Mathematics solution NCERT
Class 10 - Chapter 13: Statistics
Exercise 13.1
Question 1
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. Which method did you use for finding the mean, and why?
Solution:
The class intervals are of equal width, so we use the Direct Method to find the mean.
First, find the class marks (x) of each class.
| Class Interval | Frequency (f) | Class Mark (x) | f × x |
|---|---|---|---|
| 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 | 20 | — | 162 |
Using the formula of the Direct Method,
$$ \bar{x}=\frac{\sum fx}{\sum f} $$Substitute the values.
$$ \bar{x}=\frac{162}{20} $$ $$ \bar{x}=8.1 $$Answer:
$$ \boxed{\text{Mean number of plants per house}=8.1} $$Method Used
Direct Method is used because:
- The class intervals are equal.
- The class marks are small and easy to calculate.
- The values of f × x are simple, making the calculations straightforward.
Question 2
Consider 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.
Solution:
Since the class intervals are equal and the class marks are large, we use the Assumed Mean Method.
Take the assumed mean
$$ A=550 $$and class width
$$ h=20. $$Compute
$$ u_i=\frac{x_i-A}{h}. $$| Daily Wages (₹) | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 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 |
Using the Assumed Mean Method,
$$ \bar{x}=A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the values.
$$ \bar{x} = 550 + 20\left(\frac{-12}{50}\right) $$ $$ = 550-4.8 $$ $$ = 545.2 $$Answer:
$$ \boxed{\text{Mean daily wages}=₹545.2} $$Why is the Assumed Mean Method used?
- The class intervals are equal.
- The class marks are comparatively large.
- Taking an assumed mean makes the calculations shorter and easier.
Question 3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.
Solution:
Since the class intervals are equal, we use the Assumed Mean Method.
Take the assumed mean
$$ A=18 $$and class width
$$ h=2. $$The class marks are
$$ 12,\;14,\;16,\;18,\;20,\;22,\;24. $$| Daily Pocket Allowance (₹) | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 11–13 | 7 | 12 | -3 | -21 |
| 13–15 | 6 | 14 | -2 | -12 |
| 15–17 | 9 | 16 | -1 | -9 |
| 17–19 | 13 | 18 | 0 | 0 |
| 19–21 | f | 20 | 1 | f |
| 21–23 | 5 | 22 | 2 | 10 |
| 23–25 | 4 | 24 | 3 | 12 |
| Total | $$ 44+f $$ | — | — | $$ f-20 $$ |
Using the formula of the Assumed Mean Method,
$$ \bar{x}=A+h\left(\frac{\sum fu}{\sum f}\right) $$Since the mean is ₹18,
$$ 18=18+2\left(\frac{f-20}{44+f}\right) $$Subtract 18 from both sides.
$$ 0=2\left(\frac{f-20}{44+f}\right) $$ $$ \frac{f-20}{44+f}=0 $$A fraction is zero only when its numerator is zero.
$$ f-20=0 $$ $$ f=20 $$Verification:
$$ \sum f=44+20=64 $$ $$ \sum fu=20-20=0 $$ $$ \bar{x}=18+2\left(\frac{0}{64}\right)=18 $$Hence, the given mean is satisfied.
Answer:
$$ \boxed{f=20} $$Question 4
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.
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=75.5, $$class width
$$ h=3. $$The class marks are
$$ 66.5,\;69.5,\;72.5,\;75.5,\;78.5,\;81.5,\;84.5. $$| Heartbeats per Minute | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 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 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the given values.
$$ \bar{x} = 75.5 + 3\left(\frac{4}{30}\right) $$ $$ = 75.5 + 0.4 $$ $$ = 75.9 $$Answer:
$$ \boxed{\text{Mean heartbeats per minute}=75.9} $$Method Used
Step-Deviation Method is used because:
- All the class intervals are equal.
- The class width is constant.
- The calculations become simpler by using coded deviations instead of large class marks.
Question 5
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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?
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=57, $$and class width
$$ h=3. $$The class marks are
$$ 51,\;54,\;57,\;60,\;63. $$| Number of Mangoes | Number of Boxes (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 50–52 | 15 | 51 | -2 | -30 |
| 53–55 | 110 | 54 | -1 | -110 |
| 56–58 | 135 | 57 | 0 | 0 |
| 59–61 | 115 | 60 | 1 | 115 |
| 62–64 | 25 | 63 | 2 | 50 |
| Total | 400 | — | — | 25 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the given values.
$$ \bar{x} = 57 + 3\left(\frac{25}{400}\right) $$ $$ = 57 + 0.1875 $$ $$ = 57.1875 $$Therefore,
$$ \bar{x}\approx57.19 $$Answer:
$$ \boxed{\text{Mean number of mangoes per packing box}\approx57.19} $$Method Used
Step-Deviation Method is used because:
- The class intervals are equal.
