Statistics: 4 marks Questions (CBSE Class 10)

14 different 4 marks questions on Statistics from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (69)2 marks (15)3 marks (27)4 marks (14)5 marks (73)

Q374 marksCase StudyStatistics CBSE 2026 · Basic 430/4/1

CENTRAL POLLUTION CONTROL BOARD'S AIR QUALITY STANDARDS
AIR QUALITY INDEX (AQI) | CATEGORY
0-50 | Good
51-100 | Satisfactory
101-200 | Moderate
201-300 | Poor
301-400 | Very Poor
401-500 | Severe
The Air Quality Index (AQI) is a scale from 0 to 500 that indicates air quality, with higher numbers signifying more pollution and greater health concerns.
Mansi collected the daily data of AQI of her city for a month and presented it as given below :
AQI Range : 1 – 100 | 101 – 200 | 201 – 300 | 301 – 400 | 401 – 500
Number of Days : 3 | 9 | 12 | 4 | 2
(i) Convert the data to continuous frequency distribution. (1)
(ii) What is the quality of air in most of the days of the month ? (1)
(iii) (a) Using table formed in part (i), find mode of the data. (2)
OR
(b) Using table formed in part (i), find median of the data. (2)

Show answer & solution

Answer: (i) 0.5 – 100.5, 100.5 – 200.5, 200.5 – 300.5, 300.5 – 400.5, 400.5 – 500.5 with frequencies 3, 9, 12, 4, 2 (ii) Poor (iii) (a) Mode =

200.5 + \frac{300}{11} \approx 227.77

(b) Median = 225.5

(i) Subtract 0.5 from each lower limit and add 0.5 to each upper limit: 0.5 – 100.5 (3), 100.5 – 200.5 (9), 200.5 – 300.5 (12), 300.5 – 400.5 (4), 400.5 – 500.5 (2).
(ii) The highest frequency (12 days) is in the class 201 – 300, which is the 'Poor' category.
(iii) (a) Modal class 200.5 – 300.5: l = 200.5, $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 4$ , h = 100.
Mode = $l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 200.5 + \frac{3}{11} \times 100 \approx 200.5 + 27.27 = 227.77$ .
(iii) (b) n = 30, $\frac{n}{2} = 15$ ; cumulative frequencies 3, 12, 24, 28, 30, so median class is 200.5 – 300.5 with cf = 12, f = 12.
Median = $l + \frac{\frac{n}{2} - c f}{f} \times h = 200.5 + \frac{15 - 12}{12} \times 100 = 225.5$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q384 marksCase StudyStatistics CBSE 2025 · Standard 30/3/1

The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results.
The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.
Rainfall (mm): 200 – 400, 400 – 600, 600 – 800, 800 – 1000, 1000 – 1200, 1200 – 1400
No. of Sub-divisions: 3, 4, 7, 4, 3, 3
Based on the information given above, answer the following questions :
(i) Write the modal class. (1)
(ii) (a) Find the median of the given data. (2)
OR
(b) Find the mean rainfall in the season. (2)
(iii) If a sub-division having at least 800 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall ? (1)

Show answer & solution

Answer: (i) 600 – 800 (ii) (a)

\frac{5200}{7} \approx 742.86

mm OR (b) 775 mm (iii) 10

(i) Highest frequency is 7, for class 600 – 800; modal class is 600 – 800.
(ii)(a) $N = 24$ , $\frac{N}{2} = 12$ . Cumulative frequencies: 3, 7, 14, 18, 21, 24; median class 600 – 800.
(ii)(a) $l = 600$ , $c f = 7$ , $f = 7$ , $h = 200$ : Median $= 600 + \frac{12 - 7}{7} \times 200 = 600 + \frac{1000}{7} \approx 742.86$ mm.
(ii)(b) Class marks: 300, 500, 700, 900, 1100, 1300; $\sum f_{i} x_{i} = 900 + 2000 + 4900 + 3600 + 3300 + 3900 = 18600$ .
(ii)(b) Mean $= \frac{18600}{24} = 775$ mm.
(iii) Sub-divisions with at least 800 mm $= 4 + 3 + 3 = 10$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q364 marksCase StudyStatistics CBSE 2024 · Basic 430/1/2

