Lesson: Statistics and Probability
EXERCISE 14.1
Write the correct answer in each of the following:
Question: 1
The class mark of the class
90 120
is:
(A) 90
(B) 105
(C) 115
(D) 120
Solution
B
Class mark
upper class limit lower class limit
2
120 90 210
105
22
Question: 2
The range of the data:
25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11,
20 is
(A) 10
(B) 15
(C) 18
(D) 26
Solution
D
Range of the data
highest value
lowest value
32 6 26
Question: 3
In a frequency distribution, the mid value of a class is
10 and the width of the class is 6. The lower limit of
the class is:
(A) 6
(B) 7
(C) 8
(D) 12
Solution
B
Lower limit of class interval
Mid value of class
width of class
2
6
10 10 3 7
2
Question: 4
The width of each of five continuous classes in a
frequency distribution is 5 and the 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
Solution
C
The upper limit of highest class
Lower limit of lowest class
(Number of
continuous classes
width of each class)
10 5 5 10 25 35
Question: 5
Let m be the mid-point and l be the upper class limit of
a class in a continuous frequency distribution. The
lower class limit of the class is:
(A)
2ml
(B)
2ml
(C)
ml
(D)
2ml
Solution
B
Mid value
Upper class limit +lower class limit
2
2ml
Lower class limit
2m l
Lower class limit
Question: 6
The class marks of a frequency distribution are given
as follows: 15, 20, 25...
The class corresponding to the class mark 20 is:
(A)
12.5 17.5
(B)
17.5 22.5
(C)
18.5 21.5
(D)
19.5 20.5
Solution
B
We know that,
Upper limit Lower limit
Class mark
2
Therefore,
. .
Class mark
225 175 40.0
20
22
This is not true for the rest of the options.
Question: 7
In the class intervals
10 20, 20 30
, the number 20 is
included in:
(A)
10 20
(B)
20 30
(C) Both the intervals
(D) None of these intervals
Solution
B
Because, class interval includes values greater than or
equal to lower limit and less than upper limits.
Question: 8
A grouped frequency table with class intervals of
equal sizes, using
250 270
(270 not included in this interval) as one of
the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342,
310, 290, 300, 320, 319, 304, 402, 318,
406, 292, 354, 278, 210, 240, 330, 316,
406, 215, 258, 236.
The frequency of the class
310 330
is:
(A) 4
(B) 5
(C) 6
(D) 7
Solution
C
Frequency of a class is the number of members in the
class. So, as per the given data, 310, 310, 320, 319,
318, 316 are the members of the class
310 330
.
Question: 9
A grouped frequency distribution table with classes of
equal sizes using
63 72
(72 included) as one of the class is constructed
for the following data:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,
102, 110, 88, 74, 112, 14, 34, 44.
The number of classes in the distribution will be:
(A) 9
(B) 10
(C) 11
(D) 12
Solution
B
Classes are
13 22, 23 32, 33 42, 43 52, 53 62, 63 72,
73 82, 83 92, 93 102
and
103 112
.
Question: 10
To draw a histogram to represent the following
frequency distribution:
Class
interval
5 10
10 15
15 25
25 45
45 75
Frequenc
y
6
12
10
8
15
the adjusted frequency for the class
25 45
is:
(A) 6
(B) 5
(C) 3
(D) 2
Solution
D
Minimum class width
5
Width of the interval
25 45 45 25 20
Corresponding frequency
25 45 8
So, adjusted frequency
Frequency
Class width

Minimum class width
8
52
20
Question: 11
The mean of five numbers is 30. If one number is
excluded, their mean becomes 28.
The excluded number is:
(A) 28
(B) 30
(C) 35
(D) 38
Solution
D
Let
1 2 3 4
,,,a a a a
and
5
a
be five numbers and the
excluded number be
5
a
.
Given, Mean
30
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
30
5
150
150 ... (i)
a a a a a
a a a a a
a a a a a

Given, mean of four numbers
28
1 2 3 4
1 2 3 4
28
4
112.......(ii)
a a a a
a a a a

From (i) and (ii),
5
5
150 112
150 112 38
a
a

Question: 12
If the mean of the observations:
, 3, 5, 7, 10x x x x x
is 9, the mean of the last three observations is
(A)
1
10
3
(B)
2
10
3
(C)
1
11
3
(D)
2
11
3
Solution
C
Given,
3 5 7 10
9
5
x x x x x
5 25
9
5
5 25 45
5 20
4



x
x
x
x
So, the values are
4, 4 3 7 , 4 5 9 , 4 7 11
and
4 10 14
Now, mean of last 3 observations
9 11 14 34 1
11
3 3 3

Question: 13
If
x
represents the mean of n observations
12
, ,...,
n
x x x
, then value of
1
()
n
i
i
xx
is:
(A)
1
(B) 0
(C) 1
(D)
1n
Solution
B
Given, mean
1
1
... (i)
n
i
i
n
i
i
x
x
n
x nx

