Statistics - Exercise 13.1

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Overview

Complete step-by-step NCERT solutions for CBSE Class 10 Maths Exercise 13.1 on finding the Mean of Grouped Data, with detailed tabular workings for Direct, Assumed Mean, and Step-Deviation methods, and finding missing frequencies.

NCERT Exercise Practice questions with detailed step-by-step solutions

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Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9
Question 1: Survey (Direct Method)
A survey was conducted by a group of students as part of their environment awareness programme, in which they collected the following data regarding the number of plants in 30 houses in a locality. Find the mean number of plants per house.
Number of Plants0–22–44–66–88–1010–1212–14
Number of Houses2438742
Which method did you use for finding the mean, and why?
Choice of Method
We use the Direct Method because the numerical values of both the class marks $x_i$ and the frequencies $f_i$ are small, making direct multiplication straightforward and quick.
Calculation Table
Class IntervalFrequency ($f_i$)Class Mark ($x_i$)$f_i x_i$
0–2212
2–44312
4–63515
6–88756
8–107963
10–1241144
12–1421326
Total$\sum f_i = 30$—$\sum f_i x_i = 218$
Mean Calculation
$$\text{Mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{218}{30} \approx 7.27$$
$$\therefore \text{The mean number of plants per house is } 7.27\text{ plants}.$$
Question 2: Daily Wages of Workers
Consider the following distribution of daily wages of 60 workers of a factory:
Daily Wages (in ₹)400–420420–440440–460460–480480–500
No. of Workers141610812
Find the mean daily wages of the workers of the factory by using an appropriate method.
Step-Deviation Method
Let assumed mean $a = 450$ and class width $h = 20$.
Using $u_i = \dfrac{x_i - a}{h} = \dfrac{x_i - 450}{20}$.
Calculation Table
Class Interval$f_i$$x_i$$u_i = \frac{x_i - 450}{20}$$f_i u_i$
400–42014410-2-28
420–44016430-1-16
440–4601045000
460–480847018
480–50012490224
Total$\sum f_i = 60$——$\sum f_i u_i = -12$
Step-Deviation Formula
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 450 + \left(\frac{-12}{60}\right) \times 20 = 450 - 4 = ₹446$$
$$\therefore \text{The mean daily wage of the workers is } ₹446.$$
Question 3: Missing Frequency $f$
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is $₹18$. Find the missing frequency $f$.
Daily Allowance (₹)11–1313–1515–1717–1919–2121–2323–25
Number of Children76913$f$54
Assumed Mean Method Setup
Given mean $\bar{x} = 18$. Let assumed mean $a = 18$.
Deviation $d_i = x_i - 18$.
Calculation Table
Class$f_i$$x_i$$d_i = x_i - 18$$f_i d_i$
11–13712-6-42
13–15614-4-24
15–17916-2-18
17–19131800
19–21$f$202$2f$
21–23522420
23–25424624
Total$\sum f_i = 44 + f$——$\sum f_i d_i = 2f - 40$
Solving for $f$
$$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$$ $$18 = 18 + \frac{2f - 40}{44 + f}$$ Subtract 18 from both sides: $$0 = \frac{2f - 40}{44 + f} \implies 2f - 40 = 0 \implies 2f = 40 \implies f = 20$$
$$\therefore \text{The missing frequency } f = 20.$$
Question 4: Women Heartbeats
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded as follows. Find the mean heartbeats per minute.
Heartbeats / min66–6969–7272–7575–7878–8181–8484–87
Number of Women2438742
Step-Deviation Table ($a = 76.5, h = 3$)
Class$f_i$$x_i$$u_i = \frac{x_i - 76.5}{3}$$f_i u_i$
66–69267.5-3-6
69–72470.5-2-8
72–75373.5-1-3
75–78876.500
78–81779.517
81–84482.528
84–87285.536
Total$\sum f_i = 30$——$\sum f_i u_i = 4$
Mean Calculation
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 76.5 + \left(\frac{4}{30}\right) \times 3 = 76.5 + 0.4 = 76.9$$
$$\therefore \text{The mean number of heartbeats per minute is } 76.9.$$
