NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 14.1 — Mean of Grouped Data
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.
Answer
Classes: 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14. Frequencies: 1, 2, 1, 5, 6, 2, 3.
Ans. Using direct method: Σfᵢxᵢ/Σfᵢ
xᵢ (mid-values): 1, 3, 5, 7, 9, 11, 13. Σfᵢxᵢ = 1+6+5+35+54+22+39 = 162. Σfᵢ = 20.
Mean = 162/20 = 8.1 plants
Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages.
Answer
500-520: 12, 520-540: 14, 540-560: 8, 560-580: 6, 580-600: 10.
Ans. Using assumed mean method with a = 550:
dᵢ = xᵢ – 550: –40, –20, 0, 20, 40. fᵢdᵢ: –480, –280, 0, 120, 400. Σfᵢdᵢ = –240.
Mean = 550 + (–240/50) = 550 – 4.8 = ₹545.20
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.
Answer
11-13: 7, 13-15: 6, 15-17: 9, 17-19: 13, 19-21: f, 21-23: 5, 23-25: 4.
Ans. Using assumed mean a = 18: Σfᵢ = 44 + f. Σfᵢdᵢ = –2f + ...
After calculation: Mean = 18
⇒ –8 + 2f = 0... wait: let me compute properly.
xᵢ: 12,14,16,18,20,22,24. dᵢ: –6,–4,–2,0,2,4,6.
fᵢdᵢ: –42,–24,–18,0,2f,20,24. Σfᵢdᵢ = –40 + 2f.
18 = 18 + (–40+2f)/(44+f)
⇒ 0 = (–40+2f)/(44+f)
⇒ 2f = 40
⇒ f = 20
Exercise 14.2 — Mode of Grouped Data
The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data.
Answer
5-15: 6, 15-25: 11, 25-35: 21, 35-45: 23, 45-55: 14, 55-65: 5.
Ans. Modal class = 35-45 (highest frequency = 23). l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10.
Mode = l + [(f₁–f₀)/(2f₁–f₀–f₂)] × h = 35 + [(23–21)/(46–21–14)] × 10 = 35 + (2/11) × 10 = 35 + 1.82 = 36.82 years
Mean (using assumed mean a = 40): Σfᵢdᵢ = –370. Mean = 40 + (–370/80) = 40 – 4.625 = 35.375 years
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table. Find the mode of the data.
Answer
0-10: 7, 10-20: 14, 20-30: 13, 30-40: 12, 40-50: 20, 50-60: 11, 60-70: 15, 70-80: 8.
Ans. Modal class = 40-50 (f = 20). l = 40, f₁ = 20, f₀ = 12, f₂ = 11, h = 10.
Mode = 40 + [(20–12)/(40–12–11)] × 10 = 40 + (8/17) × 10 = 40 + 4.71 = 44.71 cars
Exercise 14.3 — Median of Grouped Data
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode.
Answer
65-85: 4, 85-105: 5, 105-125: 13, 125-145: 20, 145-165: 14, 165-185: 8, 185-205: 4.
Ans. N = 68, N/2 = 34. CF: 4, 9, 22, 42, 56, 64, 68. Median class = 125-145 (CF first exceeds 34).
l = 125, CF = 22, f = 20, h = 20.
Median = 125 + [(34–22)/20] × 20 = 125 + 12 = 137 units
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age.
Answer
Below 20: 2, Below 25: 6, Below 30: 24, Below 35: 45, Below 40: 78, Below 45: 89, Below 50: 92, Below 55: 98, Below 60: 100.
Ans. Convert to class intervals. N = 100, N/2 = 50. CF: 2,6,24,45,78... Median class = 35-40 (CF crosses 50).
l = 35, CF = 45, f = 33, h = 5.
Median = 35 + [(50–45)/33] × 5 = 35 + 25/33 = 35 + 0.76 = 35.76 years
Exercise 14.4 — Ogive (Cumulative Frequency Curve)
The following distribution gives the daily income of 50 workers of a factory. Convert it into a less than type cumulative frequency distribution and draw its ogive.
Answer
100-120: 12, 120-140: 14, 140-160: 8, 160-180: 6, 180-200: 10.
Ans. Less than type CF:
Less than 120: 12, Less than 140: 26, Less than 160: 34, Less than 180: 40, Less than 200: 50.
Plot points: (120,12), (140,26), (160,34), (180,40), (200,50). Join by smooth curve.
Median from ogive: N/2 = 25. Draw horizontal line at CF = 25, meets ogive at approximately ₹138.57.
The following table gives production yield per hectare of wheat of 100 farms of a village. Change the distribution to a more than type distribution, and draw its ogive.
Answer
50-55: 2, 55-60: 8, 60-65: 12, 65-70: 24, 70-75: 38, 75-80: 16.
Ans. More than type CF:
More than 50: 100, More than 55: 98, More than 60: 90, More than 65: 78, More than 70: 54, More than 75: 16.
Plot and join. Median: N/2 = 50
⇒ approximately 70.83 kg/hectare
Important Extra Questions
Write the formulas for Mean, Mode and Median of grouped data.
Answer
Mean (Direct): x̄ = Σfᵢxᵢ / Σfᵢ
Mean (Assumed mean): x̄ = a + Σfᵢdᵢ / Σfᵢ (where dᵢ = xᵢ – a)
Mean (Step deviation): x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h (where uᵢ = (xᵢ–a)/h)
Mode: Mode = l + [(f₁–f₀)/(2f₁–f₀–f₂)] × h
where l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class before, f₂ = frequency of class after, h = class size
Median: Median = l + [(N/2 – CF)/f] × h
where l = lower limit of median class, CF = cumulative frequency of class before median class, f = frequency of median class
Empirical relationship: 3 Median = Mode + 2 Mean
What is an ogive? How do you find the median from an ogive?
Answer
An ogive (cumulative frequency curve) is a graph plotted using cumulative frequencies against upper/lower class boundaries. There are two types:
Less than ogive: Plot cumulative frequency against upper class boundaries (rising curve).
More than ogive: Plot cumulative frequency against lower class boundaries (falling curve).
Finding median: (i) From one ogive: locate N/2 on y-axis, draw horizontal line to curve, then vertical to x-axis — that x-value is the median. (ii) From both ogives: the x-coordinate of their point of intersection is the median.