Now,
1 1 1
()
n n n
ii
i i i
x x x x
1
n
i
nx x

(From (i))
0nx nx
Question: 14
If each observation of the data is increased by 5, then
their mean
(A) Remains the same
(B) Becomes 5 times the original mean
(C) Is decreased by 5
(D) Is increased by 5
Solution
D
Let
1 2 3
, , ,...,
n
a a a a
be the n observations.
Then, original mean, m (say)
12
...
... (i)
n
a a a
n
Now, adding 5 in each observation, we get
1 2 3
5, 5, 5 ... , 5
n
a a a a
as the new observation.
So, new mean, M (say)
12
12
12
5 5 ... 5
(5 5 ... 5)
...
times
... 5
n
n
n
a a a
n
a a a
n
n
a a a n
n
12
5
5
...
n
a a a n
m
nn
(Using (i))
So, the mean is increased by 5.
Question: 15
Let
x
be the mean of
1 2 3
, , , ...
n
x x x x
and
y
be the
mean of
1 2 3
, , , ...
n
y y y y
. If
z
is the mean of
1 2 3 1 2 3
, , , ... , , , , ...
nn
x x x x y y y y
,
then
z
is equal to:
(A)
xy
(B)
2
xy
(C)
xy
n
(D)
2
xy
n
Solution
B
Given,
+ +
1 2 3
1 2 3
1 2 3
1 2 3
...
... ... (i)
...
... ... (ii)
n
n
n
n
x x x x
x
n
nx x x x x
y y y y
y
n
ny y y y y

Now,
1 2 3 1 2 3
( ... ) ( ... )
nn
x x x x y y y y
z
nn
Now, using (i) and (ii), in above equation,
()
2
2
nx ny n x y
z
n n n
xy
z


Question: 16
If
x
is the mean of
,
1 2 3
, , ,...
n
x x x x
, then for
0a
,
the
mean of
12
12
, ,..., , , ,...,
n
n
x x x
ax ax ax
a a a
is:
(A)
1
ax
a



(B)
1
2
x
a
a



(C)
1 x
a
an



(D)
1
2
ax
a
n



Solution
B
Given,
12
...
n
x x x
x
n
...(i)
12
...
n
nx x x x
Let us suppose the mean of
12
12
, ,..., , , ,...,
n
n
x x x
ax ax ax
a a a
is
y
.
Now,
12
12
1 2 1 2
... ...
1
( ... ) ( ... )
2
n
n
nn
x x x
ax ax ax
a a a
y
nn
a x x x x x x
a
n
Now using (i) in above equation,
11
. ( )
22
1
()
2
anx nx a nx
aa
y
nn
ax
a
y


Question: 17
If
1 2 3
, , ,...,
n
x x x x
are the means of n groups with
12
, ,...,
n
n n n
number of observations respectively, then
the mean x of all the groups taken together is given by
:
(A)
1
n
ii
i
nx
(B)
1
2
n
ii
i
nx
n
(C)
1
1
n
ii
i
n
i
i
nx
n
(D)
1
2
n
ii
i
nx
n
Solution
C
Given,
1 2 3
, , ,...,
n
x x x x
are the means of n groups with
12
, ,...,
n
n n n
number of observations respectively.
Then,
2
1
1
11
12
12
, , ...,
n
n
n
n
j
i
j
i
n
n
x
xx
x x x
n n n


12
1 1 2 2
1 1 1
, ,...,
n
n
nn
i j n n
ij
n x x n x x n x x
Now, the mean of all the groups taken together is
given by
12
1 1 1
12
1 1 2 2
1
1
1
...
...
...
n
n
nn
ij
ij
n
nn
n
i
i
n
ii
i
n
i
i
x x x
x
n n n
n x n x n x
x
n
nx
x
n


Question: 18
The mean of 100 observations is 50. If one of the
observations which was 50 is replaced by 150, the
resulting mean will be:
(A) 50.5
(B) 51
(C) 51.5
(D) 52
Solution
B
Let
1 2 100
, , ... ,x x x
be the 100 observations and
100
x
be
50 which is replaced by 150.
Then, original mean
50
(Given)
1 2 99 100
1 2 99
1 2 99
...
50
100
... 50 5000
... 4950 ... (i)
x x x x
x x x
x x x

Now, new mean
1 2 99 100
...
100
x x x x
1 2 99 100
( ... ) 4950 150
100 100
x x x x

(From (i))
New mean
5100
51
100

Question: 19
There are 50 numbers. Each number is subtracted
from 53 and the mean of the numbers so obtained is
found to be
3.5
. The mean of the given numbers is:
(A) 46.5
(B) 49.5
(C) 53.5
(D) 56.5
Solution
D
Given
50n
, then mean,
1 2 50
1 2 50
...
50
50 ... ... (i)
x x x
x
x x x x
Now, according to the question,
1 2 50
53 , 53 , ... ,53x x x
is the new observation
with mean
3.5
.
So,
1 2 50
1 2 50
1 2 50
53 53 ... 53
3.5
50
53 53 ... 53
3.5 50 ( ) ( ... )
50 times
175 50 53 ( ... ) ... (ii)
x x x
x x x
x x x


Now, using (i) and (ii), we get
175 2650 50
50 2650 175 2825
2825
56.5
50
x
x
x
Question: 20
The mean of 25 observations is 36. Out of these
observations if the mean of first 13 observations is 32
and that of the last 13 observations is 40, the 13
th
observation is:
(A) 23
(B) 36
(C) 38
(D) 40
Solution
B
Given, mean of 25 observations
36
1 2 25
1 2 25
1 2 25
...
36
25
... 25 36
... 900 ... (i)
x x x
x x x
x x x

