April 10, 2026
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Chapter 13: Statistics
Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)
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Exercise 13.1 (Page 260) – Mean of Grouped Data
13 marks
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants: 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14
Number of houses: 1, 2, 1, 5, 6, 2, 3
Which method did you use for finding the mean, and why?
Number of plants: 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14
Number of houses: 1, 2, 1, 5, 6, 2, 3
Which method did you use for finding the mean, and why?
Answer:
Using direct method (small numbers):
Class marks (xi): 1, 3, 5, 7, 9, 11, 13.
Σfi = 1+2+1+5+6+2+3 = 20. Σfixi = 1×1 + 2×3 + 1×5 + 5×7 + 6×9 + 2×11 + 3×13 = 1+6+5+35+54+22+39 = 162.
Mean = Σfixi/Σfi = 162/20 = 8.1.
Mean number of plants = 8.1
Using direct method (small numbers):
Class marks (xi): 1, 3, 5, 7, 9, 11, 13.
Σfi = 1+2+1+5+6+2+3 = 20. Σfixi = 1×1 + 2×3 + 1×5 + 5×7 + 6×9 + 2×11 + 3×13 = 1+6+5+35+54+22+39 = 162.
Mean = Σfixi/Σfi = 162/20 = 8.1.
Mean number of plants = 8.1
23 marks
Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in ₹): 500-520, 520-540, 540-560, 560-580, 580-600
Number of workers: 12, 14, 8, 6, 10
Find the mean daily wages of the workers.
Daily wages (in ₹): 500-520, 520-540, 540-560, 560-580, 580-600
Number of workers: 12, 14, 8, 6, 10
Find the mean daily wages of the workers.
Answer:
Using direct method: Class marks: 510, 530, 550, 570, 590.
Σfi = 50. Σfixi = 12×510 + 14×530 + 8×550 + 6×570 + 10×590 = 6120 + 7420 + 4400 + 3420 + 5900 = 27260.
Mean = 27260/50 = 545.2.
Mean daily wage = ₹545.20
Using direct method: Class marks: 510, 530, 550, 570, 590.
Σfi = 50. Σfixi = 12×510 + 14×530 + 8×550 + 6×570 + 10×590 = 6120 + 7420 + 4400 + 3420 + 5900 = 27260.
Mean = 27260/50 = 545.2.
Mean daily wage = ₹545.20
33 marks
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.
Daily pocket allowance (in ₹): 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25
Number of children: 7, 6, 9, 13, f, 5, 4
Daily pocket allowance (in ₹): 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25
Number of children: 7, 6, 9, 13, f, 5, 4
Answer:
Class marks: 12, 14, 16, 18, 20, 22, 24.
Σfi = 7+6+9+13+f+5+4 = 44+f.
Σfixi = 7×12=84, 6×14=84, 9×16=144, 13×18=234, f×20=20f, 5×22=110, 4×24=96.
Sum = 84+84+144+234+110+96 = 752 + 20f.
Mean = (752+20f)/(44+f) = 18 → 752+20f = 792+18f → 2f = 40 → f = 20.
Missing frequency f = 20
Class marks: 12, 14, 16, 18, 20, 22, 24.
Σfi = 7+6+9+13+f+5+4 = 44+f.
Σfixi = 7×12=84, 6×14=84, 9×16=144, 13×18=234, f×20=20f, 5×22=110, 4×24=96.
Sum = 84+84+144+234+110+96 = 752 + 20f.
Mean = (752+20f)/(44+f) = 18 → 752+20f = 792+18f → 2f = 40 → f = 20.
Missing frequency f = 20
43 marks
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Number of heartbeats 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
Number of heartbeats 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
Answer:
Using step-deviation method. Let a = mid of 75.5 (approx), h = 3.
Class marks: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.
ui = (xi - a)/h: -3, -2, -1, 0, 1, 2, 3.
fiui = 2×(-3)=-6, 4×(-2)=-8, 3×(-1)=-3, 8×0=0, 7×1=7, 4×2=8, 2×3=6. Σfiui = 4.
Mean = a + h×(Σfiui/Σfi) = 75.5 + 3×(4/30) = 75.5 + 0.4 = 75.9.
Mean = 75.9 beats per minute
Using step-deviation method. Let a = mid of 75.5 (approx), h = 3.
Class marks: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.
ui = (xi - a)/h: -3, -2, -1, 0, 1, 2, 3.
fiui = 2×(-3)=-6, 4×(-2)=-8, 3×(-1)=-3, 8×0=0, 7×1=7, 4×2=8, 2×3=6. Σfiui = 4.
Mean = a + h×(Σfiui/Σfi) = 75.5 + 3×(4/30) = 75.5 + 0.4 = 75.9.
Mean = 75.9 beats per minute
5-9Key Results
Selected answers from remaining problems in Exercise 13.1:
Q5: Mean = 62.4 marks.
Q6: Missing frequency f = 8.
Q7: Mean = 220.5 (using assumed mean method).
Q8: Mean = 55.2 km/h.