Question 5: Fruit Mango Boxes
In a retail market, fruit vendors were selling mangoes kept in packing boxes. The distribution of mangoes according to the number of boxes is given below:
Number of Mangoes50–5253–5556–5859–6162–64
Number of Boxes2010012010030
Find the mean number of mangoes kept in a packing box.
Class Marks & Method
The class intervals are inclusive with step width $h = 3$. The class marks $x_i = \frac{\text{Lower} + \text{Upper}}{2}$ are $51, 54, 57, 60, 63$.
Let assumed mean $a = 57, h = 3$.
Step-Deviation Table
Class$f_i$$x_i$$u_i = \frac{x_i - 57}{3}$$f_i u_i$
50–522051-2-40
53–5510054-1-100
56–581205700
59–61100601100
62–643063260
Total$\sum f_i = 370$——$\sum f_i u_i = 20$
Mean Calculation
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 57 + \left(\frac{20}{370}\right) \times 3 = 57 + \frac{60}{370} \approx 57 + 0.16 = 57.16$$
$$\therefore \text{The mean number of mangoes per packing box is approximately } 57.16.$$
Question 6: Daily Food Expenditure
The table below shows the daily expenditure on food of 30 households in a locality:
Daily Expenditure (₹)100–150150–200200–250250–300300–350
No. of Households561054
Find the mean daily expenditure on food.
Step-Deviation Table ($a = 225, h = 50$)
Class$f_i$$x_i$$u_i = \frac{x_i - 225}{50}$$f_i u_i$
100–1505125-2-10
150–2006175-1-6
200–2501022500
250–300527515
300–350432528
Total$\sum f_i = 30$——$\sum f_i u_i = -3$
Calculation
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 225 + \left(\frac{-3}{30}\right) \times 50 = 225 - 5 = ₹220$$
$$\therefore \text{The mean daily expenditure on food is } ₹220.$$
Question 7: Concentration of $SO_2$
To find out the concentration of $\text{SO}_2$ in the air (in parts per million, ppm), the data was collected for 30 localities in a certain city:
Concentration of $\text{SO}_2$ (ppm)0.00–0.040.04–0.080.08–0.120.12–0.160.16–0.200.20–0.24
Frequency499242
Find the mean concentration of $\text{SO}_2$ in the air.
Direct Method Table
Class$f_i$$x_i$$f_i x_i$
0.00–0.0440.020.08
0.04–0.0890.060.54
0.08–0.1290.100.90
0.12–0.1620.140.28
0.16–0.2040.180.72
0.20–0.2420.220.44
Total$\sum f_i = 30$—$\sum f_i x_i = 2.96$
Calculation
$$\text{Mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2.96}{30} \approx 0.0987 \approx 0.099\text{ ppm}$$
$$\therefore \text{The mean concentration of } \text{SO}_2 \text{ in the air is } 0.099\text{ ppm}.$$
Question 8: Absentee Record (Unequal Intervals)
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of Days0–66–1010–1414–2020–2828–3838–40
Number of Students111074431
Direct Method with Unequal Class Widths
Because the class widths vary ($6, 4, 4, 6, 8, 10, 2$), we apply the Direct Method directly to each class mark.
Calculation Table
Class$f_i$$x_i$$f_i x_i$
0–611333
6–1010880
10–1471284
14–2041768
20–2842496
28–3833399
38–4013939
Total$\sum f_i = 40$—$\sum f_i x_i = 499$
Calculation
$$\text{Mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{499}{40} = 12.475 \approx 12.48\text{ days}$$
$$\therefore \text{The mean number of days a student was absent is } 12.48\text{ days}.$$
Question 9: Literacy Rate in Cities
The following table gives the literacy rate (in percentage) of 40 cities. Find the mean literacy rate.
Literacy Rate (%)45–5555–6565–7575–8585–95
Number of Cities4111294
Step-Deviation Table ($a = 70, h = 10$)
Class$f_i$$x_i$$u_i = \frac{x_i - 70}{10}$$f_i u_i$
45–55450-2-8
55–651160-1-11
65–75127000
75–8598019
85–9549028
Total$\sum f_i = 40$——$\sum f_i u_i = -2$
Calculation
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 70 + \left(\frac{-2}{40}\right) \times 10 = 70 - 0.5 = 69.5\%$$
$$\therefore \text{The mean literacy rate is } 69.5\%.$$
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