Now, out of these observations, if the mean of first 13
observations
32
i.e.,
1 2 13
1 2 13
1 2 13
...
32
13
... 32 13
... 416 ... (ii)
x x x
x x x
x x x
and
13 14 25
13 14 25
13 14 25
14 25 13
...
40
13
... 40 13
... 520
( ... ) 520 ... (iii)
x x x
x x x
x x x
x x x
Now, from (i), we have
1 2 25
1 2 13 14 25
1 2 13 14 25
... 900
... ... 900
( ... ) ( ... ) 900
x x x
x x x x x
x x x x x
Now using (ii) and (iii) in above equation, we get
13
13
13
13
416 520 900
936 900
936 900
36
x
x
x
x

Question: 21
The median of the data
78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
(A) 45
(B) 49.5
(C) 54
(D) 56
Solution
C
Let us arrange the data in ascending order i.e.
22, 34, 39, 45, 54, 54, 56, 68, 78 and 84
Total no. of observations,
10n
which is an even
number
So, Median
Average of
( )th
2
n
and
( 1)th
2
n
observations
Average of
th
10
( )th 5
2
and
6
th
10
( 1)th
2

observations
Average of 54 and 54 = 54 (
54 54
54
2
)
Median
54
Question: 22
For drawing a frequency polygon of a continuous
frequency distribution, we plot the points whose
ordinates are the frequencies of the respective classes
and abscissas respectively:
(A) Upper limits of the classes
(B) Lower limits of the classes
(C) Class marks of the classes
(D) Upper limits of preceding classes
Solution
C
Question: 23
Median of the following numbers:
4, 4, 5, 7, 6, 7, 7, 12, 3 is
(A) 4
(B) 5
(C) 6
(D) 7
Solution
C
Let us arrange the data in ascending order i.e.
3, 4, 4, 5, 6, 7, 7, 7 and 12
Total no. of observations,
9n
which is an odd
number
So, Median
1
( )th
2
n
observation
th
91
( )th 5
2

observation
Median
6
Question: 24
Mode of the data
15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15
is
(A) 14
(B) 15
(C) 16
(D) 17
Solution
B
Let us arrange the data in ascending order
14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20
We can see that 15 occurs most frequently i.e., 5
times.
Mode = 15
Question: 25
In a sample study of 642 people, it was found that 514
people have a high school certificate. If a person is
selected at random, the probability that the person has
a high school certificate is:
(A) 0.5
(B) 0.6
(C) 0.7
(D) 0.8
Solution
D
Total number of people,
642nS
The number of people, who have a high school
certificate,
514nE
The probability that the person has a high school
certificate
( ) 514
0.8
( ) 642
nE
nS
Hence, the probability is 0.8.
Question: 26
In a survey of 364 children aged 19-36 months, it was
found that 91 liked to eat potato chips. If a child is
selected at random, the probability that he/she does
not like to eat potato chips is:
(A) 0.25
(B) 0.50
(C) 0.75
(D) 0.80
Solution
C
Total number of children,
364nS
The number of children who liked potato chips
91
The number of children, who do not like potato
chips,
364 91 273.nE
The probability that the selected child does not like to
eat potato chips is
( ) 273
0.75
( ) 364
nE
nS
Hence, the probability is 0.75.
Question: 27
In a medical examination of students of a class, the
following blood groups are recorded:
Blood group
A
AB
B
O
Number of
students
10
13
12
5
A student is selected at random from the class. The
probability that he/she has blood group B, is:
(A)
1
4
(B)
13
40
(C)
3
10
(D)
1
8
Solution
C
Total number of students in the medical examination,
40nS
The number of students with blood group B,
12nE
. So, the probability that the student has blood group
( ) 12 3
( ) 40 10
nE
nS
B
Question: 28
Two coins are tossed 1000 times and the outcomes are
recorded as below:
Number of
heads
2
1
0
Frequency
200
550
250
Based on this information, the probability for at most
one head is
(A)
1
5
(B)
1
4
(C)
4
5
(D)
3
4
Solution
C
Total number of times the coins were tossed,
1000nS
The number of times in which at most one head
occurs,
550 250 800nE
So, the probability of getting at most one head
( ) 800 4
( ) 1000 5
nE
nS
Question: 29
80 bulbs are selected at random from a lot and their
life time (in hrs) is recorded in the form of a frequency
table given below:
Life time
(in hours)
300
500
700
900
1100
Frequency
10
12
23
25
10
One bulb is selected at random from the lot. The
probability that its life is 1150 hours, is
(A)
1
80
(B)
7
16
(C) 0
(D) 1
Solution
C
According to the given table, there is no bulb with life
time as 1150 hours.
So, probability of getting a bulb with life time of 1150
hours
0
Question: 30
Refer to Q. 29 above.
The probability that bulbs selected randomly from the
lot have life less than 900 hours is:
(A)
11
40
(B)
5
16
(C)
7
16
(D)
9
16
Solution:
D
Total number of bulbs,
80nS
The number of bulbs with life times less than 900 h,
10 12 23 45nE
So, the probability that bulbs selected randomly from
the lot have life less than 900 hours
( ) 45 9
( ) 80 16
nE
nS
EXERCISE 14.2
Question: 1
The frequency distribution:
Marks
0-20
20-40
40-60
60-100
Number
of
students
10
15
20
25
has been repesented graphically as follows:
Do you think this representation is correct? Why?
Solution
The given representation is not correct as the widths of
the rectangles are varying.
So, we need to adjust the frequencies so that areas of
the rectangles are proportional to the frequencies.
Marks
No. Of students
(Frequency)
Class
Size
Adjusted
Frequency
0 20
10
20
10
20 10
20

