Question 1
Find the mean number of plants per house for intervals 0–2, 2–4, …, 12–14 with frequencies 1,2,1,5,6,2,3.
Steps only
Midpoints: 1,3,5,7,9,11,13
Σf = 20; Σfx = 162
Mean = 162/20 = 8.1
Answer
8.1 plants per house.
CLASS 10 · MATHEMATICS · CHAPTER 13
Statistics
Calculate mean, mode and median for grouped data, including cumulative and continuous frequency tables.
22 worked answersAll 3 exercise sets
What you will learn
- Choose a suitable measure of central tendency.
- Use class marks and cumulative frequencies correctly.
- Interpret grouped-data estimates in context.
22 worked answers · short steps first, explanations when you need them
Exercise 13.1
Textbook pages 181–183
Question 2
For wages ₹500–520 through ₹580–600 with frequencies 12,14,8,6,10, state a suitable method and find the mean.
Steps only
Take assumed mean A = 550 and class width h = 20
Σfu = −12; Σf = 50
Mean = 550 + 20(−12/50) = 545.2
Answer
₹545.20.
Question 3
The mean pocket allowance is ₹18 for classes 11–13 through 23–25 with frequencies 7,6,9,13,f,5,4. Find f.
Steps only
Known frequency = 44; known Σfx = 752
(752 + 20f)/(44 + f) = 18
752 + 20f = 792 + 18f ⇒ f = 20
Answer
The missing frequency is 20.
Question 4
Find the mean heartbeat rate for classes 65–68 through 83–86 with frequencies 2,4,3,8,7,4,2.
Steps only
Midpoints: 66.5,69.5,72.5,75.5,78.5,81.5,84.5
Σf = 30; Σfx = 2277
Mean = 2277/30 = 75.9
Answer
75.9 heartbeats per minute.
Question 5
Find the mean mangoes per box for inclusive classes 50–52, 53–55, 56–58, 59–61, 62–64 with frequencies 15,110,135,115,25.
Steps only
Class marks: 51,54,57,60,63
Σf = 400; Σfx = 22875
Mean = 22875/400 = 57.1875
Answer
About 57.19 mangoes per box.
Question 6
Find the mean daily food expenditure for classes ₹100–150 through ₹300–350 with frequencies 4,5,12,2,2.
Steps only
Midpoints: 125,175,225,275,325
Σf = 25; Σfx = 5275
Mean = 5275/25 = 211
Answer
₹211.
Question 7
Find the mean SO₂ concentration for 0.00–0.04 through 0.20–0.24 ppm with frequencies 4,9,9,2,4,2.
Steps only
Midpoints: 0.02,0.06,0.10,0.14,0.18,0.22
Σf = 30; Σfx = 2.96
Mean = 2.96/30 = 0.0987
Answer
Approximately 0.099 ppm.
Question 8
Find the mean days absent for intervals 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40 with frequencies 11,10,7,4,4,3,1.
Steps only
Midpoints: 3,8,12,17,24,33,39
Σf = 40; Σfx = 499
Mean = 499/40 = 12.475
Answer
12.475 days, about 12.48 days.
Question 9
Find the mean literacy rate for 45–55 through 85–95 percent with frequencies 3,10,11,8,3.
Steps only
Midpoints: 50,60,70,80,90
Σf = 35; Σfx = 2430
Mean = 2430/35 = 69.4286
Answer
Approximately 69.43%.
Exercise 13.2
Textbook pages 186–187
Question 1
For patient ages 5–15 through 55–65 with frequencies 6,11,21,23,14,5, find mode and mean and compare them.
Steps only
Modal class 35–45: mode = 35 + (23−21)/(46−21−14)×10 = 36.82
Σfx/Σf = 2830/80 = 35.375
Answer
Mode ≈ 36.82 years; mean = 35.375 years.
Question 2
Find the modal lifetime for 0–20 through 100–120 hours with frequencies 10,35,52,61,38,29.