Q9: Mean = ₹50.35.
Q6: Missing frequency f = 8.
Q7: Mean = 220.5 (using assumed mean method).
Q8: Mean = 55.2 km/h.
Q9: Mean = ₹50.35.
Exercise 13.2 (Page 267) – Mode of Grouped Data
13 marks
The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65
Number of patients: 6, 11, 21, 23, 14, 5
Find the mode and mean of the data. Compare and interpret the two measures of central tendency.
Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65
Number of patients: 6, 11, 21, 23, 14, 5
Find the mode and mean of the data. Compare and interpret the two measures of central tendency.
Answer:
Modal class: 35-45 (frequency 23). l=35, h=10, f1=23, f0=21, f2=14.
Mode = l + [(f1-f0)/(2f1-f0-f2)] × h = 35 + [(23-21)/(46-21-14)]×10 = 35 + (2/11)×10 = 35 + 20/11 = 35 + 1.82 = 36.82 years.
Mean = (class mark method): Σfi = 80, Σfixi = 10×6+20×11+30×21+40×23+50×14+60×5 = 60+220+630+920+700+300 = 2830. Mean = 2830/80 = 35.375 years.
Mode ≈ 36.8 years, Mean ≈ 35.4 years.
Modal class: 35-45 (frequency 23). l=35, h=10, f1=23, f0=21, f2=14.
Mode = l + [(f1-f0)/(2f1-f0-f2)] × h = 35 + [(23-21)/(46-21-14)]×10 = 35 + (2/11)×10 = 35 + 20/11 = 35 + 1.82 = 36.82 years.
Mean = (class mark method): Σfi = 80, Σfixi = 10×6+20×11+30×21+40×23+50×14+60×5 = 60+220+630+920+700+300 = 2830. Mean = 2830/80 = 35.375 years.
Mode ≈ 36.8 years, Mean ≈ 35.4 years.
23 marks
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetime (in hours): 0-20, 20-40, 40-60, 60-80, 80-100, 100-120
Frequency: 10, 35, 52, 61, 38, 29
Determine the modal lifetimes of the components.
Lifetime (in hours): 0-20, 20-40, 40-60, 60-80, 80-100, 100-120
Frequency: 10, 35, 52, 61, 38, 29
Determine the modal lifetimes of the components.
Answer:
Modal class: 60-80 (max freq 61). l=60, h=20, f1=61, f0=52, f2=38.
Mode = 60 + [(61-52)/(122-52-38)]×20 = 60 + (9/32)×20 = 60 + 5.625 = 65.625 hours.
Modal lifetime ≈ 65.6 hours
Modal class: 60-80 (max freq 61). l=60, h=20, f1=61, f0=52, f2=38.
Mode = 60 + [(61-52)/(122-52-38)]×20 = 60 + (9/32)×20 = 60 + 5.625 = 65.625 hours.
Modal lifetime ≈ 65.6 hours
33 marks
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure. Also, find the mean monthly expenditure.
Expenditure (in ₹): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000
Number of families: 24, 40, 33, 28, 30, 22, 16, 7
Expenditure (in ₹): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000
Number of families: 24, 40, 33, 28, 30, 22, 16, 7
Answer:
Modal class: 1500-2000 (max freq 40). l=1500, h=500, f1=40, f0=24, f2=33.
Mode = 1500 + [(40-24)/(80-24-33)]×500 = 1500 + (16/23)×500 = 1500 + 347.83 = 1847.83.
Mean using step-deviation: assume a = 2750, h = 500. Σfi = 200, Σfiui = -30. Mean = 2750 + 500×(-30/200) = 2750 - 75 = 2675.
Mode ≈ ₹1847.83, Mean = ₹2675
Modal class: 1500-2000 (max freq 40). l=1500, h=500, f1=40, f0=24, f2=33.
Mode = 1500 + [(40-24)/(80-24-33)]×500 = 1500 + (16/23)×500 = 1500 + 347.83 = 1847.83.
Mean using step-deviation: assume a = 2750, h = 500. Σfi = 200, Σfiui = -30. Mean = 2750 + 500×(-30/200) = 2750 - 75 = 2675.
Mode ≈ ₹1847.83, Mean = ₹2675
4-6Key Results
Selected answers from remaining problems in Exercise 13.2:
Q4: Modal class: 30-40, Mode = 36.8 years.
Q5: Modal class: 35-40, Mode = 36.36 marks.
Q6: Modal class: 30-40, Mode = 32.5 kg (maximum number of students).
Q5: Modal class: 35-40, Mode = 36.36 marks.
Q6: Modal class: 30-40, Mode = 32.5 kg (maximum number of students).
Exercise 13.3 (Page 271) – Median of Grouped Data
13 marks
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205
Number of consumers: 4, 5, 13, 20, 14, 8, 4
Monthly consumption (in units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205
Number of consumers: 4, 5, 13, 20, 14, 8, 4
Answer:
N=68, N/2=34. Median class: 125-145 (cumulative frequency 4+5+13=22, +20=42). l=125, cf=22, f=20, h=20.
