Statistics Questions
Statistics questions based on CBSE syllabus and NCERT guidelines are provided here for students. These questions cover the syllabus of Statistics concept for Class 9, 10 and 11. Practising these problems will help students to score better on this topic. Also, it will help them to participate in competitive exams. Also, find probability questions here at BYJU’S.
Definition: Statistics is a branch of mathematics concerned with collecting meaningful data or information.
Also, read: Statistics
| Collection of Data: The data collected to define a state or condition. |
Questions on Statistics with Answers
1. Give any two examples of collecting data from day-to-day life.
Solution:
A. Increase in population of our country in the last two decades.
B. Number of tables and chairs in a classroom
| Presentation of Data: After the collection of data, when we represent them in the form of table or chart or any other means, which help us to have a quick glance over the data, it is said to be its presentation. It also means a rearrangement of raw data in a particular order. |
2. If marks obtained by students in a class test is given as per below:
55 36 95 73 60 42 25 78 75 62
Then arrange the marks from lowest to highest.
Solution: We need to arrange the marks obtained by each student in ascending order:
25 36 42 55 60 62 73 75 78 95
3. Check the following frequency distribution table, consisting of weights of 38 students of a class:
| Weights (in kg) | Number of students |
| 31 – 35
36 – 40 41 – 45 46 – 50 51 – 55 56 – 60 61 – 65 66 – 70 71 – 75 |
9
5 14 3 1 2 2 1 1 |
(i) What is class-interval for classes 31 – 35?
(ii) How many students are there in the range of 41-45 kgs?
Solution:
(i) Class interval = Upper class limit – lower class limit
= 35-31
= 4
(ii) For the 41-45 range, there are 14 students.
| Bar Graph: The pictorial representation of data, in the form of vertical or horizontal rectangular bars.
Histogram: A set of rectangles with bases along with the intervals between class boundaries and with areas proportional to frequencies in the corresponding classes. Frequency Polygons: It is used to compare sets of data or to show a cumulative frequency distribution. It uses a line graph to represent quantitative data. Class-interval (range) = Upper class limit – lower class limit Class-mark = (Upper class limit + lower class limit)/2 |
4. A family with a monthly income of ` 20,000 had planned the following expenditures per month under various heads:
| Heads | Expenditure (in thousand rupees) |
| Grocery | 4 |
| Rent | 5 |
| Children’s Education | 5 |
| Medicine | 2 |
| Fuel | 2 |
| Entertainment | 1 |
| Miscellaneous | 1 |
Draw a bar graph for the data above.
Solution:
5. In a city, the weekly observations made in a study on the cost of the living index are given in the following table:
| Cost of living index | Number of weeks |
| 140-150 | 5 |
| 150-160 | 10 |
| 160-170 | 20 |
| 170-180 | 9 |
| 180-190 | 6 |
| 190-200 | 2 |
| Total | 52 |
Draw a frequency polygon for the data above.
Solution: Class- mark = (Upper class limit-lower class limit)/2 = (150 + 140)/2 = 290/2 = 145
Thus we can create a new table with class-mark.
| Cost of living index | Class mark | Number of weeks |
| 140-150 | 145 | 5 |
| 150-160 | 155 | 10 |
| 160-170 | 165 | 20 |
| 170-180 | 175 | 9 |
| 180-190 | 185 | 6 |
| 190-200 | 195 | 2 |
| Total | 52 |
Now with these class marks we can plot the frequency polygon as shown below.
| Measure of central tendency (Mean, Mode and Median)
Mean: The average of number of observations given. Mode: The mode is the value of the observation occuring most frequently or repeating. An observation with the maximum frequency is called the mode. Median: The median which divides the given observation into exactly two parts. |
6. Consider a small unit of a factory where there are 5 employees : a supervisor and four labourers. The workers earn a salary of Rs. 5,000 per month each while the supervisor gets Rs. 15,000 per month. Calculate the mean, median and mode of the salaries.
