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Chapter 13: Statistics
Exercise & In-Text Questions & Answers

1A 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 & Reference:
We'll use the direct method to find the mean.
Number of plants Number of houses (fi) Class mark (xi) fixi
0-2 1 1 1
2-4 2 3 6
4-6 1 5 5
6-8 5 7 35
8-10 6 9 54
10-12 2 11 22
12-14 3 13 39
Total Σfi = 20 Σfixi = 162
Mean (x̄) = Σfixi / Σfi = 162 / 20 = 8.1
We used the direct method because the values of fi and xi are small. (Page 181, Exercise 13.1 Q1)
2Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer & Reference:
We'll use the step-deviation method since the values are large.
Let a = 550 (assumed mean), h = 20 (class size)
Daily wages (₹) Number of workers (fi) Class mark (xi) ui = (xi-a)/h fiui
500-520 12 510 -2 -24
520-540 14 530 -1 -14
540-560 8 550 0 0
560-580 6 570 1 6
580-600 10 590 2 20
Total 50 -12
Mean (x̄) = a + (Σfiui/Σfi) × h = 550 + (-12/50) × 20 = 550 - 4.8 = ₹545.20
(Page 181, Exercise 13.1 Q2)
3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Answer & Reference:
Given mean = 18
Allowance (₹) Number of children (fi) Class mark (xi) fixi
11-13 7 12 84
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
Total 44 + f 752 + 20f
Mean = (752 + 20f)/(44 + f) = 18
752 + 20f = 18(44 + f)
752 + 20f = 792 + 18f
2f = 40
f = 20
(Page 181, Exercise 13.1 Q3)
4Thirty 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.
Answer & Reference:
We'll use the step-deviation method.
Let a = 75.5, h = 3
Heartbeats/min Number of women (fi) Class mark (xi) ui = (xi-a)/h fiui
65-68 2 66.5 -3 -6
68-71 4 69.5 -2 -8
71-74 3 72.5 -1 -3
74-77 8 75.5 0 0
77-80 7 78.5 1 7
80-83 4 81.5 2 8
83-86 2 84.5 3 6
Total 30 4
Mean (x̄) = a + (Σfiui/Σfi) × h = 75.5 + (4/30) × 3 = 75.5 + 0.4 = 75.9
(Page 182, Exercise 13.1 Q4)
5In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Answer & Reference:
We'll use the step-deviation method because the class intervals are not continuous.
First, we make the classes continuous by adding 0.5 to upper limit and subtracting 0.5 from lower limit.
New classes: 49.5-52.5, 52.5-55.5, 55.5-58.5, 58.5-61.5, 61.5-64.5
Let a = 57, h = 3
Number of mangoes Number of boxes (fi) Class mark (xi) ui = (xi-a)/h fiui
49.5-52.5 15 51 -2 -30
52.5-55.5 110 54 -1 -110
55.5-58.5 135 57 0 0
58.5-61.5 115 60 1 115
61.5-64.5 25 63 2 50
Total 400 25
Mean (x̄) = a + (Σfiui/Σfi) × h = 57 + (25/400) × 3 = 57 + 0.1875 = 57.19
We chose step-deviation method because it simplifies calculations with large frequencies. (Page 182, Exercise 13.1 Q5)
6The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer & Reference:
Finding Mode:
Modal class = 35-45 (highest 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 + 1.818 = 36.82 years
Finding Mean:
Using direct method:
Age (years) Number of patients (fi) Class mark (xi) fixi
5-15 6 10 60
15-25 11 20 220
25-35 21 30 630
35-45 23 40 920
45-55 14 50 700
55-65 5 60 300
Total 80 2830
Mean = 2830/80 = 35.375 years
Comparison: The mode (36.82) is higher than the mean (35.38). This indicates that while the most common age group is 35-45, the average age is slightly lower due to more patients in younger age groups. (Page 186, Exercise 13.2 Q1)
7The 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.
Answer & Reference:
Finding Median:
n = 68, n/2 = 34
Cumulative frequency just greater than 34 is 42, so median class = 125-145
l = 125, cf = 22, f = 20, h = 20
Median = l + [(n/2 - cf)/f] × h = 125 + [(34-22)/20] × 20 = 125 + 12 = 137 units
Finding Mean:
Using direct method:
Consumption fi xi fixi
65-85 4 75 300
85-105 5 95 475
105-125 13 115 1495
125-145 20 135 2700
145-165 14 155 2170
165-185 8 175 1400
185-205 4 195 780
Total 68 9320
Mean = 9320/68 = 137.06 units
Finding Mode:
Modal class = 125-145 (highest frequency = 20)
l = 125, h = 20, f1 = 20, f0 = 13, f2 = 14
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] × h
= 125 + [(20-13)/(40-13-14)] × 20
= 125 + (7/13) × 20 = 125 + 10.77 = 135.77 units
Comparison: Mean (137.06) ≈ Median (137) > Mode (135.77). This indicates a slightly symmetrical distribution with a small negative skew. (Page 198, Exercise 13.3 Q1)
FImportant Formulas for Statistics

Mean of Grouped Data

Direct Method: x̄ = Σfixi / Σfi (Page 172)

Assumed Mean Method: x̄ = a + Σfidi / Σfi, where di = xi - a (Page 175)

Step Deviation Method: x̄ = a + (Σfiui / Σfi) × h, where ui = (xi - a)/h (Page 177)

Mode of Grouped Data

Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] × h (Page 184)

Where l = lower limit of modal class, h = class size, f1 = frequency of modal class, f0 = frequency of class preceding modal class, f2 = frequency of class succeeding modal class

Median of Grouped Data

Median = l + [(n/2 - cf) / f] × h (Page 193)

Where l = lower limit of median class, n = total number of observations, cf = cumulative frequency of class preceding median class, f = frequency of median class, h = class size

Empirical Relationship

3 Median = Mode + 2 Mean (Page 197)

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