Median = 125 + [(34-22)/20]×20 = 125 + (12/20)×20 = 125 + 12 = 137 units.
Mean = (Σfixi)/68 = (4×75+5×95+13×115+20×135+14×155+8×175+4×195)/68 = (300+475+1495+2700+2170+1400+780)/68 = 9320/68 = 137.05.
Mode: modal class 125-145, Mode = 125 + [(20-13)/(40-13-14)]×20 = 125 + (7/13)×20 = 125 + 10.77 = 135.77.
Median=137, Mean≈137.05, Mode≈135.77
N=68, N/2=34. Median class: 125-145 (cumulative frequency 4+5+13=22, +20=42). l=125, cf=22, f=20, h=20.
Median = 125 + [(34-22)/20]×20 = 125 + (12/20)×20 = 125 + 12 = 137 units.
Mean = (Σfixi)/68 = (4×75+5×95+13×115+20×135+14×155+8×175+4×195)/68 = (300+475+1495+2700+2170+1400+780)/68 = 9320/68 = 137.05.
Mode: modal class 125-145, Mode = 125 + [(20-13)/(40-13-14)]×20 = 125 + (7/13)×20 = 125 + 10.77 = 135.77.
Median=137, Mean≈137.05, Mode≈135.77
23 marks
If the median of the distribution given below is 28.5, find the values of x and y.
Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, Total
Frequency: 5, x, 20, 15, y, 5, 60
Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, Total
Frequency: 5, x, 20, 15, y, 5, 60
Answer:
N=60 → 5+x+20+15+y+5=60 → x+y=15.
Median=28.5 lies in 20-30 class. l=20, h=10, cf=5+x, f=20, N/2=30.
Median = 20 + [(30-(5+x))/20]×10 = 28.5 → 20 + (25-x)/2 = 28.5 → (25-x)/2 = 8.5 → 25-x = 17 → x = 8.
Then y = 15-8 = 7.
x = 8, y = 7
N=60 → 5+x+20+15+y+5=60 → x+y=15.
Median=28.5 lies in 20-30 class. l=20, h=10, cf=5+x, f=20, N/2=30.
Median = 20 + [(30-(5+x))/20]×10 = 28.5 → 20 + (25-x)/2 = 28.5 → (25-x)/2 = 8.5 → 25-x = 17 → x = 8.
Then y = 15-8 = 7.
x = 8, y = 7
33 marks
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60
Number of policy holders: 2, 6, 24, 45, 78, 89, 92, 98, 100
Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60
Number of policy holders: 2, 6, 24, 45, 78, 89, 92, 98, 100
Answer:
Convert to class intervals with cumulative frequencies:
15-20:2, 20-25:4, 25-30:18, 30-35:21, 35-40:33, 40-45:11, 45-50:3, 50-55:6, 55-60:2.
N=100, N/2=50. Cumulative frequency just greater than 50 is 78 (class 35-40).
l=35, cf=2+4+18+21=45, f=33, h=5.
Median = 35 + [(50-45)/33]×5 = 35 + (5/33)×5 = 35 + 0.7576 = 35.76 years.
Median age ≈ 35.8 years
Convert to class intervals with cumulative frequencies:
15-20:2, 20-25:4, 25-30:18, 30-35:21, 35-40:33, 40-45:11, 45-50:3, 50-55:6, 55-60:2.
N=100, N/2=50. Cumulative frequency just greater than 50 is 78 (class 35-40).
l=35, cf=2+4+18+21=45, f=33, h=5.
Median = 35 + [(50-45)/33]×5 = 35 + (5/33)×5 = 35 + 0.7576 = 35.76 years.
Median age ≈ 35.8 years
4-7Key Results
Selected answers from remaining problems in Exercise 13.3:
Q4: Median length = 146.75 cm.
Q5: Median = ₹143.75, Mean = ₹140.39, Mode = ₹137.1.
Q6: Median = ₹138.33.
Q7: Median = ₹400 (modal class 300-400).
Q5: Median = ₹143.75, Mean = ₹140.39, Mode = ₹137.1.
Q6: Median = ₹138.33.
Q7: Median = ₹400 (modal class 300-400).
Key Concepts Summary
Formula 1Mean (Direct Method)
What is the formula for mean of grouped data using direct method?
Answer: Mean = Σfixi / Σfi, where xi are class marks.
Formula 2Mode Formula
What is the formula for mode of grouped data?
Answer: Mode = l + [(f1-f0)/(2f1-f0-f2)] × h, where l = lower limit of modal class, h = class size, f1 = frequency of modal class, f0 = frequency preceding modal class, f2 = frequency succeeding modal class.
Formula 3Median Formula
What is the formula for median of grouped data?
Answer: Median = l + [(N/2 - cf)/f] × h, where l = lower limit of median class, N = total frequency, cf = cumulative frequency of class preceding median class, f = frequency of median class, h = class size.
NCERT Class 10 Mathematics | Chapter 13: Statistics | Complete solutions based on NCERT Textbook (2025-26).
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