Gurpreet is very fond of doing research on plants. She collected some leaves from different plants and measured their lengths in mm.
The data obtained is represented in the following table :
Length (in mm) : 70-80, 80-90, 90-100, 100-110, 110-120, 120-130, 130-140
Number of leaves : 3, 5, 9, 12, 5, 4, 2
Based on the above information, answer the following questions :
(i) Write the median class of the data. (1)
(ii) How many leaves are of length equal to or more than 10 cm ? (1)
(iii) (a) Find median of the data. (2)
OR
(b) Write the modal class and find the mode of the data. (2)

Show answer & solution

Answer: (i) 100-110 (ii) 23 (iii) (a) 102.5 mm OR (b) modal class 100-110, mode = 103 mm

Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40; $N = 40$ .
(i) $\frac{N}{2} = 20$ lies in the class with cf 29, so the median class is 100-110.
(ii) 10 cm = 100 mm; leaves of length 100 mm or more $= 12 + 5 + 4 + 2 = 23$ .
(iii) (a) Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 100 + \frac{20 - 17}{12} \times 10 = 100 + 2.5 = 102.5$ mm.
(iii) (b) Highest frequency is 12, so the modal class is 100-110; $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 5$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 100 + \frac{3}{24 - 9 - 5} \times 10 = 100 + 3 = 103$ mm.

Also asked in: 2024 Basic 430/1/3

Q374 marksCase StudyStatistics CBSE 2024 · Basic 430/2/1

Heart Rate : The heart rate is one of the ‘vital signs’ of health in the human body. It measures the number of times per minute that the heart contracts or beats. While a normal heart rate does not guarantee that a person is free of health problems, it is a useful benchmark for identifying a range of health issues.
Thirty women were examined by doctors of AIIMS and the number of heart beats per minute were recorded and summarized as follows :
Number of heart beats per minute : 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86
Number of Women : 2, 4, 3, 8, 7, 4, 2
Based on the above information, answer the following questions :
(i) How many women are having heart beat in the range 68 – 77 ? (1)
(ii) What is the median class of heart beats per minute for these women ? (1)
(iii) Find the modal value of heart beats per minute for these women. (2)
OR (iii) Find the median value of heart beats per minute for these women. (2)

Show answer & solution

Answer: (i) 15 (ii) 74–77 (iii) Mode = 76.5; OR (iii) Median = 76.25

(i) Classes 68–71, 71–74, 74–77: $4 + 3 + 8 = 15$ women
(ii) Cumulative frequencies: 2, 6, 9, 17, 24, 28, 30; $\frac{N}{2} = 15$ , which first lies in cf 17, so median class is 74–77
(iii) Modal class 74–77: $l = 74$ , $f_{1} = 8$ , $f_{0} = 3$ , $f_{2} = 7$ , $h = 3$
Mode $= 74 + \frac{8 - 3}{16 - 3 - 7} \times 3 = 74 + \frac{5}{6} \times 3 = 76.5$
OR (iii) Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 74 + \frac{15 - 9}{8} \times 3 = 74 + 2.25 = 76.25$

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q364 marksCase StudyStatistics CBSE 2024 · Basic 430/4/1

Mutual Fund : A mutual fund is a type of investment vehicle that pools money from multiple investors to invest in securities like stocks, bonds or other securities. Mutual funds are operated by professional money managers, who allocate the fund’s assets and attempt to produce capital gains or income for the fund’s investors.
Net Asset Value (NAV) represents a fund’s per share market value. It is the price at which the investors buy fund shares from a fund company and sell them to a fund company.
The following table shows the Net Asset Value (NAV) per unit of mutual fund of ICICI mutual funds :
NAV (in ₹): 0–5, 5–10, 10–15, 15–20, 20–25
Number of mutual funds: 13, 16, 22, 18, 11
Based on the above information, answer the following questions :
(i) What is the upper limit of modal class of the data ? (1)
(ii) What is the median class of the data ? (1)
(iii) (a) What is the mode NAV of mutual funds ? (2)
OR (b) What is the median NAV of mutual funds ? (2)

Show answer & solution

Answer: (i) 15 (ii) 10–15 (iii) (a) ₹13 OR (b) ₹12.5

(i) Highest frequency 22 is for 10–15, so the modal class is 10–15 and its upper limit is 15.
(ii) $N = 13 + 16 + 22 + 18 + 11 = 80$ , $\frac{N}{2} = 40$ . Cumulative frequencies: 13, 29, 51, 69, 80. The median class is 10–15.
(iii)(a) Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 10 + \frac{22 - 16}{44 - 16 - 18} \times 5 = 10 + \frac{6}{10} \times 5 = 13$ . Mode NAV = ₹13.
(iii)(b) Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 10 + \frac{40 - 29}{22} \times 5 = 10 + 2.5 = 12.5$ . Median NAV = ₹12.5.