Steps only
Modal class 60–80
Mode = 60 + (61−52)/(122−52−38)×20
= 65.625
Answer
65.625 hours.
Question 3
For monthly expenditure ₹1000–1500 through ₹4500–5000 with frequencies 24,40,33,28,30,22,16,7, find mode and mean.
Steps only
Modal class 1500–2000: mode = 1500 + 16/46×500 = 1847.83
Mean = 532500/200 = 2662.5
Answer
Mode ≈ ₹1847.83; mean = ₹2662.50.
Question 4
For ratios 15–20 through 50–55 with state counts 3,8,9,10,3,0,0,2, find mode and mean.
Steps only
Modal class 30–35: mode = 30 + (10−9)/(20−9−3)×5 = 30.625
Mean = 1022.5/35 = 29.2143
Answer
Mode ≈ 30.63; mean ≈ 29.21 students per teacher.
Question 5
Find the mode for run intervals 3000–4000 through 10000–11000 with batsman counts 4,18,9,7,6,3,1,1.
Steps only
Modal class 4000–5000
Mode = 4000 + (18−4)/(36−4−9)×1000
≈ 4608.70
Answer
Approximately 4608.70 runs.
Question 6
Find the mode for car-count intervals 0–10 through 70–80 with frequencies 7,14,13,12,20,11,15,8.
Steps only
Modal class 40–50
Mode = 40 + (20−12)/(40−12−11)×10
≈ 44.71
Answer
Approximately 44.71 cars per three-minute period.
Exercise 13.3
Textbook pages 198–200
Question 1
For monthly electricity use 65–85 through 185–205 with frequencies 4,5,13,20,14,8,4, find median, mean and mode.
Steps only
Median = 125 + (34−22)/20×20 = 137
Mean = 9320/68 ≈ 137.06
Mode = 125 + (20−13)/(40−13−14)×20 ≈ 135.77
Answer
Median = 137 units; mean ≈ 137.06 units; mode ≈ 135.77 units.
Question 2
The median is 28.5 for classes 0–10 through 50–60 with frequencies 5,x,20,15,y,5 and total 60. Find x and y.
Steps only
x + y = 15
28.5 = 20 + (30−5−x)/20×10
x = 8; y = 7
Answer
x = 8 and y = 7.
Question 3
Convert the less-than cumulative age table for 100 policy holders to class frequencies and find the median age.
Steps only
Frequencies: 2,4,18,21,33,11,3,6,2
Median class 35–40; l=35, cf=45, f=33, h=5
Median = 35 + (50−45)/33×5 ≈ 35.76
Answer
Approximately 35.76 years.
Question 4
Leaf lengths measured to the nearest millimetre have classes 118–126 through 172–180 and frequencies 3,5,9,12,5,4,2. Find the median.
Steps only
Use continuous class 144.5–153.5
N=40, cf=17, f=12, h=9
Median = 144.5 + (20−17)/12×9 = 146.75
Answer
146.75 mm.
Question 5
Find the median lifetime for 1500–2000 through 4500–5000 hours with frequencies 14,56,60,86,74,62,48.
Steps only
N=400; median class 3000–3500
l=3000, cf=130, f=86, h=500
Median = 3000 + (200−130)/86×500 ≈ 3406.98
Answer
Approximately 3406.98 hours.
Question 6
For surname lengths 1–4 through 16–19 with frequencies 6,30,40,16,4,4, find median, mean and mode.
Steps only
Median = 7 + (50−36)/40×3 = 8.05
Mean = 832/100 = 8.32
Mode = 7 + (40−30)/(80−30−16)×3 ≈ 7.88
Answer
Median = 8.05 letters; mean = 8.32 letters; mode ≈ 7.88 letters.
Question 7
Find the median for weight classes 40–45 through 70–75 kg with frequencies 2,3,8,6,6,3,2.
Steps only
N=30; median class 55–60
l=55, cf=13, f=6, h=5
Median = 55 + (15−13)/6×5 = 56.67
Answer
Approximately 56.67 kg.