NCERT Solutions for Class 10 Maths Chapter 14: Statistics
Class 10 Maths Chapter 14 deals with one of the most important topics included in the Class 10 curriculum, which imparts the basics of concepts in Data Interpretation and Analysis. This chapter caters to ways of summarizing and analyzing data, amongst them, measures of central tendency like mean, median, and mode. It focuses on applying these concepts to real-life problems to enable students to apply statistical measures to practical problems.
Download PDF For NCERT Solutions for Maths Statistics
The NCERT Solutions for Class 10 Maths Chapter 14: Statistics are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 10 Maths Chapter 14: Statistics
Students can access the NCERT Solutions for Class 10 Maths Chapter 14: Statistics. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 14.1
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.
|
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?
|
No. of plants (Class interval) |
No. of houses (fi) |
Mid-point (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 |
|
|
Sum fi = 20 |
|
Sum fixi = 162 |
Mean = x̄ = ∑fixi /∑fi = 162/20 = 8.1
We would use a direct method because the numerical value of fi and xi are small.
Consider the following distribution of daily wages of 50 workers of a factory.
|
Daily wages (in Rs.) |
100-120 |
120-140 |
140-160 |
160-180 |
180-200 |
|
Number of workers |
12 |
14 |
8 |
6 |
10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui = (xi - A)/h = ui = (xi - 150)/20
|
Daily wages (Class interval) |
Number of workers frequency (fi) |
Mid-point (xi) |
ui = (xi - 150)/20 |
fiui |
|
100-120 |
12 |
110 |
-2 |
-24 |
|
120-140 |
14 |
130 |
-1 |
-14 |
|
140-160 |
8 |
150 |
0 |
0 |
|
160-180 |
6 |
170 |
1 |
6 |
|
180-200 |
10 |
190 |
2 |
20 |
|
Total |
Sum fi = 50 |
|
|
Sum fiui = -12 |
Mean = x̄ = A + h∑fiui /∑fi =150 + (20 × -12/50) = 150 - 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Here, the value of mid-point (xi) mean x̄ = 18
|
Class interval |
Number of children (fi) |
Mid-point (xi) |
fixi |
|
11-13 |
7 |
12 |
84 |
|
13-15 |
6 |
14 |
84 |
|
15-17 |
9 |
16 |
144 |
|
17-19 |
13 |
18 = A |
234 |
|
19-21 |
f |
20 |
20f |
|
21-23 |
5 |
22 |
110 |
|
23-25 |
4 |
24 |
96 |
|
Total |
fi = 44+f |
|
Sum fixi = 752+20f |
Mean = x̄ = ∑fixi /∑fi = (752+20f)/(44+f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 - 752 = 20f - 18f
⇒ 40 = 2f
⇒ f = 20
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows:
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5
|
Class Interval |
Number of women (fi) |
Mid-point (xi) |
ui = (xi - 75.5)/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 |
3 |
8 |
|
83-86 |
2 |
84.5 |
3 |
6 |
|
|
Sum fi= 30 |
|
|
Sum fiui = 4 |
Mean = x̄ = A + h∑fiui /∑fi = 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number 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?
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
|
Class Interval |
Number of boxes (fi) |
Mid-point (xi) |
di = xi - A |
fidi |
|
49.5-52.5 |
15 |
51 |
-6 |
90 |
|
52.5-55.5 |
110 |
54 |
-3 |
-330 |
|
55.5-58.5 |
135 |
57 = A |
0 |
0 |
|
58.5-61.5 |
115 |
60 |
3 |
345 |
|
61.5-64.5 |
25 |
63 |
6 |
150 |
|
|
Sum fi = 400 |
|
|
Sum fidi = 75 |
Mean = x̄ = A + ∑fidi /∑fi = 57 + (75/400) = 57 + 0.1875 = 57.19
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.
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
Let is assume the mean (A) = 225
Class size (h) = 50
= 225+50(-7/25)
= 225-14
= 211
Therefore, the mean daily expenditure on food is 211
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 and is presented below:
Find the mean concentration of SO2 in the air.
The formula to find out the mean is
Mean = x̄ = ∑fixi /∑fi
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 in air is 0.099 ppm.