Solution:
Mean = (5000 + 5000 + 5000 + 5000 + 15000)/5 = 35000/5 = 7000
So, the mean salary is Rs. 7000 per month
To obtain the median, let us arrange the salaries in ascending order:
5000, 5000, 5000, 5000, 15000
Median = (n+1)/2 = (5+1)/2 = 6/2 = 3rd observation
Median = Rs. 5000/-
Mode = Number of times an observation is repeated = Rs.5000/-
| Mean of grouped data \(\bar{x}=\frac{f_{1} x_{1}+f_{2} x_{2}+\cdots+f_{n} x_{n}}{f_{1}+f_{2}+\cdots+f_{n}}\) \(\bar{x}=\frac{\sum_{i=1}^{n} f_{i} x_{i}}{\sum_{i=1}^{n} f_{i}}\) |
7. The distribution in the table below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets using the correct method. What does the mean signify?
| Number of wickets | 20-60 | 60-100 | 100-150 | 150-250 | 250-350 | 350-450 |
| Number of bowlers | 7 | 5 | 16 | 12 | 2 | 3 |
Solution: Here, the class size varies, and the class marks (xi) are large. Apply the step deviation method for a = 200 and h = 20.
| No. of wck taken | No. of bowlers | xi | di=xi-200 | ui=di/20 | uifi |
| 20-60 | 7 | 40 | -160 | -8 | -56 |
| 60-100 | 5 | 80 | -120 | -6 | -30 |
| 100-150 | 16 | 125 | -75 | -3.75 | -60 |
| 150-250 | 12 | 200 | 0 | 0 | 0 |
| 250-300 | 2 | 300 | 100 | 5 | 10 |
| 350-450 | 3 | 400 | 200 | 10 | 30 |
| Total | 45 | -106 |
So, ʉ = -106/45
x̄ = 200+20(-106/45) = 200 – 47.11 = 152.89
Hence, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89.
| Mode of grouped data \(\text { Mode }=i+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h\) |
8. A survey conducted on 20 houses in an area by a group of people resulted in the subsequent frequency table for the number of family members in a house:
| Size of family | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 |
| No. of families | 7 | 8 | 2 | 2 | 1 |
Find the mode of this data.
Solution: Here the maximum class frequency = 8,
Class corresponding to this frequency = 3 – 5.
So, the modal class = 3 – 5.
Modal class = 3 – 5, lower limit (l) of modal class = 3, class size (h) = 2
Frequency (f1) of the modal class = 8
Frequency (f0) of class preceding the modal class = 7,
Frequency (f2) of class succeeding the modal class = 2.
Now, let us put these values in the formula :
\(3+(\frac{8-7}{2\times 8-7-2})\times 2\)Therefore, the mode of the data above is 3.286.
| Median of Grouped Data
\(\text { Median }=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) \times h\)
l = lower limit of median class, n = number of observations, cf = cumulative frequency of class preceding the median class, f = frequency of median class, h = class size |
9. A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data were obtained:
| Height (in cm) | Number of girls |
| >140 | 4 |
| >145 | 11 |
| >150 | 29 |
| >155 | 40 |
| >160 | 46 |
| >165 | 51 |
Find the median height.
Solution: Observe the below table:
| Height (in cm) | Frequency | Cumulative frequency |
| >140 | 4 | 4 |
| >145 | 7 | 11 |
| >150 | 18 | 29 |
| >155 | 11 | 40 |
| >160 | 6 | 46 |
| >165 | 5 | 51 |
n = 51
n/2 = 51/2 = 25.5
This observation lies in the class 145 – 150.
l (the lower limit) = 145,
cf (the cumulative frequency of the class preceding 145 – 150) = 11
f (the frequency of the median class 145 – 150) = 18,
h (the class size) = 5
Using the formula,
\(\begin{aligned} \operatorname{Median} &=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) \times h, \\ \text { Median } &=145+\left(\frac{25.5-11}{18}\right) \times 5 \\ &=145+\frac{72.5}{18}=149.03 \end{aligned}\)So, the median height of the girls is 149.03 cm.