20 40
15
20
15
20 15
20

40 60
20
20
20
20 20
20

60 100
25
40
25
20 12.5
40

So, the correct histogram with varying width is given
below:
Question: 2
In a diagnostic test in mathematics given to students,
the following marks (out of 100) are recorded:
46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44.
Which ‘average’ will be a good representative of the
above data and why?
Solution
Median will be a good representative of the data,
because the data has extreme values.
Now, let us arrange observations in ascending order as
11, 40, 41, 44, 46, 48, 52, 53, 54, 62, 96 and 98
No. of observations
12
(even)
Median
Average of
12
th
2



and
12
1 th
2



observations
Median
Average of 6
th
and 7
th
observations
Median

48 52
50
2
, this is an appropriate
central tendency.
Question: 3
A child says that the median of 3, 14, 18, 20 and 5 is
18. What doesn’t the child understand about finding
the median?
Solution
The child does not understand that the observations
have to be arranged in ascending/ descending order
before finding the median.
We arrange the observations in ascending order: 3, 5,
14, 18 and 20.
Number of observations
5
, which is an odd number
So, the median
rd
51
3
2




observation.
Thus, median
14
.
Question: 4
A football player scored the following number of
goals in the 10 matches:
1, 3, 2, 5, 8, 6, 1, 4, 7 and 9.
Since the number of matches is 10 (an even number),
therefore, the median
th th
5 observation 6 observation
2
86
7
2

Is it the correct answer and why?
Solution
No, because first the data have to be arranged in
ascending (or descending) order before finding the
median.
The observations in ascending order: 1, 1, 2, 3, 4, 5, 6,
7, 8, 9.
Total no. of observations
10
(Even number)
Then, median
average of
10
th
2



and
10
1 th
2



observation
Average of 5
th
and 6
th
observation
4 5 9
4.5
22
Question: 5
Is it correct to say that in a histogram, the area of each
rectangle is proportional to the class size of the
corresponding class interval? If not, correct the
statement.
Solution
No. The correct statement is that the area of each
rectangle, in a histogram, is proportional to the
corresponding frequency of its class.
Question: 6
The class marks of a continuous distribution are:
1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and 1.64
Is it correct to say that the last interval will be 1.55 -
1.73? Justify your answer.
Solution:
Given, class marks 1.04, 1.14, 1.24, 1.34, 1.44, 1.54,
and 1.64.
So, class size
1.14 1.04 0.10
and half of class size
0.10
0.05
2

Therefore,
Lower limit of last class interval
1.64 0.05 1.59,
Upper limit of last class interval
1.64 0.05 1.69
So, last class Interval is
1.59 1.69
.
Therefore
1.55 1.73
is not the last interval.
Hence, the given statement is incorrect.
Question: 7
30 children were asked about the number of hours
they watched TV programmes last week. The results
are recorded as under:
Number of
hours
0-5
5-10
10-15
15-20
Frequency
8
16
4
2
Can we say that the number of children who watched
TV for 10 or more hours a week is 22? Justify your
answer.
Solution
Number of children who watched TV for 10 or more
hours = Number of children who watched TV for 10-
15 hours and 15-20 hours
42
6

