CBSE Class 10 Maths Chapter-14 Statistics – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 14 – Statistics prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-14 Statistics Important Questions
CBSE Class 10 Maths Important Questions Chapter 14 – Statistics
2 Mark Questions
1. The following data gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of students:
| Age (in years) | 15 | 16 | 17 | 18 | 19 | 20 |
| No. of student | 3 | 8 | 10 | 10 | 5 | 4 |
Ans. We have
| Age (in years) (x) | No. of students (f) | fx |
| 15 | 3 | 45 |
| 16 | 8 | 128 |
| 17 | 10 | 170 |
| 18 | 10 | 180 |
| 19 | 5 | 95 |
| 20 | 4 | 80 |
 |  |
Mean 
years
2. For the following grouped frequency distribution, find the mode.
| Class | 3-6 | 6-9 | 9-12 | 12-15 | 15-18 | 18-21 | 21-24 |
| Frequency | 2 | 5 | 10 | 23 | 21 | 12 | 3 |
Ans. Since the maximum frequency = 23 and it corresponds to the class 12-15
Modal class = 12-15
3. Construct the cumulative frequency distribution of the following distribution:
| Class | 12.5-17.5 | 17.5-22.5 | 22.5-27.5 | 27.5-32.5 | 32.5-37.5 |
| Frequency | 2 | 22 | 19 | 14 | 13 |
Ans. The required cumulative frequency distribution of the given distribution is given below:
| Class | Frequency | Cumulative frequency |
| 12.5-17.5 | 2 | 2 |
| 17.5-22.5 | 22 | 24 |
| 22.5-27.5 | 19 | 43 |
| 27.5-32.5 | 14 | 57 |
| 32.5-37.5 | 13 | 70 |
4. The median and mode of a distribution are 21.2 and 21.4 respectively, find its mean.
Ans. We know that Mean = Mode + 
(Median – Mode)
5. The marks distribution of 30 students in a mathematics examination are given below
| Class Interval | 10-25 | 25-40 | 40-55 | 55-70 | 70-85 | 85-100 |
| No. of students | 2 | 3 | 7 | 6 | 0 | 6 |
Ans. Since the maximum frequency = 7 and it corresponds to the class 40-55.
The modal class= 40-55
Here,
We know that mode Mo is given by
Mo = 
Thus, Mode marks = 52
6. Find the mode of this data.
Construct the cumulative frequency distribution of following distribution:
| Marks | 39.5-49.5 | 49.5-59.5 | 59.5-69.5 | 69.5-79.5 | 79.5-89.5 | 89.5-99.5 |
| Students | 5 | 10 | 20 | 30 | 20 | 15 |
Ans. The required cumulative frequency distribution of the given distribution is given below.
| Marks | No. of Students | Cumulative Frequency |
| 39.5-49.5 | 5 | 5 |
| 49.5-59.5 | 10 | 15 |
| 59.5-69.5 | 20 | 35 |
| 69.5-79.5 | 30 | 65 |
| 79.5-89.5 | 20 | 85 |
| 89.5-99.5 | 15 | 100 |
 |
7. If the values of mean and mode are respectively 30 and 15, then median =
(a) 22.5
(b) 24.5
(c) 25
(d) 26
Ans. Median = Mode
(Mean – Mode)
8. If the mean of the following data is 18.75. find the value of P.
 | 10 | 15 | P | 25 | 30 |
 | 5 | 10 | 7 | 8 | 2 |
Ans. We have
 |  |  |
| 10 | 5 | 50 |
| 15 | 10 | 150 |
| P | 7 | 7P |
| 25 | 8 | 200 |
| 30 | 2 | 60 |
 |  |
Now mean 
= 18.75
9. Find the mean of the following data.
| Classes | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 5 | 8 | 13 | 15 | 9 |
Ans. We have
| Classes | Mid-value | Frequency |  |
| 10-20 | 15 | 5 | 75 |
| 20-30 | 25 | 8 | 200 |
| 30-40 | 35 | 13 | 455 |
| 40-50 | 45 | 15 | 675 |
| 50-60 | 55 | 9 | 495 |
 |  |
Now mean 
Hence, mean 
10. The following data gives the information observed life times (in hours) of 225 electrical components. Determine the modal life times of the components.
