class 10 maths Statistics questions with solutions

Q 1 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'.

Numbers announced Number of times

0-15 8

15-30 9

30-45 10

45-60 12

60-75 9

Based on the above information, answer the following:

(i) Write the median class.

(30-45)

Numbers announced	Number of times	cf
0-15	8	8
15-30	9	17
30-45	10	27
45-60	12	39
60-75	9	48

Here, n = 48,

then n/2 = 48/2 = 24

$∴$ Median class : 30 – 45

Q 2 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'.

Numbers announced Number of times

0-15 8

15-30 9

30-45 10

45-60 12

60-75 9

Based on the above information, answer the following:

(ii) When first ball was picked up, what was the probability of calling out an even number?

(37/75)

The number of even numbers between 1 to 75 is 37, [i.e., (75 – 1) ÷ 2 = 37]

Prob.(calling out an even number) = 37/75

Q 3 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'.

Numbers announced Number of times

0-15 8

15-30 9

30-45 10

45-60 12

60-75 9

Based on the above information, answer the following:

(iii) Find median of the given data.

(40.5)

$l = 30, c f = 17, f = 10, h = 15$

$Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$

$\Rightarrow Median = 30 + (\frac{24 - 17}{10}) \times 15 = 30 + \frac{7 \times 15}{10} = 30 + 10.5 = 40.5$

Q 4 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'.

Numbers announced Number of times

0-15 8

15-30 9

30-45 10

45-60 12

60-75 9

Based on the above information, answer the following:

(iv) Find mode of the given data.

(51)

Since the highest frequency is 12 which belongs to 45 – 60, therefore modal class is 45 – 60

Here, $l = 45, f_{0} = 10, f_{1} = 12, f_{2} = 9, h = 15$

$Mode = l + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h = 45 + \frac{12 - 10}{24 - 10 - 9} \times 15 = 45 + \frac{2}{5} \times 15 = 45 + 6 = 51$

Q 5 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said ‘BINGO’.

Numbers announced

Number of times

0 – 15

8

15 – 30

9

30 – 45

10

45 – 60

12

60 – 75

9

Based on the above information, answer the following questions:

(i) Write the median class.

Numbers announced	Number of times
0 – 15	8
15 – 30	9
30 – 45	10
45 – 60	12
60 – 75	9

30 – 45
15 – 30
45 – 60
60 – 75

1

2

3

4

(1)

We have,

Numbers announced	Number of times	Cumulative Frequency
0 – 15	8	8
15 – 30	9	17
30 – 45	10	27
45 – 60	12	39
60 – 75	9	48

$H e r e, n = 48 \Rightarrow n / 2 = 48 / 2 = 24 ∴ T h e m e d i a n c l a s s i s 30 - 45$

Q 6 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said ‘BINGO’.

Numbers announced

Number of times

0 – 15

8

15 – 30

9

30 – 45

10

45 – 60

12

60 – 75

9

Based on the above information, answer the following questions:

(ii) What is the modal class?

Numbers announced	Number of times
0 – 15	8
15 – 30	9
30 – 45	10
45 – 60	12
60 – 75	9

15 – 35
45 – 60
60 – 75
0 – 15

1

2

3

4

(2)

Since 12 is the maximum frequency, 45 – 60 is the modal class

Q 7 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said ‘BINGO’.

Numbers announced

Number of times

0 – 15

8

15 – 30

9

30 – 45

10

45 – 60

12

60 – 75

9

Based on the above information, answer the following questions:

(iii) (a) Find median of the given data.

Numbers announced	Number of times
0 – 15	8
15 – 30	9
30 – 45	10
45 – 60	12
60 – 75	9

40.5
35.5
36.5
39.5

1

2

3

4

(1)

The median class is 30 – 45.

$H e r e, n / 2 = 24, l = 30, c f = 17, f = 10, h = 15 ∴ M e d i a n = l + (\frac{\frac{n}{2} - c f}{f}) \times h = 30 + (\frac{24 - 17}{10}) \times 15 = 30 + \frac{105}{10} = 40.5 T h e m e d i a n o f t h e g i v e n d a t a i s 40.5 .$

Q 8 :

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said ‘BINGO’.

Numbers announced

Number of times

0 – 15

8

15 – 30

9

30 – 45

10

45 – 60

12

60 – 75

9

Based on the above information, answer the following questions:

(iii) (b) Find mode of the given data

Numbers announced	Number of times
0 – 15	8
15 – 30	9
30 – 45	10
45 – 60	12
60 – 75	9

50
45
46
51

1

2

3

4

(4)

Here, the maximum frequency is 12 and the class corresponding to this frequency is 45 – 60

$S o, l = 45, f_{1} = 12, f_{0} = 10, f_{2} = 9, h = 15 ∴ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 45 + (\frac{12 - 10}{2 \times 12 - 10 - 9}) \times 15 = 45 + \frac{30}{5} = 51 T h e m o d e o f t h e g i v e n d a t a i s 51 .$

Q 9 :

A stopwatch was used to time a 100-metre race run by a group of students. The sports teacher carefully recorded their times in seconds, which are presented in the table below.

Time (in seconds)

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

Number of students

8

10

13

6

3

Based on the above information, answer the following questions:

(i) Find the lower limit of the modal class.

Time (in seconds)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Number of students	8	10	13	6	3

40
30
25
15

1

2

3

4

(1)

Highest frequency = 13
∴ Modal class = 40 – 60

So, lower limit of the modal class = 40

Q 10 :

A stopwatch was used to time a 100-metre race run by a group of students. The sports teacher carefully recorded their times in seconds, which are presented in the table below.

Time (in seconds)

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

Number of students

8

10

13

6

3

Based on the above information, answer the following questions:

(ii) How many students finished the race after 1 minute?