Hence, the given statement is incorrect.
Question: 8
Can the experimental probability of an event be a
negative number? If not, why?
Solution
No. The experimental probability of an event cannot
be a negative number because the number of favorable
trials cannot be negative and the total number of trials
is always positive.
Question: 9
Can the experimental probability of an event be
greater than 1? Justify your answer.
Solution
No. Because,
Probability of an event
Number of trials in which event happens
Total number of trials
The number of trials in which an event can happen,
cannot be greater than the total number of trials.
Question: 10
As the number of tosses of a coin increases, the ratio
of the number of heads to the total number of tosses
will be
1
2
. Is it correct? If not, write the correct one.
Solution
No. As the number of tosses of a coin increases, the
ratio of the number of heads to the total number of
tosses will be closer to
1
2
, but not exactly
1
2
.
EXERCISE 14.3
Question: 1
The blood groups of 30 students are recorded as
follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB,
B, A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
Solution
A frequency distribution table for the data is as given
below:
Blood Group
Tally Marks
Frequency
A
12
B
8
AB
4
O
6
Total
30
Question: 2
The value of
up to 35 decimal places is given below:
3.14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after
the decimal point.
Solution:
A frequency distribution table for the data is as given
below:
Digit
Tally Marks
Frequency
0
1
1
2
2
5
3
6
4
3
5
4
6
3
7
2
8
5
9
4
Total
35
Question: 3
The scores (out of 100) obtained by 33 students in a
mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84
66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency
distribution.
Solution:
A frequency distribution table for the data is as given
below:
Scores
Tally Marks
Frequency
48
3
58
3
64
4
66
7
69
6
71
3
73
2
81
1
83
2
84
2
Total
33
Question: 4
Prepare a continuous grouped frequency distribution
from the following data:
Mid-point
Frequency
5
15
25
35
45
4
8
13
12
6
Also find the size of class intervals.
Solution:
Class Size
difference between the two midpoints
15 5 10
Lower limit of the first interval
2
class size
midpoint
10
5 5 5 0
2
Upper Limit of the first interval
2
class size
midpoint
10
5 5 5 10
2
So, first class interval
0 10.
Now, a continuous grouped frequency distribution
table is as follows:
Mid Value
Class Interval
Frequency
5
0-10
4
15
10-20
8
25
20-30
13
35
30-40
12
45
40-50
6
Total
43
The size of class interval
10.
Question: 5
Convert the given frequency distribution into a
continuous grouped frequency distribution:
Class interval
Frequency
150-153
154-157
158-161
162-165
166-169
170-173
7
7
15
10
5
6
In which intervals would 153.5 and 157.5 be included?
Solution
The given table is clearly in discontinuous form.
So, we will convert this to continuous form.
For this, let us consider the classes:
150 153
and
154 157
Upper limit of first class
153
and lower limit of second class
154.
So, the difference
154 153 1
Hence, half of the difference
0.5
Now, we construct the continuous grouped frequency distribution by subtracting 0.5
from each lower limit and adding 0.5 to each upper limit.
Class intervals
Frequency
149.5-153.5
153.5-157.5
157.5-161.5
161.5-165.5
165.5-169.5
169.5-173.5
7
7
15
10
5
6
Hence, 153.5 and 157.5 lie in the class interval 153.5-157.5 and 157.5-161.5,
respectively.
Question: 6
The expenditure of a family on different heads in a
month is given below:
Head
Food
Educ
ation
Cloth
ing
House
Rent
Others
Savings
Expen
diture
(in Rs)
4000
2500
1000
3500
2500
1500
Draw a bar graph to represent the data above.
Solution:
The bar graph for given data is shown below by
representing expenditure on y-axis (
unit Rs. 1 500
)
and different heads on x-axis.
Question: 7
Expenditure on education of a country during a five
year period (2002-2006), in crores of rupees, is given
below:
Elementary Education
Secondary Education
University Education
Teacher’s Training
Social Education
Other Educational Programmes
Cultural Programmes
Technical Education
240
120
190
20
10
115
25
125
Represent the information above by a bar graph.
Solution:
The bar graph for given data is shown below by
representing Expenditure on y-axis (
unit Rs 1 25
crore) and different heads on x-axis.
Question: 8
The following table gives the frequencies of most
commonly used letters a, e, i, o, r, t, u from a page of a
book:
Letters
a
e
i
o
r
t
U
Frequency
75
125
80
70
80
95
75
Represent the information above by a bar graph.
Solution:
The bar graph for given data is shown below by
representing frequency on y-axis (
unit1 15
frequency) and letters on x-axis.
Question: 9
If the mean of the following data is 20.2, find the
value of p:
x
10
15
20
25
30
f
6
8
p
10
6
Solution:
Let us re-arrange the table as follows:
x
i
f
i
( f
i
x
i
)
10
15
20
25
30
6
8
p
10
6
60
120
20p
250
180
Total
5
1
30
i
i
fp

5
1
610 20
ii
i
f x p
It is given that mean of the data
20.2
5
1
5
1
20.2
ii
i
i
i
fx
f

610 20
20.2
30
p
p

610 20 606 20.2pp
4 0.2p
4 40
20
0.2 2
p
Question: 10
Obtain the mean of the following distribution:
Frequency
Variable
4
8
14
11
3
4
6
8
10
12
Solution
Let us re-arrange the data table as follows:
x
i
f
i
( f
i
x
i
)
4
4
16
6
8
48
8
14
112
10
11
110
12
3
36
Total
5
1
40
i
i
f


5
1
322
ii
i
fx
Mean

5
1
5
1
322
8.05
40
ii
i
i
i
fx
f
Question: 11
A class consists of 50 students out of which 30 are
girls. The mean of marks scored by girls in a test is 73
(out of 100) and that by boys is 71. Determine the
mean score of the whole class.
Solution:
Sum of the marks scored by the girls
mean no of girls.
73 30
2190
Sum of the marks scored by the boys
mean
no. of boys
71 20
1420
sum of marks scored by all students
no. of students
Mean
2190 1420
50
3610
50
72.2
Hence, the mean score of the class is 72.2.
Question: 12
Mean of 50 observations was found to be 80.4. But
later on, it was discovered that 96 was misread as 69 at
one place. Find the correct mean.
Solution:
Sum of 50 observations
mean
no. of observations
80.4 50 4020
Now,
Correct sum of 50 observations
4020 69 96 4047
sum of all observations
Correct mean
no. of observation
4047
50
80.94
Hence, the correct mean is 80.94.
Question: 13
Ten observations
6, 14, 15, 17, 1, 2 13, 30, 32, 34,43xx
are written
in an ascending order. The median of the data is 24.
Find the value of x.
Solution:
Given observations are:
6, 14, 15, 17, 1, 2 13, 30, 32, 34, 43xx
Total no. of observations
10
(even)
Median
Average of
10
2
th



and
10
1
2
th



observations
Average of
5
th
and
6
th
observations
1 2 13 3 12
22
x x x

But, Median
24
(Given)
3 12
24
2
x

3 12 48x
3 48 12 60x
20x
Question: 14
The points scored by a basketball team in a series of
matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28
Find the median and mode for the data.
Solution:
First, let us arrange the observations in ascending
order
2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48
Mode
10
(frequency
3
)
Total no. of observations is
16
, which is an even
number.
So, Median
Average of
16
2
th



