class 10 maths Statistics questions with solutions

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

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1

2

3

4

Enter Answer here

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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