, 800-900, 900-100
Frequency: 36, 32, 32, 20, 30
Answer: \( N = 150, \frac{N}{2} = 75 \). Cumulative frequencies: 36, 68, 100, 120, 150. Median class is 700-800.
\( l = 700, cf = 68, f = 32, h = 100 \)
Median = \( l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h = 700 + \left( \frac{75 - 68}{32} \right) \times 100 \)
= \( 700 + \frac{700}{32} = 700 + 21.88 = 721.88 \)

Question. Size of agricultural holdings in a survey of 200 families is given in the following table:
Size (in hectare): 0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35
Number of families: 10, 15, 30, 80, 40, 20, 5
Compute median and modal class of the holdings.
Answer: \( N = 200, \frac{N}{2} = 100 \). Median class is 15-20 (CF = 135).
Median = \( 15 + \left( \frac{100 - 55}{80} \right) \times 5 = 15 + \frac{45}{16} = 15 + 2.81 = 17.81 \) hec.
Modal class is the one with highest frequency (80), which is 15-20.

Question. If median = 137 units and mean = 137.05 units, then find the mode.
Answer: Using empirical formula: Mode = 3(Median) - 2(Mean)
= \( 3(137) - 2(137.05) = 411 - 274.10 = 136.90 \) units.

Long Answer Type Questions

Question. The weights (in kg) of 50 wrestlers are recorded in the following table.
Weight (in kg): 100-110, 110-120, 120-130, 130-140, 140-150
Number of wrestlers: 4, 14, 21, 8, 3
Find the mean weight of the wrestlers.
Answer: Assumed mean method with \( a = 125, h = 10 \).
Mid-values \( x_i \): 105, 115, 125, 135, 145
Deviations \( d_i \): -20, -10, 0, 10, 20
\( f_i d_i \): -80, -140, 0, 80, 60
\( \Sigma f_i = 50, \Sigma f_i d_i = -80 \)
Mean = \( a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 125 + \frac{-80}{50} = 125 - 1.6 = 123.4 \) kg.

Question. If mode of the following series is 54, then find the value of f.
Class interval: 0-15, 15-30, 30-45, 45-60, 60-75, 75-90
Frequency: 3, 5, f, 16, 12, 7
Answer: Mode = 54, which lies in class 45-60. Modal class is 45-60.
\( l = 45, f_1 = 16, f_0 = f, f_2 = 12, h = 15 \)
\( 54 = 45 + \left( \frac{16 - f}{2(16) - f - 12} \right) \times 15 \)
\( 9 = \left( \frac{16 - f}{20 - f} \right) \times 15 \)
\( 3(20 - f) = 5(16 - f) \Rightarrow 60 - 3f = 80 - 5f \Rightarrow 2f = 20 \Rightarrow f = 10 \)

Question. Find the mode of the following distribution.
Classes: 0-20, 20-40, 40-60, 60-80, 80-100
Frequency: 10, 8, 12, 16, 4
Answer: Highest frequency is 16, modal class is 60-80.
\( l = 60, f_1 = 16, f_0 = 12, f_2 = 4, h = 20 \)
Mode = \( 60 + \left( \frac{16 - 12}{2(16) - 12 - 4} \right) \times 20 \)
= \( 60 + \frac{4 \times 20}{32 - 16} = 60 + \frac{80}{16} = 60 + 5 = 65 \)

Question. The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years): 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Number of patients: 60, 42, 55, 70, 53, 20
Form (i) less than type cumulative frequency distribution, (ii) more than type cumulative frequency distribution.
Answer:
(i) Less than type:
Less than 10: 0
Less than 20: 60
Less than 30: 102
Less than 40: 157
Less than 50: 227
Less than 60: 280
Less than 70: 300

(ii) More than type:
More than or equal to 10: 300
More than or equal to 20: 240
More than or equal to 30: 198
More than or equal to 40: 143
More than or equal to 50: 73
More than or equal to 60: 20

Question. Find the unknown entries a, b, c, d, e and f in the following distribution of heights of students in a class:
Height (in cm) | Frequency | Cumulative frequency
150-155 | 12 | a
155-160 | b | 25
160-165 | 10 | c
165-170 | d | 43
170-175 | e | 48
175-180 | 2 | f
Total | 50
Answer: By comparing frequency and cumulative frequency:
a = 12
12 + b = 25 \( \Rightarrow \) b = 13
25 + 10 = c \( \Rightarrow \) c = 35
35 + d = 43 \( \Rightarrow \) d = 8
43 + e = 48 \( \Rightarrow \) e = 5
48 + 2 = f \( \Rightarrow \) f = 50

Question. The maximum bowling speeds (in km/h) of 33 players at a cricket coaching centre are given as follows:
Speed (in km/h): 85-100, 100-115, 115-130, 130-145
Number of players: 11, 9, 8, 5
Calculate the median bowling speed.
Answer: \( N = 33, \frac{N}{2} = 16.5 \). Median class is 100-115 (CF=20).
\( l = 100, f = 9, cf = 11, h = 15 \)
Median = \( 100 + \left( \frac{16.5 - 11}{9} \right) \times 15 = 100 + \frac{5.5 \times 15}{9} = 100 + 9.17 = 109.17 \) km/h.

