Q 1 – The mean of x + 2, x + 3, x + 4 and x – 2 is:
a. (x + 7)/4
b. (2x + 7)/4
c.(3x + 7)/4
d. (4x + 7)/4
d. (4x + 7)/4
Q 2 – The median of the data: 17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28 is:
a.10
b. 24
c.12
d. 8
c.12
Q 3 – Find the range of the following data: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20.
a.10
b. 15
c.18
d. 26
d. 26
Q 4 – Find the value of x, if the arithmetic mean of 4, 5, 6, 7, 8 and x is 7.
a.4
b. 6
c. 8
d. 12
d. 12
Q 5 – Find the maximum value if the range is 38 and the minimum value is 82.
a.60
b. 76
c.120
d. 82
c.120
Q 6 – In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The upper limit of the class is :
a.6
b. 7
c. 10
d. 13
b. 7
Q 7 – The width of each o five continuous classes in a frequency distribution is 5 and lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
a.15
b. 25
c.35
d. 40
c.35
Q 8 – In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:
Observe the bar graph given above and answer the following questions:
(i) How many students were born in the month of November?
(ii) In which month were the maximum number of students born?
(i) 4 students were born in the month of November.
(ii) In Maximum number of students were born in the month of August.
Q 9 – Consider the marks, out of 100, obtained by 51 students of a class in a test, given in Table
Marks | Number of Students |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 | 5 10 4 6 7 3 2 2 3 9 |
Total | 51 |
Draw a frequency polygon corresponding to this frequency distribution table.
Q 10 – There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be –3.5. What is the mean of given numbers?
Q 11 – The following table gives the distribution of students of two sections according to the marks obtained by them:
Section A | Section B | ||
Marks | Frequency | Marks | Frequency |
0 – 10 | 3 | 0 – 10 | 5 |
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Marks | Class−mark | Section A frequency | Section B frequency |
0−10 | 5 | 3 | 5 |
10−20 | 15 | 9 | 19 |
20−30 | 25 | 17 | 15 |
30−40 | 35 | 12 | 10 |
40−50 | 45 | 9 | 1 |
Since, in the graph section A is to the right of section B, more students of section A has highest marks.
Q 12 – 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
(ii) Maximum number of surnames lie in the interval 6-8
Q 14 – 5 people were asked about the time in a week they spend in doing social work in their community. They said 10, 7, 13, 20 and 15 hours, respectively. Find the mean (or average) time in a week devoted by them for social work.
Given data
10,7,13,20,15
Here n=5
∴ Mean 
∑ xi/n = (10 + 7 + 13 + 20 + 15)/5 = 65/5 = 13
Therefore, the mean time in a week devoted by them for social work is 13 hours.
Q 15 – The points scored by a Kabaddi team in a series of matches are as follows:
17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28
Find the median of the points scored by the team.
Let’s arrange the given data in descending order:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Here, n = 16 when is even
∴ Mean (x)
Ʃ xi/ n = (2 + 5 + 7 + 7 + 8 + 8 + 10 + 10 + 14 + 15 + 17 + 18 + 24 + 27 + 28 + 48)/15
= 248/16 = 15.5
And,
Median = ½ [(16/2)th term + (16/2 + 1)th term]
= ½ (8th term + 9th term)
= ½ (10 + 14)
= ½ x 24 = 12
Therefore, the mean and the median of the points scored by the Kabaddi team are 15.5 and 14 respectively.
Q 16 – Numbers 50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are written in descending order and their median is 25, find x.
50, 42, 35, 2x + 10, 2x − 8, 12, 11, 8 are written descending order
Here n = 8 which is even
Since number of observation is 6 , Median is the average of the middle data
The middle data = 2x+10, 2x−8
50 = 4x + 2
4x = 48 Hence x = 12
Q 17 – Consider a small unit of a factory where there are 5 employees : a supervisor and four labourers. The labourers draw 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 of this unit of the factory.
Mean = Rs 7000
Median = Rs 5000
Mode = Rs 5000
Salary of supervisor in rupees = 15000
Salary of 4 labourers in rupees = 5000
So, the salaries of all employees in ascending order can be represented as
5000,5000,5000,5000,15000
Mean=Sum of all observations/Total number of observations
Mean = 5000+5000+5000+5000+150005=350005=Rs 7000
Median can be obtained by finding the middle term.
Here, Middle terms are (n+12)th term that is third term.
So, Median= Rs 5000
In the given data frequency of 5000 is maximum as 4 employees has 5000 salary.
So, Mode = Rs 5000
Q 18 – The following data given the weight (in grams) of 30 oranges picked from a basket:
106 107 76 109 187 95 125 92 70 139 128 100 88 84 99 113 204 141 136 123 90
115 110 97 90 107 75 80 118 82
Construct a grouped frequency distribution table taking class width equal to 20 in such a way that the mid-value of first class in 70.
Q 19 – The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
Since,
Mean of 200 items = 50
sum of 200 items = mean of x no of items
50 × 200 =10,000
wrong sum of 200 items = 10,000
Correct sum = 10,000+192+88-92-8
= 10,280-100
=10,180
∴ correct mean = 
= 50.9
Q 20 – Find the missing frequency(p) for the following distribution whose mean is 7.68.
x | 3 | 5 | 7 | 9 | 11 | 13 |
f | 6 | 8 | 15 | p | 8 | 4 |
The given distribution in tabulated form is
We have to find the value of p using the information that the mean of the distribution is 7.68.
Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing . 
Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,
The mean is
Hence, we have
Q 21 – The following table presents the number of illiterate females in the age group (10-34) in a town.
Age group: | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 |
No. of Females | 300 | 980 | 800 | 580 | 290 |
Draw a histogram to represent the above data.