Time (in seconds)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Number of students	8	10	13	6	3

8
9
7
5

1

2

3

4

(2)

Number of students who finished the race after 1 minute
= 6 + 3 = 9

Q 11 :

A stopwatch was used to time a 100-metre race run by a group of students. The sports teacher carefully recorded their times in seconds, which are presented in the table below.

Time (in seconds)

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

Number of students

8

10

13

6

3

Based on the above information, answer the following questions:

(iii) (a) Find the mean time taken by students to complete the race.

Time (in seconds)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Number of students	8	10	13	6	3

43
44
42
41

1

2

3

4

(1)

We have,

Time	No. of students (f?)	Class Marks (x?)	f?x?
0 – 20	8	10	80
20 – 40	10	30	300
40 – 60	13	50	650
60 – 80	6	70	420
80 – 100	3	90	270
Total	Σf? = 40		Σf?x? = 1720

So, $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{1720}{40} = 43$

$∴ The mean time taken by students to complete the race is 43 seconds.$

Q 12 :

A stopwatch was used to time a 100-metre race run by a group of students. The sports teacher carefully recorded their times in seconds, which are presented in the table below.

Time (in seconds)

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

Number of students

8

10

13

6

3

Based on the above information, answer the following questions:

(iii) (b) Find the modal time taken by students to complete the race.

Time (in seconds)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Number of students	8	10	13	6	3

45
46
42
43

1

2

3

4

(2)

Here l = 40, $f ₁ = 13, f ₀ = 10, f ₂ = 6, h = 20$

$∴ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 40 + (\frac{13 - 10}{2 \times 13 - 10 - 6}) \times 20 = 40 + \frac{3}{10} \times 20 = 40 + 6 = 46$

Q 13 :

A farmer recorded the yield of 200 chilli plants, and the data is presented in the following table:

Chillies per plant

1–7

7–13

13–19

19–25

25–31

Number of plants

15

14

75

87

9

To reduce the excessive use of chemicals on his farm, he plans to buy organic manure, but only if the average number of chillies per plant is less than 19.

Based on the above information, answer the following questions:

(i) How many plants are there whose yield is less than 19 chillies per plant?

Chillies per plant	1–7	7–13	13–19	19–25	25–31
Number of plants	15	14	75	87	9

101
102
100
104

1

2

3

4

(4)

Number of plants whose yield is less than 19 chillies per plant
= 15 + 14 + 75
= 104

Q 14 :

A farmer recorded the yield of 200 chilli plants, and the data is presented in the following table:

Chillies per plant

1–7

7–13

13–19

19–25

25–31

Number of plants

15

14

75

87

9

To reduce the excessive use of chemicals on his farm, he plans to buy organic manure, but only if the average number of chillies per plant is less than 19.

Based on the above information, answer the following questions:

(ii) To help the farmer make his decision, which measure of central tendency (mean, median or mode) will you find out? Calculate its actual value.

Chillies per plant	1–7	7–13	13–19	19–25	25–31
Number of plants	15	14	75	87	9

17.83
17.50
17.60
17.25

1

2

3

4

(1)

Farmer will buy organic manure only if the overall number of chillies per plant is less than 19, i.e., mean number of chillies per plant is less than 19.
We need to find mean of the given data.

Chillies per plant	Number of plants (f?)	Class mark (x?)	f?x?
1–7	15	4	60
7–13	14	10	140
13–19	75	16	1200
19–25	87	22	1914
25–31	9	28	252
Total	Σf? = 200		Σf?x? = 3566

$M e a n (\bar{x}) - \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{3566}{200} = 17.83$

Q 15 :

A farmer recorded the yield of 200 chilli plants, and the data is presented in the following table:

Chillies per plant

1–7

7–13

13–19

19–25

25–31

Number of plants

15

14

75

87

9

To reduce the excessive use of chemicals on his farm, he plans to buy organic manure, but only if the average number of chillies per plant is less than 19.

Based on the above information, answer the following questions:

(iii) (a) Find median of the given data.

Chillies per plant	1–7	7–13	13–19	19–25	25–31
Number of plants	15	14	75	87	9

18.65
18.66
17.66
18.68

1

2

3

4

(4)

Median

We prepare a cumulative frequency table:

Chillies per plant	Number of plants (f?)	Cumulative frequency
1–7	15	15
7–13	14	29
13–19	75	104
19–25	87	191
25–31	9	200 = n

$H e r e, n = 200 \Rightarrow n / 2 = 100 M e d i a n c l a s s = 13 - 19 l = 13, c f = 29, f = 75, h = 6 Now, median = l + (\frac{\frac{n}{2} - c f}{f}) \times h = 13 + (\frac{100 - 29}{75}) \times 6 = 13 + \frac{71}{25} \times 2 = 13 + \frac{142}{25} = 13 + 5.68 = 18.68$

Q 16 :

A farmer recorded the yield of 200 chilli plants, and the data is presented in the following table:

Chillies per plant

1–7

7–13

13–19

19–25

25–31

Number of plants

15

14

75

87

9

To reduce the excessive use of chemicals on his farm, he plans to buy organic manure, but only if the average number of chillies per plant is less than 19.

Based on the above information, answer the following questions:

(iii) (b) Find mode of the given data.

Chillies per plant	1–7	7–13	13–19	19–25	25–31
Number of plants	15	14	75	87	9

18.5
19.4
19.8
19.1

1

2

3

4

(3)

Mode

Highest frequency = 87
Modal class = 19–25

$l = 19, f ₁ = 87, f ₀ = 75, f ₂ = 9, h = 6 Now, mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 19 + (\frac{87 - 75}{2 \times 87 - 75 - 9}) \times 6 = 19 + \frac{12}{90} \times 6 = 19 + 0.8 = 19.8$

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