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 days |
0-6 |
6-10 |
10-14 |
14-20 |
20-28 |
28-38 |
38-40 |
|
Number of students |
11 |
10 |
7 |
4 |
4 |
3 |
1 |
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
The mean formula is,
Mean = x̄ = ∑fixi /∑fi
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
|
Literacy rate (in %) |
45-55 |
55-65 |
65-75 |
75-85 |
85-98 |
|
Number of cities |
3 |
10 |
11 |
8 |
3 |
|
Class Interval |
Frequency (fi) |
(xi) |
di = xi - a |
ui = di/h |
fiui |
|
45-55 |
3 |
50 |
-20 |
-2 |
-6 |
|
55-65 |
10 |
60 |
-10 |
-1 |
-10 |
|
65-75 |
11 |
70 |
0 |
0 |
0 |
|
75-85 |
8 |
80 |
10 |
1 |
8 |
|
85-95 |
3 |
90 |
20 |
2 |
6 |
|
|
Sum fi = 35 |
|
|
|
Sum fiui = -2 |
Mean = x̄ = a + (∑fiui /∑fi) х h
= 70 + (-2/35) х 10 = 69.42
Exercise 14.2
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 the mean of the data given above. Compare and interpret the two
measures of central tendency.
To find out the modal class, let us the consider the class interval with high frequency
Here, the greatest frequency = 23, so the modal class = 35 – 45,
l = 35,
class width (h) = 10,
fm = 23,
f1 = 21 and f2 = 14
The formula to find the mode is
Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h
Substitute the values in the formula, we get
Mode = 35+[(23-21)/(46-21-14)]×10
Mode = 35+(20/11) = 35+1.8
Mode = 36.8 year
So the mode of the given data = 36.8 year
Calculation of Mean:
First find the midpoint using the formula, xi = (upper limit +lower limit)/2
The mean formula is
Mean = x̄ = ∑fixi /∑fi
= 2830/80
= 35.37 years
Therefore, the mean of the given data = 35.37 years.
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.
From the given data the modal class is 60–80.
l = 60,
The frequencies are:
fm = 61, f1 = 52, f2 = 38 and h = 20
The formula to find the mode is
Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h
Substitute the values in the formula, we get
Mode =60+[(61-52)/(122-52-38)]×20
Mode = 60+((9 x 20)/32)
Mode = 60+(45/8) = 60+ 5.625
Therefore, modal lifetime of the components = 65.625 hours.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :
|
Expenditure |
Number of families |
|
1000-1500 |
24 |
|
1500-2000 |
40 |
|
2000-2500 |
33 |
|
2500-3000 |
28 |
|
3000-3500 |
30 |
|
3500-4000 |
22 |
|
4000-4500 |
16 |
|
4500-5000 |
7 |
For Mode:
Given data:
Modal class = 1500-2000,
l = 1500,
Frequencies:
fm = 40 f1 = 24, f2 = 33 and
h = 500
Mode formula:
Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h
Substitute the values in the formula, we get
Mode =1500+[(40-24)/(80-24-33)]×500
Mode = 1500+((16×500)/23)
Mode = 1500+(8000/23) = 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83
Calculation for mean:
First find the midpoint using the formula, xi =(upper limit +lower limit)/2
Let us assume a mean, A be 2750
The formula to calculate the mean,
Mean = x̄ = a +(∑fiui /∑fi)×h
Substitute the values in the given formula
= 2750+(-35/200)×500
= 2750-87.50
= 2662.50
So, the mean monthly expenditure of the families = Rupees 2662.50
Exercise 14.3
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.
If the median of the distribution given below is 28.5, then find the values of x and y.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter and data obtained is represented in the following table. Find the median length of the leaves.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day cricket matches:
Find mode of the data.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below:
Find the mode of the data.
Exercise 14.4
The following distribution gives the daily income of 50 workers of a factory:
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Now, by drawing the points on the graph,
i.e., (120, 12); (140, 26); 160, 34); (180, 40); (200, 50)
During the medical checkup of 35 students of a class, their weights were recorded as follows:
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Hence, the points for graph are:
(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)
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.
The points for the graph are:
(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16)
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