Given below are the Class 10 Maths Important Questions for Statistics
a) Concepts questionsFill in the blank
(a) The mean of 50 numbers is 18, the new mean will be …. if each observation is increased by 4 ( 22/24/20)
(b) The sum of deviations of a set of values {a,b,c,d,e,f,g.h.i……} n items measured from 26 is 10 and the sum of deviations of the values from 20 is 50. The value of n is …(10/12/9) And mean of the items is ……. ( 19/18/17/25)
(c) For a given data with 110 observations the ‘less than ogive’ and the ‘more then ogive’ intersect at (18, 20). The median of the data is …… (18/20/19)
(d) The curve drawn by taking upper limits along xaxis and cumulative frequency along yaxis is ……….( less than ogive /more than ogive)
(e) The mean of five numbers is 40. If one number is excluded, their mean becomes
28. The excluded number is……. (68 / 88)
(f) The modal class of the grouped size frequency table given below is
55.2 
5.25.4 
5.45.6 
5.65.8 
5.86.0 
34 
4 
4 
4 
6 
…. ……
Solution
(a) 24
So M=22
(b) 10,25
Subtracting ,we get
6n=60 => n=10
So $\sum x=250$
(c) 18. Median is the point of intersection of ogive curves
(d) less than ogive curve
(e) 88
Sum of five number=5Xmean=200
Sum of four number=4Xmean=112
Subtracting, we get the number=88
f) Modal class is 55.2
The following shows the class interval and respective frequency
Class interval 
515 
1525 
2535 
3545 
4555 
5565 
Frequency 
6 
11 
21 
23 
14 
4 
a) The mean is 33
b) The modal class is 3545
c) The mode is 34
d) The Frequency for less than is 35 is 38
e) The median is
Solution
Lets find the mean value first
Class interval 
515 
1525 
2535 
3545 
4555 
5565 
Frequency 
6 
11 
21 
23 
14 
5 
Class mark 
10 
20 
30 
40 
50 
60 
F_{i}x_{i} 
60 
220 
630 
920 
700 
300 
Mean is given by
The class having highest frequency is called the Modal class, so Modal class is 3545
Where
l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f_{1} = frequency of the modal class,
f_{0} = frequency of the class preceding the modal class,
f_{2} = frequency of the class succeeding the modal class.
Now l=35 h=10 f_{1}=23 f_{0}=21 and f_{2}=14
So substituting all the values we get
M_{0}=36.81
Now
3 Median=Mode +2 Mean
So median =Mode +2 Mean)/3=35.853
a) False. I
b) True
c) False
d) True.
e) False
3) While computing mean of grouped data, we assume that the frequencies are
a) centred at the upper limits of the classes
b) centred at the lower limits of the classes
c) centred at the classmarks of the classes
d) evenly distributed over all the classes
Solution (c)
Question 4
Which of the following is the measure of central tendency?
a) Mean
b)Mode
c) Median
d) Range
Solution (a) ,(b),(c)
Question 5
For drawing a frequency polygon of a continuous frequency distribution, we plot the Points whose ordinates are the frequencies of the respective classes and abscissae are respectively :
a) class marks of the classes
b) upper limits of preceding classes
c) lower limits of the classes
d) upper limits of the classes
Solution (a)
Question 6
Find the mean of 32 numbers given mean of ten of them is 12 and the mean of other 20 is 9. And last 2 number is 10
a) 10
b) 12
c) 13
d) 14
Solution (a)
Mean of 10 number =12
Mean of 20 number =9
Mean of 2 number =10
Mean of 32 number= Sum of all number/32=(120+180+20)/32=10
Question 7
The median and mean of the first 10 natural numbers.
a) 5.5,5.5
b) 5.5,6
c) 5,6
d) None of these
Solution (a)
Mean =5.5
Median is mean of 5 and 6 th term, So 5
Question 8
Anand says that the median of 3, 14, 19, 20, 11 is 19. What doesn’t the Anand understand about finding the median?
a) The dataset should be ascending order
b) Highest no in the dataset is the median
c) Average of lowest and highest is the median
d) None of these
Solution (a)
Question 9
The following observations are arranged in ascending order :
20, 23, 42, 53, x, x + 2, 70, 75, 82, 96
If the median is 63, find the value of x.
a) 62
b) 64
c) 60
d) None of these
Solution (a)
Median is mean of 5 and 6 term
So x+1=63
X=62
Question 10
The mean of 20 observations was 60. It was detected on rechecking that the value of 125 was wrongly copied as 25 for computation of mean. Find the correct mean
a) 67
b) 66
c) 65
d) None of the above
Solution
Let x be the sum of observation of 19 numbers leaving 125,
Then
X+25=20*60=1200
Now
X+125=20*y=20y
Subtracting
12525=20y1200
20y=1300
y=65
Question 11
Compute the Median for the given data
Class –interval 
100110 
110120 
120130 
130140 
140150 
150160 
Frequency 
6 
35 
48 
72 
100 
4 
Solution
First calculate the Class mark and cumulative frequency of the data
Class –interval 
100110 
110120 
120130 
130140 
140150 
150160 
Frequency 
6 
35 
48 
72 
100 
5 
Class Mark 
105 
115 
125 
135 
145 
155 
Cummulative Frequency 
6 
41 
89 
161 
261 
266 
We have N=266, So N/2=133,Cummulative frequency first greater than 133,lies in class 130140
So median class is 130140
Now
Median is calculated as
Where
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 (assuming class size to be equal)
Here l=130 n=266 cf=89 f=72 h=10
Substituting these
M_{m}= 130+(13389)x10/72=136.11
A histogram 
is the diagram showing a system of connections or interrelations between two or more things by using bars 
Discontinuous Frequency Distribution. 
A frequency distribution in which the upper limit of one class coincides from the lower limit of the succeeding class 
Continuous Frequency Distribution. 
Is the bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval. 
Ogive is the graph of 
Is a set of adjacent rectangles whose areas are proportional to the frequencies of a given continuous frequency distribution? 

lower/upper limits and cumulative frequency 