CBSE Test Paper 01

Chapter 14 Statistics

1. =. (1)

a. Mode
b. Median
c. Mean
d. None of these

2. Construction of cumulative frequency table is useful to determine (1)

a. mean
b. all the three
c. median
d. mode

3. For the following distribution

Class Below 10 Below 20 Below 30 Below 40 Below 50 Below 60

Frequency 3 12 27 57 75 80

the modal class is (1)

a. 50 - 60
b. 40 - 50
c. 20 - 30
d. 30 - 40

4. The mean of the first 10 natural numbers is (1)

a. 4.5
b. 5
c. 6
d. 5.5
 5. The marks obtained by 9 students in Mathematics are 59, 46, 30, 23, 27, 44, 52, 40 and
29. The median of the data is (1)

a. 35
b. 29
c. 30
d. 40

6. Find the mode of the given data 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4. (1)

7. If the median of a series exceeds the mean by 3, find by what number the mode
exceeds its mean? (1)

8. If the values of mean and median are 26.4 and 27.2, what will be the value of mode?
(1)

9. In the following frequency distribution, find the median class. (1)

Height (in cm) 140 -145 145-150 150-155 155 -160 160 -165 165 -170

Frequency 5 15 25 30 15 10

10. Find median of the data, using an empirical relation when it is given that Mode = 12.4
and Mean = 10.5. (1)

11. Find the mode of the following distribution. (2)

Class
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
interval

Frequency 5 8 7 12 28 20 10 10

12. Convert the following data into 'more than type' distribution: (2)

Class 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 75 - 80

Frequency 2 8 12 24 38 16

13. Calculate the mean of the following data, using direct method: (2)
 Class 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75

Frequency 6 10 8 12 4

14. If the median of the following frequency distribution is 46, find the missing
frequencies. (3)

Variable 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total

Frequency 12 30 ? 65 ? 25 18 229

15. Find median for the following data: (3)

Wages(in Rs) Number of workers

More than 150 Nil

More than 140 12

More than 130 27

More than 120 60

More than 110 105

More than 100 124

More than 90 141

More than 80 150

16. Draw a pie-chart for the following data of expenditure on various items in a family.

Item Education Food Rent Clothing Others

Expenditure (in Rs.) 1600 3200 4000 2400 3200

17. Find the mean and mode of the following frequency distribution: (3)

Classes 0 - 10 10 -20 20 -30 30 -40 40 -50 50 -60 60 -70

Frequency 3 8 10 15 7 4 3
18. From the following frequency distribution, prepare the 'more than' ogive. (4)

Score Number of candidates

400 - 450 20

450 - 500 35

500 - 550 40

550 - 600 32

600 - 650 24

650 - 700 27

700 - 750 18

750 - 800 34

Total 230

Also, find the median.

19. Find the mean marks of students from the following cumulative frequency
distribution: (4)

Marks Number of students

0 and above 80

10 and above 77

20 and above 72

30 and above 65

40 and above 55

50 and above 43

60 and above 28

70 and above 16

80 and above 10
 90 and above 8

100 and above 0

20. The marks obtained by 100 students of a class in an examination are given below:

Marks Number of students

0-5 2

5 - 10 5

10 - 15 6

15 - 20 8

20 - 25 10

25 - 30 25

30 - 35 20

35 - 40 18

40 - 45 4

45 - 50 2

Draw cumulative frequency curves by using (i) 'less than' series and (ii) 'more than'
series.
Hence, find the median. (4)
 CBSE Test Paper 01
Chapter 14 Statistics

Solution

1. b. Median
Explanation: Since, 3 Median = Mode + 2 Mean
Median =
Median =
Median =

2. c. median
Explanation: A cumulative frequency distribution is the sum of the class and
all classes below it in a frequency distribution. Construction of cumulative
frequency table is useful to determine Median.

3. d. 30 – 40
Explanation: According to the question,

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

Freq 3 9 15 30 18 5

Here Maximum frequency is 30.

Therefore, the modal class is 30 – 40.

4. d. 5.5
Explanation: The first 10 natural numbers are 1, 2, 3, …………, 10
Mean =
=
= = 5.5

5. d. 40
Explanation: Arranging the given data in ascending order: 23, 27, 29, 30, 40, 44,
46, 52, 59
Here n = 9, which is even.

Median =

=
 = 5th term = 40
6.

Value x 3 4 5 6 7 8 9

Frequency f 5 2 4 2 2 1 2

We observe that the value 3 has the maximum frequency i.e 5 .