Question. Obtain the median for the following frequency distribution.
x: 1, 2, 3, 4, 5, 6, 7, 8, 9
y: 8, 10, 11, 16, 20, 25, 15, 9, 6
Answer: \( N = 120 \) (even). Median = Average of \( (\frac{120}{2})th \) and \( (\frac{120}{2} + 1)th \) terms = Average of 60th and 61st terms.
Cumulative frequencies: 8, 18, 29, 45, 65, 90, 105, 114, 120.
Both 60th and 61st terms lie in the range where x = 5. Median = 5.

Question. Weekly income of 600 families is tabulated below:
Weekly income (in ₹): 0-1000, 1000-2000, 2000-3000, 3000-4000, 4000-5000, 5000-6000
Number of families: 250, 190, 100, 40, 15, 5
Compute the median income.
Answer: \( N = 600, \frac{N}{2} = 300 \). CF values: 250, 440, 540, 580, 595, 600. Median class is 1000-2000.
\( l = 1000, f = 190, cf = 250, h = 1000 \)
Median = \( 1000 + \left( \frac{300 - 250}{190} \right) \times 1000 = 1000 + \frac{5000}{19} = 1000 + 263.15 = 1263.15 \) ₹.

Question. A survey regarding the heights (in cm) of 51 boys of Class X of a school was conducted and the following data was obtained:
Heights (in cm): Less than 140, 145, 150, 155, 160, 165
Number of boys: 4, 11, 29, 40, 46, 51
Find the median height.
Answer: Frequency distribution:
Below 140: 4
140-145: 7
145-150: 18
150-155: 11
155-160: 6
160-165: 5
\( N = 51, \frac{N}{2} = 25.5 \). Median class is 145-150.
\( l = 145, f = 18, cf = 11, h = 5 \)
Median = \( 145 + \left( \frac{25.5 - 11}{18} \right) \times 5 = 145 + \frac{72.5}{18} = 145 + 4.03 = 149.03 \) cm.

Question. Find the missing frequencies in the following frequency distribution table, if N = 100 and median is 32. CBSE 2019
Marks obtained: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
Number of students: 10, ?, 25, 30, ?, 10
Answer: Let missing frequencies be \( f_1 \) and \( f_2 \).
Sum = \( 75 + f_1 + f_2 = 100 \Rightarrow f_1 + f_2 = 25 \).
Median 32 is in 30-40. \( l = 30, f = 30, cf = 35 + f_1, h = 10 \).
\( 32 = 30 + \left( \frac{50 - (10 + f_1 + 25)}{30} \right) \times 10 \)
\( 2 = \frac{15 - f_1}{3} \Rightarrow 6 = 15 - f_1 \Rightarrow f_1 = 9 \).
\( f_2 = 25 - 9 = 16 \). Missing frequencies are 9 and 16.

Question. The table below shows the salaries of 280 persons.
Salary (in ₹ thousand): 5-10, 10-15, 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50
Number of persons: 49, 133, 63, 15, 6, 7, 4, 2, 1
Calculate (i) median of the data, (ii) mode of the data.
Answer: (i) \( N = 280, \frac{N}{2} = 140 \). Median class is 10-15.
Median = \( 10 + \left( \frac{140 - 49}{133} \right) \times 5 = 10 + 3.421 = 13.421 \) thousand = ₹ 13421.
(ii) Modal class is 10-15 (freq 133).
Mode = \( 10 + \left( \frac{133 - 49}{2(133) - 49 - 63} \right) \times 5 = 10 + 2.727 = 12.727 \) thousand = ₹ 12727.

Case Based Questions

The men’s 200 m race event at the 2020 Tokyo Olympic took place 3 and 4 August. A stopwatch was used to find the time that it took a group of Athletes to run 200 m.
Time (in seconds): 0-20, 20-40, 40-60, 60-80, 80-100
Number of Students: 8, 10, 13, 6, 3

Question. Estimate the mean time taken by a student to finish the race.
Answer: Total \( f_i = 40 \). Mid-values \( x_i \): 10, 30, 50, 70, 90. \( f_i x_i \): 80, 300, 650, 420, 270. Total \( \Sigma f_i x_i = 1720 \).
Mean = \( \frac{1720}{40} = 43 \) s.

Question. What is the sum of lower limits of median class and modal class.
Answer: Highest frequency is 13, so modal class is 40-60. Lower limit = 40.
\( N/2 = 20 \). CF values: 8, 18, 31... Median class is 40-60. Lower limit = 40.
Sum = 40 + 40 = 80.

Question. How many students finished the race within 1 min?
Answer: Within 1 min means less than 60 seconds.
Students = (0-20) + (20-40) + (40-60) = 8 + 10 + 13 = 31.