and
16
1
2
th



observations
Average of
8
th
and
9
th
observations
10 14 24
12
22
Hence, median of the data and mode .12 10
Question: 15
In the Fig. 14.2, there is a histogram depicting daily
wages of workers in a factory.
Construct the frequency distribution table.
Fig. 14.2
Solution:
The frequency distribution table is as shown below.
Class Interval
Frequency
150-200
50
200-250
30
250-300
35
300-350
20
350-400
10
Total
145
Question: 16
A company selected 4000 households at random and
surveyed them to find out a relationship between the
income level and the number of television sets in a
home. The information so obtained is listed in the
following table:
Monthly
income
(in Rs)
Number of
Televisions/household
0
1
2
Above 2
10000
10000-14999
15000-19999
20000-24999
25000 and
above
20
10
0
0
0
80
240
380
520
1100
10
60
120
370
760
0
0
30
80
220
Find the probability:
(i) Of a household earning Rs 10000-Rs 14999 per
year and having exactly one television
(ii) Of a household earning Rs 25000 and more per
year and owning 2 televisions
(iii) Of a household not having any television
Solution:
Total number of the households selected by the
company,
4000nS
.
(i) Number of households earning Rs 10000-Rs 14999
per year and having exactly one television,
1
240.nE
So, Required Probability
1
( ) 240 6
0.06
( ) 4000 100
nE
nS
(ii) Number of household earning Rs 25000 and more
per year and owning 2 televisions,
2
760.nE
So, Required Probability
2
) 760 76
0.19
( ) 4000 400
nE
nS
=
(iii) Number of household not having any television,
3
20 10 30nE
So, Required Probability
3
( ) 30 3
0.0075
( ) 4000 400
nE
nS
Question: 17
Two dices are thrown simultaneously for 500 times.
Each time the sum of two numbers appearing on their
tops is noted and recorded as given in the following
table:
Sum
Frequency
2
3
4
5
6
7
8
9
10
14
30
42
55
72
75
70
53
46
11
12
28
15
If the dices are thrown once more, what is the
probability of getting a sum of
(i) 3?
(ii) More than 10?
(iii)Less than or equal to 5?
(iv) Between 8 and 12?
Solution:
Total number of times the two dices are thrown
simultaneously,
500S n
(i) Number of times of getting a sum 3,
1
( ) 30nE
Required Probability
1
( ) 30 3
0.06
( ) 500 50
nE
nS
(ii) Number of times of getting a sum more than 10,
2
( ) 28 15 43nE
Required Probability
2
( ) 43
0.086
( ) 500
nE
nS
(iii)Number of times of getting a sum less than or
equal to 5,
3
( ) 14 30 42 55 141nE
Required Probability
3
( ) 141
0.282
( ) 500
nE
nS
(iv) Number of times of getting a sum between 8 and
12,
4
( ) 53 46 28 127nE
Required Probability
4
( ) 127
0.254
( ) 500
nE
nS
Question: 18
Bulbs are packed in cartons each containing 40 bulbs.
Seven hundred cartons were examined for defective
bulbs and the results are given in the following table:
Number
of
defective
bulbs
0
1
2
3
4
5
6
More
than 6
Frequenc
y
400
18
0
48
41
18
8
3
2
One carton was selected at random. What is the
probability that it has
(i) No defective bulb?
(ii) Defective bulbs from 2 to 6?
(iii) Defective bulbs less than 4?
Solution:
Total no. of cartons,
700nS
(i) Number of cartons which have no defective bulb
1
( ) 400nE
.
Required Probability
1
( ) 400 4
( ) 700 7
nE
nS
(ii) Number of cartons which have defective bulbs from 2 to 6,
2
( ) 48 41 18 8 3 118nE
. Required Probability
2
( ) 118 59
( ) 700 350
nE
nS
(iii) Number of cartons which have defective bulbs
less than 4,
3
( ) 400 180 48 41 669nE
.
Required Probability
3
( ) 669
( ) 700
nE
nS