The mode of data is 3.

7. Given,

Since,

Mode = Mean + 9
Hence Mode exceeds Mean by 9.

8. We know that
Mode = 3 median -2 mean
= 3(27.2) - 2(26.4)
= 81.6 - 52.8 = 28.8
Mode = 28.8

9.

Height Frequency c.f.

f = 100
 th term = = 50th term
Hence, Median class is 155 - 160.

10. Mode = 3 median - 2 mean

Mode = and mean =
Median = Mode + Mean
= (12.4) + (10.5)

So, median is 11.13.

11.

Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency 5 8 7 12 28 20 10 10

Here the maximum frequency is 28 then the corresponding class 40 - 52 is the modal
class
l = 40, h = 50 - 40 = 10, f = 28, f1 = 12, f2 = 20

Mode

= 40 + 6.67
= 46.67
12.

Class Frequency Cumulative Frequency

More than 50 2 98 + 2 = 100

More than 55 8 90 + 8 = 98

More than 60 12 78 + 12 = 90

More than 65 24 54 + 24 = 78

More than 70 38 16 + 38 = 54
 More than 75 16 16

13.

Class Interval Frequency Class mark

25 - 35 6 30 180

35 - 45 10 40 400

45 - 55 8 50 400

55 - 65 12 60 720

65 - 75 4 70 280

from table ,
,
we know that,

mean =

14. Let the frequency of the class 30 - 40 be f1 and that of the class 50 - 60 be f2. The total

frequency is 229.
12 + 30 + f1 + 65 + f2 + 25 + 18 =229

f1 + f2 =79

It is given that the median is 46

Clearly, 46 lies in the class 40 - 50. So, 40 - 50 is the median class.
and

N=229
 Since ,

=45
Hence, f1 = 34 and f2 =45

15.

C.I. f c.f.

80 - 90 9 9

90 - 100 17 26

100 - 110 19 45

110 - 120 45 90

120 - 130 33 123

130 - 140 15 138

140 - 150 12 150

Median Class

we know that, Median =

16.

Item Expenditure (Ei) Central angle =

Education 1600

Food 3200

Rent 4000

Clothing 2400
 Others 3200

17.

Class interval

5 3

15 8

25 10

35 15

45 7

55 4

65 3

Mean =

Mean = 32.8
For Mode, Modal class = 30 - 40
and

Mode =

=
 =
=
=
Mean of given data is 32.8 and mode is 33.85.
18. More than series:

Score Number of candidates

More than 400 230

More than 450 210

More than 500 175

More than 550 135

More than 600 103

More than 650 79

More than 700 52

More than 750 34

plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700,
52), (750, 34).

Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw
PM x-axis intersecting x-axis at M
 Hence, median
19. Here we have, the cumulative frequency distribution.
So, first we convert it into an ordinary frequency distribution.
We observe that there are 80 students getting marks greater than or equal to 0 and 77
students have secured 10 and more marks.
Therefore, the number of students getting marks between 0 and 10 is 80 - 77 = 3.
Similarly, the number of students getting marks between 10 and 20 is 77 - 72 = 5 and
so on.

Marks Mid-value (xi) Frequency (fi) fiui

0-10 5 3 -5 -15

10-20 15 5 -4 -20

20-30 25 7 -3 -21

30-40 35 10 -2 -20

40-50 45 12 -1 -12

50-60 55 15 0 0

60-70 65 12 1 12

70-80 75 6 2 12

80-90 85 2 3 6

90-100 95 8 4 32

Total

Let assumed mean (a) = 55.

We have,
a = 55 and h = 10

Therefore, the mean number of marks is

20. i. Less than series:
 Marks Number of students

Less than 5 2

Less than 10 7

Less than 15 13

Less than 20 21

Less than 25 31

Less than 30 56

Less than 35 76

Less than 40 94

Less than 45 98

Less than 50 100

Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45,
98) and (50, 100).
Join these points free hand to get the "less than" cumulative curve.
ii. 'more than' series:

Marks Number of students

More than 45 2

More than 40 6

More than 35 24

More than 30 44

More than 25 69

More than 20 79

More than 15 87

More than 10 93

More than 5 98

More than 0 100

Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93),
(15, 87), (20, 79), (25, 69), (30, 44), (35, 24) , (40, 6) and (45, 2)

N = 100
Two curves intersect at Point P(28, 50)
Hence, median = 28