NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 provide step-by-step answers to problems related to data handling and representation. This exercise focuses on calculating measures of central tendency, including mean, median, and mode, for grouped and ungrouped data, as covered in the CBSE Class 10 Maths syllabus
These NCERT solutions explain concepts clearly with step-by-step methods, enhancing understanding and problem-solving skills. These solutions will help you interpret and analyse data effectively.
| 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?
Solution:
To find the mean value, we will use the direct method because the numerical value of f i and x i are small. Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2
|
No. of plants (Class interval) |
No. of houses Frequency (f i ) |
Mid-point (x i ) |
f i x i |
|
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 f i = 20 |
Sum f i x i = 162 |
The formula to find the mean is: Mean = x̄ = ∑f i x i /∑f i = 162/20 = 8.1. Therefore, the mean number of plants per house is 8.1.
|
Daily wages (in Rs.) |
500-520 |
520-540 |
540-560 |
560-580 |
580-600 |
|
Number of workers |
12 |
14 |
8 |
6 |
10 |
Find the mean daily wages of the workers of the factory by using an appropriate method .
Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 550. Class interval (h) = 20 So, u i = (x i – a)/h u i = (x i – 550)/20 Substitute and find the values as follows:
|
Daily wages (Class interval) |
Number of workers frequency (f i ) |
Mid-point (x i ) |
u i = (x i – 550)/20 |
f i u i |
|
500-520 |
12 |
510 |
-2 |
-24 |
|
520-540 |
14 |
530 |
-1 |
-14 |
|
540-560 |
8 |
550 = a |
0 |
0 |
|
560-580 |
6 |
570 |
1 |
6 |
|
580-600 |
10 |
590 |
2 |
20 |
|
Total |
Sum f i = 50 |
Sum f i u i = -12 |
So, the formula to find out the mean is: Mean = x̄ = a + h(∑f i u i /∑f i ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20 Thus, mean daily wage of the workers = Rs. 545.20
|
Daily Pocket Allowance(in c) |
11-13 |
13-15 |
15-17 |
17-19 |
19-21 |
21-23 |
23-35 |
|
Number of children |
7 |
6 |
9 |
13 |
f |
5 |
4 |
Solution: To find out the missing frequency, use the mean formula. Given, mean x̄ = 18
|
Class interval |
Number of children (f i ) |
Mid-point (x i ) |
f i x i |
|
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 |
f i = 44+f |
Sum f i x i = 752+20f |
The mean formula is Mean = x̄ = ∑f i x i /∑f i = (752 + 20f)/ (44 + f)
Now substitute the values and equate to find the missing frequency (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 So, the missing frequency, f = 20.
|
Number of heart beats per minute |
65-68 |
68-71 |
71-74 |
74-77 |
77-80 |
80-83 |
83-86 |
|
Number of women |
2 |
4 |
3 |
8 |
7 |
4 |
2 |
Solution: From the given data, let us assume the mean as a = 75.5 x i = (Upper limit + Lower limit)/2 Class size (h) = 3. Now, find the u i and f i u i as follows:
|
Class Interval |
Number of women (f i ) |
Mid-point (x i ) |
u i = (x i – 75.5)/h |
f i u i |
|
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 = a |
0 |
0 |
|
77-80 |
7 |
78.5 |
1 |
7 |
|
80-83 |
4 |
81.5 |
2 |
8 |
|
83-86 |
2 |
84.5 |
3 |
6 |
|
Sum f i = 30 |
Sum f i u i = 4 |
Mean = x̄ = a + h(∑f i u i /∑f i ) = 75.5 + 3 × (4/30) = 75.5 + (4/10) = 75.5 + 0.4 = 75.9 Therefore, the mean heart beats per minute for these women is 75.9
|
Number of mangoes |
50-52 |
53-55 |
56-58 |
59-61 |
62-64 |
|
Number of boxes |
15 |
110 |
135 |
115 |
25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution: The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1. Here, assumed mean (a) = 57 Class size (h) = 3 Here, the step deviation is used because the frequency values are big.
|
Class Interval |
Number of boxes (f i ) |
Mid-point (x i ) |
u i = (x i – 57)/h |
f i u i |
|
49.5-52.5 |
15 |
51 |
-2 |
-30 |
|
52.5-55.5 |
110 |
54 |
-1 |
-110 |
|
55.5-58.5 |
135 |
57 = a |
0 |
0 |
|
58.5-61.5 |
115 |
60 |
1 |
115 |
|
61.5-64.5 |
25 |
63 |
2 |
50 |
|
Sum f i = 400 |
Sum f i u i = 25 |
The formula to find out the Mean is: Mean = x̄ = a + h(∑f i u i /∑f i ) = 57 + 3(25/400) = 57 + 0.1875 = 57.19 Therefore, the mean number of mangoes kept in a packing box is 57.19
|
Daily expenditure(in c) |
100-150 |
150-200 |
200-250 |
250-300 |
300-350 |
|
Number of households |
4 |
5 |
12 |
2 |
2 |
Find the mean daily expenditure on food by a suitable method.
Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2 Let us assume the mean (a) = 225 Class size (h) = 50
|
Class Interval |
Number of households (f i ) |
Mid-point (x i ) |
d i = x i – A |
u i = d i /50 |
f i u i |
|
100-150 |
4 |
125 |
-100 |
-2 |
-8 |
|
150-200 |
5 |
175 |
-50 |
-1 |
-5 |
|
200-250 |
12 |
225 = a |
0 |
0 |
0 |
|
250-300 |
2 |
275 |
50 |
1 |
2 |
|
300-350 |
2 |
325 |
100 |
2 |
4 |
|
Sum f i = 25 |
Sum f i u i = -7 |
Mean = x̄ = a + h(∑f i u i /∑f i ) = 225 + 50(-7/25) = 225 – 14 = 211 Therefore, the mean daily expenditure on food is 211.
|
Concentration of SO 2 ( in ppm) |
Frequency |
|
0.00 – 0.04 |
4 |
|
0.04 – 0.08 |
9 |
|
0.08 – 0.12 |
9 |
|
0.12 – 0.16 |
2 |
|
0.16 – 0.20 |
4 |
|
0.20 – 0.24 |
2 |
Find the mean concentration of SO 2 in the air.
Solution: To find out the mean, first find the midpoint of the given frequencies as follows:
|
Concentration of SO 2 (in ppm) |
Frequency (f i ) |
Mid-point (x i ) |
f i x i |
|
0.00-0.04 |
4 |
0.02 |
0.08 |
|
0.04-0.08 |
9 |
0.06 |
0.54 |
|
0.08-0.12 |
9 |
0.10 |
0.90 |
|
0.12-0.16 |
2 |
0.14 |
0.28 |
|
0.16-0.20 |
4 |
0.18 |
0.72 |
|
0.20-0.24 |
2 |
0.22 |
0.44 |
|
Total |
Sum f i = 30 |
Sum (f i x i ) = 2.96 |
The formula to find out the mean is Mean = x̄ = ∑f i x i /∑f i = 2.96/30 = 0.099 ppm Therefore, the mean concentration of SO 2 in the air is 0.099 ppm.
|
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 |
Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2
|
Class interval |
Frequency (f i ) |
Mid-point (x i ) |
f i x i |
|
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 |
|
Sum f i = 40 |
Sum f i x i = 499 |
The mean formula is, Mean = x̄ = ∑f i x i /∑f i = 499/40 = 12.48 days Therefore, the mean number of days a student was absent = 12.48.
|
Literacy rate (in %) |
45-55 |
55-65 |
65-75 |
75-85 |
85-98 |
|
Number of cities |
3 |
10 |
11 |
8 |
3 |
Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 70. Class interval (h) = 10 So, u i = (x i – a)/h u i = (x i – 70)/10 Substitute and find the values as follows:
|
Class Interval |
Frequency (f i ) |
(x i ) |
u i = (x i – 70)/10 |
f i u i |
|
45-55 |
3 |
50 |
-2 |
-6 |
|
55-65 |
10 |
60 |
-1 |
-10 |
|
65-75 |
11 |
70 = a |
0 |
0 |
|
75-85 |
8 |
80 |
1 |
8 |
|
85-95 |
3 |
90 |
2 |
6 |
|
Sum f i = 35 |
Sum f i u i = -2 |
So, Mean = x̄ = a + (∑f i u i /∑f i ) × h = 70 + (-2/35) × 10 = 69.43 Therefore, the mean literacy part = 69.43%
Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:
Build Strong Concepts:
Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.
Work on Weak Areas:
Focus on difficult topics from the CBSE Class 10 Maths syllabus instead of skipping them to avoid losing marks.
Revise Formulas Daily:
Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.
Practise Regularly:
Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.
Solve Previous Year Papers:
Practising class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.