Question: 19
Over the past 200 working days, the number of
defective parts produced by a machine is given in the
following table:
Numbe
r of
defectiv
e parts
0
1
2
3
4
5
6
7
8
9
10
1
1
1
2
1
3
Days
5
0
3
2
2
2
18
1
2
1
2
1
0
1
0
10
8
6
6
2
2
Determine the probability that tomorrow’s output will
have
(i) No defective part
(ii) Atleast one defective part
(iii)Not more than 5 defective parts
(iv) More than 13 defective parts
Solution:
Total number of working days,
200nS
.
(i) Number of days in which no defective part is
produced
1
( ) 50nE
. Required
Probability
1
( ) 50 1
( ) 200 4
nE
nS
(ii) Probability of days in which at least one defective
part is produced
1
probability of days in which
no defective part is produced
13
1
44
(Using part (i))
(iii)Number of days in which not more than 5
defective parts are produced,
2
2
( ) 50 32 22 18 12 12
( ) 146
nE
nE
Required Probability
2
( ) 146
0.73
( ) 200
nE
nS
(iv) Number of days in which not more than 13
defective parts are produced is 0.
Required Probability
0
Question: 20
A recent survey found that the ages of workers in a
factory is distributed as follows:
Age (in
years)
20-29
30-39
40-49
50-59
60 and
above
Numbe
r of
worker
s
38
27
86
46
3
If a person is selected at random, find the probability
that the person is:
(i) 40 years or more
(ii) Under 40 years
(iii)Having age from 30 to 39 years
(iv) Under 60 but over 39 years
Solution:
Total no. of workers in a factory,
38 27 86 46 3 200nS
(i) Number of workers at the age of 40 years or more,
1
86 46 3 135nE
Required Probability
1
( ) 135
0.675
( ) 200
nE
nS
(ii) Number of workers under the age of 40 years,
2
38 27 65nE
Required Probability
2
( ) 65
0.325
( ) 200
nE
nS
(iii) Number of workers having age from 30 to 39
years,
3
27nE
Required Probability
3
( ) 27
0.135
( ) 200
nE
nS
(iv) Number of workers under age 60 but over 39
years,
4
86 46 132nE
Required Probability
4
( ) 132
0.66
( ) 200
nE
nS
EXERCISE 14.4
Question: 1
The following are the marks (out of 100) of 60
students in mathematics:
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16,
36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45,
62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92,
62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30
Construct a grouped frequency distribution table with
width 10 of each class starting from 0-9.
Solution:
We arrange the given data into groups like 0-9, 10-19,
20-29 etc. The class width in each case is 10.
The frequency distribution table of the given data is as
shown below:
Scores
Tally Marks
Frequency
0-9
4
10-19
7
20-29
5
30-39
10
40-49
5
50-59
8
60-69
5
70-79
8
80-89
5
90-99
3
Total
60
Question: 2
Refer to Q1 above. Construct a grouped frequency
distribution table with width 10 of each class, in such
a way that one of the classes is 10-20 (20 not
included).
Solution:
We arrange the given data into groups like 0-10 (10
not included), 10-20, 20-30 etc. The class width in
each case is 10.
The frequency distribution table of the given data is as
shown below:
Scores
Tally Marks
Frequency
0-10
4
10-20
7
20-30
5
30-40
10
40-50
5
50-60
8
60-70
5
70-80
8
80-90
5
90-100
3
Total
100
Question: 3
Draw a histogram of the following distribution:
Heights (in cm)
Number of students
150-153
7
153-156
8
156-159
14
159-162
10
162-165
6
165-168
5
Solution:
Clearly, the given frequency distribution is in
continuous form.
We represent the number of students along y-axis (
unit12
students) and heights (
unit cm13
) along
x-axis.
A histogram of the given distribution is as follows:
Questions: 4
Draw a histogram to represent the following grouped
frequency distribution:
Ages (in years)
Number of teachers
20-24
10
25-29
28
30-34
32
35-39
48
40-44
50
45-49
35
50-54
12
Solution:
The given frequency distribution is in discontinuous
form.
So, first we convert it into continuous form.
Let us consider classes 20-24 and 25-29.
Adjusting factor
25 24 1
0.5
22
Now, we subtract 0.5 from each lower limit and add
0.5 to each upper limit.
So, the table for continuous grouped frequency
distribution is as follows:
Ages (in years)
Number of teachers
19.5-24.5
10
24.5-29.5
28
29.5-34.5
32
34.5-39.5
48
39.5-44.5
50
44.5-49.5
35
49.5-54.5
12
Now, we represent the number of teachers along y-
axis (
unit15
teachers) and Ages (
unit15
years)
along x-axis.
A histogram of the given distribution is as follows:
Question: 5
The lengths of 62 leaves of a plant are measured in
millimeters and the data is represented in the
following table:
Length (in mm)
Number of leaves
118-126
8
127-135
10
136-144
12
145-153
17
154-162
7
163-171
5
172-180
3
Draw a histogram to represent the data above.
Solution:
The given frequency distribution is in discontinuous
form.
So, first we convert it into continuous form.
Let us consider classes 118-126 and 127-135.
Adjusting factor
127 126 1
0.5
22
Now, we subtract 0.5 from each lower limit and add
0.5 to each upper limit.
So, the table for continuous grouped frequency
distribution is as follows:
Length (in mm)
Number of leaves
117.5-126.5
8
126.5-135.5
10
135.5-144.5
12
144.5-153.5
17
153.5-162.5
7
162.5-171.5
5
171.5-180.5
3
Now, we represent the number of leaves along y-axis
(1 unit
2
leaves) and length (1 unit
9 mm) along
x-axis.
A histogram of the given distribution is as follows:
Question: 6
The marks obtained (out of 100) by a class of 80
students are given below:
Marks
Number of students
10-20
6
20-30
17
30-50
15
50-70
16
70-100
26
Construct a histogram to represent the data above.
Solution:
In the given table, class sizes are different.
So, we calculate adjusted frequency of each class by
using the formula
Minimum class size
frequency
class size of the class
Here, minimum class size
20 10 10
The modified table for frequency distribution is as
follows:
Marks
Number of
students (f
i
)
Adjusted
frequency
10-20
6
10
66
10