| Life time (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-200 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Ans. Since the maximum frequency = 61 and it corresponds to the class 60-80
Modal class = 60-80
Here,
We know that mode Mo is given by
Thus, modal life times = 65.625 hours
11. Construct the cumulative frequency distribution of the following distribution:
| Class Interval | 6.5-7.5 | 7.5-8.5 | 8.5-9.5 | 9.5-10.5 | 10.5-11.5 | 11.5-12.5 | 12.5-13.5 |
| Frequency | 5 | 12 | 25 | 48 | 32 | 6 | 1 |
Ans. The required cumulative frequency distribution of the given distribution is given below:
| Class Interval | Frequency | Cumulative Frequency |
| 6.5-7.5 | 5 | 5 |
| 7.5-8.5 | 12 | 17 |
| 8.5-9.5 | 25 | 42 |
| 9.5-10.5 | 48 | 90 |
| 10.5-11.5 | 32 | 122 |
| 11.5-12.5 | 6 | 128 |
| 12.5-13.5 | 1 | 129 |
 |
12. Calculate the median from the following data:
| Marks | 0-10 | 10-30 | 30-60 | 60-80 | 80-100 |
| No. of students | 5 | 15 | 30 | 8 | 2 |
Ans. We have
| Marks | No. of students (f) | C.F |
| 0-10 | 5 | 5 |
| 10-30 | 15 | 20 |
| 30-60 | 30 | 50 |
| 60-80 | 8 | 58 |
| 80-100 | 2 | 60 |
 |
Since 
which his in the class 30-60
Median class is 30-60
We know that median Me is given by
Here,
= 30 +10= 40
Hence, median = 40
13. Find the mean of the following data:
| Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 3 | 5 | 9 | 5 | 3 |
Ans. We have
| Classes | Mid-value  | Frequency  | |
| 0-10 | 5 | 3 | 15 |
| 10-20 | 15 | 5 | 75 |
| 20-30 | 25 | 9 | 225 |
| 30-40 | 35 | 5 | 175 |
| 40-50 | 45 | 3 | 135 |
 |  |
Now Mean 
14. A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household. Find the mode.
| Family size | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 |
| No. of families | 7 | 8 | 2 | 4 | 1 |
Ans. Since the maximum frequency = 8 and it corresponds to the class 3-5
Modal class = 3-5
Here,
We know that mode Mo is given by
15. Construct the cumulative frequency distribution of the following distribution:
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 5 | 3 | 10 | 6 | 4 | 2 |
Ans. The required cumulative frequency distribution of the given distribution is given below:
| Class Interval | Frequency (f) | Cumulative frequency |
| 0-10 | 5 | 5 |
| 10-20 | 3 | 8 |
| 20-30 | 10 | 18 |
| 30-40 | 6 | 24 |
| 40-50 | 4 | 28 |
| 50-60 | 2 | 30 |
| Total | N= 30 |
16. If the values of mean and median are 26.4 and 27.2, what will be the value of mode?
Ans. We know that
Mode = 3 median -2 mean
= 3(27.2) – 2(26.4)
= 81.6 – 52.8 = 28.8
Mode = 28.8
17. The marks obtained by 30 students of class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students.
Marks obtained  | 10 | 20 | 36 | 40 | 50 | 56 | 60 | 70 | 72 | 80 | 88 | 92 | 98 |
students  | 1 | 1 | 3 | 4 | 3 | 2 | 4 | 4 | 1 | 1 | 2 | 3 | 1 |
Ans.
| Marks obtained | No. of students | |
| 10 | 1 | 10 |
| 20 | 1 | 20 |
| 36 | 3 | 108 |
| 40 | 4 | 160 |
| 50 | 3 | 150 |
| 56 | 2 | 112 |
| 60 | 4 | 240 |
| 70 | 4 | 280 |
| 72 | 1 | 72 |
| 80 | 1 | 80 |
| 88 | 2 | 176 |
| 92 | 3 | 276 |
| 95 | 1 | 95 |
 |  |
Mean 
=
Thus, mean = 59.3
18. A student noted the numbers of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized in the table given below. Find the mode of the data.
| No. of cars | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Ans. Since the maximum frequency = 20
And it corresponds to the class 40-50
Modal class = 40-50
Here,
We know that mode M0 is given by
19. Construct the cumulative frequency distribution of the following distribution:
| consumption (units) | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 |
| Consumers | 4 | 5 | 12 | 20 | 14 | 8 |
Ans. The required accumulative frequency distribution of the given distribution is given below.
| Monthly consumption (in units) | No. of consumes | Cumulative frequency  |
| 65-85 | 4 | 4 |
| 85-105 | 5 | 9 |
| 105-125 | 13 | 22 |
| 125-145 | 20 | 42 |
| 145-165 | 14 | 56 |
| 165-185 | 8 | 64 |
| N = 64 |
20. If the values of mean and median are 53.6 and 55.81, what will be the value of mode?
Ans. We know that
Mode = 3 Median – 2 mean
Mean = 