- The frequencies are large.
- Using coded deviations makes the calculations much shorter and easier.
Question 6
The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=225, $$and class width
$$ h=50. $$The class marks are
$$ 125,\;175,\;225,\;275,\;325. $$| Daily Expenditure (₹) | Number of Households (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 100–150 | 4 | 125 | -2 | -8 |
| 150–200 | 5 | 175 | -1 | -5 |
| 200–250 | 12 | 225 | 0 | 0 |
| 250–300 | 2 | 275 | 1 | 2 |
| 300–350 | 2 | 325 | 2 | 4 |
| Total | 25 | — | — | -7 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the given values.
$$ \bar{x} = 225 + 50\left(\frac{-7}{25}\right) $$ $$ = 225-14 $$ $$ = 211 $$Answer:
$$ \boxed{\text{Mean daily expenditure on food}=₹211} $$Method Used
Step-Deviation Method is used because:
- The class intervals are equal.
- The class marks are comparatively large.
- Using coded deviations simplifies the calculations and reduces arithmetic work.
Question 7
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city as given below. Find the mean concentration of SO2 in the air.
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=0.10, $$and class width
$$ h=0.04. $$The class marks are
$$ 0.02,\;0.06,\;0.10,\;0.14,\;0.18,\;0.22. $$| Concentration of SO2 (ppm) | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 0.00–0.04 | 4 | 0.02 | -2 | -8 |
| 0.04–0.08 | 9 | 0.06 | -1 | -9 |
| 0.08–0.12 | 9 | 0.10 | 0 | 0 |
| 0.12–0.16 | 2 | 0.14 | 1 | 2 |
| 0.16–0.20 | 4 | 0.18 | 2 | 8 |
| 0.20–0.24 | 2 | 0.22 | 3 | 6 |
| Total | 30 | — | — | -1 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the given values.
$$ \bar{x} = 0.10 + 0.04\left(\frac{-1}{30}\right) $$ $$ = 0.10-0.00133 $$ $$ = 0.09867 $$Therefore,
$$ \boxed{\bar{x}\approx0.099\text{ ppm}} $$Method Used
Step-Deviation Method is used because:
- The class intervals are equal.
- The class width is constant.
- The values are decimal numbers, and using coded deviations makes the calculations easier and more accurate.
Question 8
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.
Solution:
Since the class intervals are of unequal widths, we use the Direct Method.
First, find the class marks (x) of each class.
| Number of Days | Frequency (f) | Class Mark (x) | f × x |
|---|---|---|---|
| 0–6 | 11 | 3 | 33 |
| 6–10 | 10 | 8 | 80 |
| 10–14 | 7 | 12 | 84 |
| 14–20 | 4 | 17 | 68 |
| 20–28 | 4 | 24 | 96 |
| 28–38 | 3 | 33 | 99 |
| 38–40 | 1 | 39 | 39 |
| Total | 40 | — | 499 |
Using the formula of the Direct Method,
$$ \bar{x}=\frac{\sum fx}{\sum f} $$Substitute the values.
$$ \bar{x}=\frac{499}{40} $$ $$ \bar{x}=12.475 $$Therefore,
$$ \boxed{\bar{x}\approx12.48} $$Answer:
$$ \boxed{\text{Mean number of days absent}=12.48\text{ days}} $$Method Used
- The class intervals are not of equal width.
- Therefore, the Direct Method is the most suitable method.
- The class marks are easy to calculate, making the computations straightforward.
Question 9
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=70, $$and class width
$$ h=10. $$The class marks are
$$ 50,\;60,\;70,\;80,\;90. $$| Literacy Rate (%) | Number of Cities (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 45–55 | 3 | 50 | -2 | -6 |
| 55–65 | 10 | 60 | -1 | -10 |
| 65–75 | 11 | 70 | 0 | 0 |
| 75–85 | 8 | 80 | 1 | 8 |
| 85–95 | 3 | 90 | 2 | 6 |
| Total | 35 | — | — | -2 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the given values.
$$ \bar{x} = 70 + 10\left(\frac{-2}{35}\right) $$ $$ = 70-\frac{20}{35} $$ $$ = 70-0.5714 $$ $$ = 69.4286 $$Therefore,
$$ \boxed{\bar{x}\approx69.43\%} $$Answer:
$$ \boxed{\text{Mean literacy rate}=69.43\%} $$Method Used
- All class intervals are equal.
- The class width is constant.
- The Step-Deviation Method simplifies the calculations by using coded deviations.