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q374 marksCase StudyStatistics CBSE 2024 · Standard 30/1/1

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.
The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'.
Numbers announced: 0-15, 15-30, 30-45, 45-60, 60-75
Number of times: 8, 9, 10, 12, 9
Based on the above information, answer the following :
(i) Write the median class. (1)
(ii) When first ball was picked up, what was the probability of calling out an even number ? (1)
(iii) (a) Find median of the given data. (2)
OR (b) Find mode of the given data. (2)

Show answer & solution

Answer: (i) 30-45 (ii)

\frac{37}{75}

(iii) (a) 40.5 OR (b) 51

Cumulative frequencies: 8, 17, 27, 39, 48; $n = 48$ , $\frac{n}{2} = 24$ .
(i) The cumulative frequency first exceeds 24 in the class 30-45, so the median class is 30-45.
(ii) There are 37 even numbers from 1 to 75, so P(even) $= \frac{37}{75}$ .
(iii) (a) $l = 30$ , $c f = 17$ , $f = 10$ , $h = 15$ . Median $= 30 + \frac{24 - 17}{10} \times 15 = 30 + 10.5 = 40.5$ .
OR (b) Modal class 45-60: $l = 45$ , $f_{1} = 12$ , $f_{0} = 10$ , $f_{2} = 9$ , $h = 15$ .
Mode $= 45 + \frac{12 - 10}{2 ( 12 ) - 10 - 9} \times 15 = 45 + \frac{2}{5} \times 15 = 51$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q374 marksCase StudyStatistics CBSE 2024 · Standard 30/3/1

Vocational training complements traditional education by providing practical skills and hands-on experience. While education equips individuals with a broad knowledge base, vocational training focuses on job-specific skills, enhancing employability thus making the student self-reliant. Keeping this in view, a teacher made the following table giving the frequency distribution of students/adults undergoing vocational training from the training institute.
Age (in years): 15-19, 20-24, 25-29, 30-34, 35-39, 40-44, 45-49, 50-54
Number of participants: 62, 132, 96, 37, 13, 11, 10, 4
From the above answer the following questions :
(i) What is the lower limit of the modal class of the above data ? (1)
(ii) Find the median class of the above data. (2)
OR Find the number of participants of age less than 50 years who undergo vocational training. (2)
(iii) Give the empirical relationship between mean, median and mode. (1)

Show answer & solution

Answer: (i) 19.5 (modal class 20-24, i.e. 19.5-24.5) (ii) 20-24 (19.5-24.5); OR 361 (iii) 3 Median = Mode + 2 Mean

(i) Highest frequency is 132, so the modal class is 20-24. Making the classes continuous (subtract 0.5 / add 0.5), it is 19.5-24.5, with lower limit 19.5.
(ii) N = 62 + 132 + 96 + 37 + 13 + 11 + 10 + 4 = 365, so N/2 = 182.5.
Cumulative frequencies: 62, 194, 290, 327, 340, 351, 361, 365.
The first cumulative frequency above 182.5 is 194, so the median class is 20-24 (19.5-24.5).
OR: Participants below 50 years = 365 - 4 = 361.
(iii) Empirical relationship: 3 Median = Mode + 2 Mean.

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q384 marksCase StudyStatistics CBSE 2024 · Standard 30/5/1

Activities like running or cycling reduce stress and the risk of mental disorders like depression. Running helps build endurance. Children develop stronger bones and muscles and are less prone to gain weight. The physical education teacher of a school has decided to conduct an inter school running tournament in his school premises. The time taken by a group of students to run 100 m, was noted as follows :
Time (in seconds): 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 8, 10, 13, 6, 3
Based on the above, answer the following questions :
(i) What is the median class of the above given data ? (1)
(ii) (a) Find the mean time taken by the students to finish the race. (2)
OR
(b) Find the mode of the above given data. (2)
(iii) How many students took time less than 60 seconds ? (1)

Show answer & solution

Answer: (i) 40–60 (ii) (a) 43 seconds OR (b) 46 seconds (iii) 31

Total $n = 40$ ; cumulative frequencies: 8, 18, 31, 37, 40.
(i) $\frac{n}{2} = 20$ lies in the class with cf 31, so the median class is 40–60.
(ii) (a) Class marks 10, 30, 50, 70, 90; $\sum f_{i} x_{i} = 80 + 300 + 650 + 420 + 270 = 1720$ ; mean $= \frac{1720}{40} = 43$ seconds.
(ii) (b) Modal class 40–60: $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 6$ , $h = 20$ ; mode $= 40 + \frac{13 - 10}{26 - 10 - 6} \times 20 = 40 + 6 = 46$ seconds.
(iii) $8 + 10 + 13 = 31$ students.

