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

)