20-30
17
10
17 17
10

30-50
15
10
15 7.5
20

50-70
16
10
16 8
20

70-100
26
10
26 8.67
30

Now, we represent the number of students along y-
axis (1 unit
2 students) and marks (1 unit
10
marks) along x-axis.
A histogram of the given distribution is as follows:
Question: 7
Following table shows a frequency distribution for the
speed of cars passing through at a particular spot on a
high way:
Class interval (km/h)
Frequency
30-40
3
40-50
6
50-60
25
60-70
65
70-80
50
80-90
28
90-100
14
Draw a histogram and frequency polygon representing
the data above.
Solution:
Clearly, the given frequency distribution is in
continuous form and has the same class size
throughout.
We represent the frequency along y-axis (1 unit
5)
and speed of cars (1 unit
10 km/h) along x-axis.
Also, we consider the imagined classes 20-30 and
100-110 with class marks 25 and 105 respectively and
each with frequency 0.
Let us draw a histogram.
Now, we join the midpoints of the top of the
rectangles taking 25 and 105 as the start and end
points respectively, to draw the frequency polygon.
Then, the curve ABCDEFGHI is the required
frequency polygon.
Question: 8
Refer to Q. 7:
Draw the frequency polygon representing the above
data without drawing the histogram.
Solution:
To draw a frequency polygon without histogram, we
need to find the class marks of the classes.
Class mark
lower limit upper limit
2
Now, we can determine the class marks by the above
formula.
Class interval
(km/h)
Class marks
Frequency
20-30
20 30
25
2
0
30-40
30 40
35
2
3
40-50
40 50
45
2
6
50-60
50 60
55
2
25
60 - 70
60 70
65
2
65
70-80
70 80
75
2
50
80-90
80 90
85
2
28
90-100
90 100
95
2
14
100-110
100 110
105
2
0
We represent the frequency along Y axis (1 unit
5)
and class marks (1 unit
10 km/h) along X axis. Now,
plotting the points
25, 0 , 35, 3 , 45, 6 , 55, 25 ,A B C D
65, 65 , 75, 50 , 85, 28 , 95, 14E F G H
And
105, 0I
and joining them through line
segments, we get curve ABCDEFGHI as the frequency
polygon.
Question: 9
Following table gives the distribution of students of
sections A and B of a class according to the marks
obtained by them.
Section A
Section B
Marks
Frequency
Marks
Frequency
0-15
5
0-15
3
15-30
12
15-30
16
30-45
28
30-45
25
45-60
30
45-60
27
60-75
35
60-75
40
75-90
13
75-90
10
Represent the marks of the students of both the
sections on the same graph by two frequency
polygons. What do you observe?
Solution:
To draw a frequency polygon without histogram, we
need to find the class marks of the classes.
Class mark
lower limit upper limit
2
Now, we can determine the class marks by the above
formula.
We represent the frequency along y-axis
(1 unit
5) and class marks (1 unit
15) along x-
axis.
Now, plotting the points for Section A,
A (7.5, 5), B (22.5, 12), C (37.5, 28), D (52.5, 30),
67.5, 35 , 82.5, 13EF
and for section B,
J 7.5, 3 , 22.5, 16 , 37.5, 25 , 52.5, 27 ,G H I
67.5, 40 , 82.5, 10 .KL
Section A
Section B
Marks
Class
marks
Frequenc
y
Marks
Class
mark
Frequen
cy
0 - 15
7.5
5
0 - 15
7.5
3
15 - 30
22.5
12
15 - 30
22.5
16
30 - 45
37.5
28
30 - 45
37.5
25
45 - 60
52.5
30
45 - 60
52.5
27
60 - 75
67.5
35
60 - 75
67.5
40
75 - 90
82.5
13
75 - 90
82.5
10
It is clear from the graph that the maximum marks of
67.5 was scored by 40 students in section B.
Question: 10
The mean of the following distribution is 50.
X
f
10
17
30
53a
50
32
70
7 11a
90
19
Find the value of a and hence the frequencies of 30
and 70.
Solution:
Let us re-arrange the given data table:
x
i
f
i
( f
i
x
i
)
10
17
170
30
53a
150 90a
50
32
1600
70
7 11a
490 770a
90
19
1710
Total
5
1
60 12
i
i
fa
5
1
2800 640
ii
i
f x a
Mean
5
=1
5
=1
ii
i
i
i
fx
f
5
1
5
1
50
2800 640
50
60 12
2800 640 3000 600
640 600 3000 2800
40 200
200
5
40
ii
i
i
i
fx
f
a
a
aa
aa
a
a



So, frequency for
30 5 3 5 5 3 28a
Frequency for
70 7 11 7 5 11 24 .a
Question: 11
The mean marks (out of 100) for boys and girls in an
examination are 70 and 73, respectively. If the mean
marks of all the students in that examination is 71,
find the ratio of the number of boys to the number of
girls.
Solution:
Let the number of boys and girls in the examination be
a and b respectively.
Then, total number of students
.ab
We know that, Sum of all observations
Mean
Number of observations
So, sum of marks obtained by boys in the examination
70 70aa
Similarly, sum of marks obtained by girls in the
examination
73 73bb
Also, it is given that,
Mean marks of all students in the examination
71
Sum of marks obtained by all students in the
examination
71 ()ab
Now,
Sum of marks obtained by all students in the
examination
Sum of marks obtained by boys in the examination
Sum of marks obtained by girls in the examination
71( ) 70 73
71 71 70 73
71 70 73 71
2
2
1
: 2:1
a b a b
a b a b
a a b b
ab
a
b
ab



Question: 12
A total of 25 patients admitted to a hospital were
tested for levels of blood sugar, (mg/dl) and the results
obtained are as follows:
87 71 83 67 85
77 69 76 65 85
85 54 70 68 80
73 78 68 85 73
81 78 81 77 75
Find the mean, median and mode (mg/dl) of the above
data.
Solution:
Let us arrange the data in ascending order:
54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77,
78, 78, 80, 81, 81, 83, 85, 85, 85, 85, 87
Total no. of observation, n = 25
Mean
54 65 67 68 68 69 70
71 73 73 75 76 77 77 78 78
80 81 81 83 85 85 85 85 87
25
1891
75.64
25

Now,
25n
(which is an odd number)
So, median
hth t
25 1
2
13




observation
Hence, median
77
Now,
Mode
observation with maximum frequency
85
(Frequency
4
)