Also asked in: 2024 Standard 30/5/2

Q374 marksCase StudyStatistics CBSE 2024 · Standard 30/5/3

Activities like running or cycling reduce stress and the risk of mental disorder like depression. Running helps build endurance. Children develop stronger bones and muscles and are less prone to gain weight. The physical education teacher of a school has decided to conduct an inter school running tournament in his school premises. The time taken by a group of students to run 100 m, was noted as follows :
Time (in seconds): 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 8, 10, 13, 6, 3
Based on the above, answer the following questions :
(i) What is the median class of the above given data ? (1)
(ii) (a) Find the mean time taken by the students to finish the race. (2)
OR
(b) Find the mode of the above given data. (2)
(iii) How many students took time less than 60 seconds ? (1)

Show answer & solution

Answer: (i) 40–60 (ii) (a) 43 seconds OR (b) 46 seconds (iii) 31

Total $n = 40$ ; cumulative frequencies: 8, 18, 31, 37, 40.
(i) $\frac{n}{2} = 20$ lies in the class with cf 31, so the median class is 40–60.
(ii) (a) Class marks 10, 30, 50, 70, 90; $\sum f_{i} x_{i} = 80 + 300 + 650 + 420 + 270 = 1720$ ; mean $= \frac{1720}{40} = 43$ seconds.
(ii) (b) Modal class 40–60: $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 6$ , $h = 20$ ; mode $= 40 + \frac{13 - 10}{26 - 10 - 6} \times 20 = 40 + 6 = 46$ seconds.
(iii) $8 + 10 + 13 = 31$ students.

Q374 marksCase StudyStatistics CBSE 2023 · Standard 30/1/1

India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country.
It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon) rainfall (mm) in 2018 :
Rainfall (mm): 200-400, 400-600, 600-800, 800-1000, 1000-1200, 1200-1400, 1400-1600, 1600-1800
Number of Sub-divisions: 2, 4, 7, 4, 2, 3, 1, 1
Based on the above information, answer the following questions :
(I) Write the modal class. (1)
(II) Find the median of the given data. (2)
OR
Find the mean rainfall in this season.
(III) If sub-division having at least 1000 mm rainfall during monsoon season, is considered good rainfall sub-division, then how many sub-divisions had good rainfall ? (1)

Show answer & solution

Answer: (I) 600-800 (II) Median ≈ 771.43 mm; OR: Mean = 850 mm (III) 7

(I) Highest frequency is 7, for 600–800. Modal class = 600–800.
(II) N = 24, $\frac{N}{2} = 12$ . Cumulative frequencies: 2, 6, 13, 17, 19, 22, 23, 24. Median class 600–800.
Median $= 600 + \frac{12 - 6}{7} \times 200 = 600 + 171.43 = 771.43$ mm (approx.).
OR: Class marks 300, 500, 700, 900, 1100, 1300, 1500, 1700. $\sum f_{i} x_{i} = 600 + 2000 + 4900 + 3600 + 2200 + 3900 + 1500 + 1700 = 20400$ .
Mean $= \frac{20400}{24} = 850$ mm.
(III) Sub-divisions with at least 1000 mm = 2 + 3 + 1 + 1 = 7.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q134 marksCase StudyStatistics CBSE 2022 · Basic 430/4/1

Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses have a range of approximately 280 km with just one charge. These buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day.
Daily distance travelled (in km): 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200
Number of buses: 12, 14, 8, 6, 10
(a) Find the 'median' distance travelled by a bus. (2)
(b) Find the 'mean (average)' distance travelled by a bus. (2)

Show answer & solution

Answer: (a) Median

\approx 138.57

km (b) Mean = 145.2 km

(a) Cumulative frequencies: 12, 26, 34, 40, 50; $N = 50$ , $\frac{N}{2} = 25$ .
Median class is 120 – 140: $l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
Median $= 120 + \frac{25 - 12}{14} \times 20 = 120 + \frac{260}{14} \approx 120 + 18.57 = 138.57$ km.
(b) Class marks: 110, 130, 150, 170, 190.
$\sum f_{i} x_{i} = 1320 + 1820 + 1200 + 1020 + 1900 = 7260$ .
Mean $= \frac{7260}{50} = 145.2$ km.

Q134 marksCase StudyStatistics CBSE 2022 · Basic 430/4/2

Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses could have a range of approximately 280 km with just one charge. These buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day.
Daily distance travelled (in km): 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200
Number of buses: 12, 14, 8, 6, 10
(a) Find the 'median' distance travelled by a bus. (2)
(b) Find the 'mean (average)' distance travelled by a bus. (2)

Show answer & solution

Answer: (a) Median

\approx 138.57

km (b) Mean = 145.2 km

(a) Cumulative frequencies: 12, 26, 34, 40, 50; $N = 50$ , $\frac{N}{2} = 25$ .
Median class is 120 – 140: $l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
Median $= 120 + \frac{25 - 12}{14} \times 20 = 120 + \frac{260}{14} \approx 120 + 18.57 = 138.57$ km.
(b) Class marks: 110, 130, 150, 170, 190.
$\sum f_{i} x_{i} = 1320 + 1820 + 1200 + 1020 + 1900 = 7260$ .
Mean $= \frac{7260}{50} = 145.2$ km.

Q134 marksCase StudyStatistics CBSE 2022 · Basic 430/4/3

Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses could have a range of approximately 280 km with just one charge. Electric buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day.
Daily distance travelled (in km): 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200
Number of buses: 12, 14, 8, 6, 10
(a) Find the 'median' distance travelled by a bus. (2)
(b) Find the 'mean (average)' distance travelled by a bus. (2)

Show answer & solution

Answer: (a) Median

\approx 138.57

km (b) Mean = 145.2 km

(a) Cumulative frequencies: 12, 26, 34, 40, 50; $N = 50$ , $\frac{N}{2} = 25$ .
Median class is 120 – 140: $l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
Median $= 120 + \frac{25 - 12}{14} \times 20 = 120 + \frac{260}{14} \approx 120 + 18.57 = 138.57$ km.
(b) Class marks: 110, 130, 150, 170, 190.
$\sum f_{i} x_{i} = 1320 + 1820 + 1200 + 1020 + 1900 = 7260$ .
Mean $= \frac{7260}{50} = 145.2$ km.

Q134 marksCase StudyStatistics CBSE 2022 · Standard 30/4/1

Yoga is an ancient practice which is a form of meditation and exercise. By practising yoga, we not even make our body healthy but also achieve inner peace and calmness. The International Yoga Day is celebrated on $2 1^{s t}$ of June every year since 2015.
To promote Yoga, Green park society in Pune organised a 7-day Yoga camp in their society. The number of people of different age groups who enrolled for this camp is given as follows :
Age Group: 15–25, 25–35, 35–45, 45–55, 55–65, 65–75, 75–85
Number of People: 8, 10, 15, 25, 40, 24, 18
Based on the above, find the following :
(a) Find the median age of people enrolled for the camp. (2)
(b) If $x$ more people of age group 65 – 75 had enrolled for the camp, the mean age would have been 58. Find the value of $x$ . (2)

Show answer & solution

Answer: (a) 58 years (b)

x = \frac{145}{6} \approx 24.17

(about 24)

(a) Cumulative frequencies: 8, 18, 33, 58, 98, 122, 140; $N = 140$ , $\frac{N}{2} = 70$ .
Median class is 55–65: $l = 55$ , $c f = 58$ , $f = 40$ , $h = 10$ .
Median $= 55 + \frac{70 - 58}{40} \times 10 = 55 + 3 = 58$ years.
(b) Class marks: 20, 30, 40, 50, 60, 70, 80.
$\sum f_{i} x_{i} = 160 + 300 + 600 + 1250 + 2400 + 1680 + 1440 = 7830$ , $\sum f_{i} = 140$ .
With $x$ more people in 65–75 (class mark 70): $\frac{7830 + 70 x}{140 + x} = 58$
$7830 + 70 x = 8120 + 58 x \Rightarrow 12 x = 290$
$x = \frac{145}{6} \approx 24.17$ , i.e. about 24 people.

Also asked in: 2022 Standard 30/4/2, 2022 Standard 30/4/3

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