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Maths 2 Marks Questions

Statistics · Maharashtra Board STD 10 (MSBSHSE - English Medium). 13 questions.

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Question 12 Marks
The monthly profits (in Rs.) of 100 shops are distributed as follows:
Profits per shop 0-50 50-100 100-150 150-200 200-250 250-300
No. of shops 12 18 27 20 17 6
Draw the frequency polygor it.
Answer
Profits per shop
No. of shops (f)
c.f.
0-50
12
12
50-100
18
30
100-150
27
57
150-200
20
77
200-250
17
94
250-300
6
100
Represent profits per shop along x-axis and no. of shop (c.f.) along y-axis.
Plot the points (50, 12), (100, 30), (150, 57), (200, 77), (250, 94) and (300, 100) on the graph and join them with ruler. This is the cumulative polygon as shown.
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Question 22 Marks
Find the mode of the following distribution.
Class-interval
25-30
30-35
35-40
40-45
45-50
50-60
Frequency
25
34
50
42
38
14
Answer
Class interval
25-30
30-35
35-40
40-45
45-50
50-60
No. of persons
25
34
50
42
38
14
Here the maximum frequency is 28 then the corresponding class 40-45 is the model class
$\text{L}=35,\text{h}=40-35=5,\text{f}=50,\text{f}_1=34,\text{f}_2=42$
$\text{Model}=\text{L}+\frac{\text{f}-\text{h}}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=35+\frac{50-34}{2(50)-34-42}\times5$
$=35+\frac{16\times5}{24}$
$=35+3.33$
$=35.33$
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Question 32 Marks
Find the mode of the following distribution:
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
Answer
Class interval
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
No.of persons
5
8
7
12
28
20
10
10
Here the maximum frequency is 28 then the corresponding class 40-50 is the model class
$\text{L}=40,\text{h}=50-40=10,\text{f}=28,\text{f}_1=12,\text{f}_2=20$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=40+\frac{28-12}{2\times28-12-20}\times10$
$=40+\frac{16\times10}{24}$
$=40+16.67=46.67$
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Question 42 Marks
n the graphical representation of a frequency distribution, if the distance between mode and mean isk times the distance between median and mean, then write the value of k.
Answer
We know that
Mode = 3 median – 2 mean ….(i)
Now mode – mean = k (median – mean), ….(ii)
But mode – mean = 3 median – 2 mean [from (i)]
⇒ Mode – mean = 3 (median – mean) ….(iii)
Comparing (ii) and (iii)
k = 3
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Question 52 Marks
Find the mode of the following distribution.
Class-interval
10-15
15-20
02-25
25-30
30-35
35-40
Frequency
30
45
75
35
25
15
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Question 62 Marks
find the class marks of classes 10-25 and 35-55.
Answer
We know that
Class mark $=\frac{\text{Sum of its limits}}{2}$
$\therefore$ Class mark of 10-25 $=\frac{10+25}{2}=\frac{35}{2}=17.5$
and class mark of 35-55 $=\frac{35+55}{2}=\frac{90}{2}=45$
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Question 72 Marks
Draw an ogive to represent the following frequency distribution:
Class-interval
0-4
5-9
10-14
15-19
20-24
No. of students
2
6
10
5
3
Answer
Firstly, prepare the cumulative frequency table.
Class interval
No. of students
Less than
Cumulative frequency
Suitable points
0-4
2
4
2
(4,2)
5-9
6
9
8
(9,8)
10-14
10
14
18
(14,18)
15-19
5
19
23
(19,23)
20-24
3
24
26
(24,26)
Now, plot the less than ogive using the suitable points.
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Question 82 Marks
Calculate the mean for the following distribution:
x 5 6 7 8 9
f 4 8 14 11 3
Answer
x f fx
5 4 20
6 8 48
7 14 98
8 11 88
9 3 27
  N = 40 $\sum\text{fx}=281$
$\text{Mean}\ \bar{\text{X}}=\frac{\sum\text{fx}}{\text{N}}$
$\bar{\text{X}}=\frac{281}{40}=7.025$
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Question 92 Marks
A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.
Answer
Median marks
Here N = 40, then $\frac{\text{N}}{2}=\frac{40}{2}=20$
From 20 on y-axis, draw a line parallel to the x-axis meeting the curve at P and from P, draw a perpendicular on x-axis meeting it at M. Then M is the median which is 50.
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Question 102 Marks
Write the median class of the following distribution:
Classes
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
4
4
8
10
12
8
4
Answer
Consider the following distribution table.
Class
Frequency
C.F.
0-10
4
4
10-20
4
8
20-30
8
16
30-40
10
26
40-50
12
38
50-60
8
46
60-70
4
50
 
N = 50
  
Here,
N = 50
$\frac{\text{N}}{2}=25$
The cumulative frequency just greater than 25 is 26.
So, the meian class is 30-40
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Question 112 Marks
Write the median class for the following frequency distribution:
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
Answer
We are given the following table.
Class interval
Frequency
Cumulative Frequency
0-10
5
5
10-20
8
13
20-30
7
20
30-40
12
32
40-50
28
60
50-60
20
80
60-70
10
90
70-80
10
100
 
N = 100
  
Here, N = 100
$\therefore\frac{\text{N}}{2}=50$
The cumulative frequency just greater than 50 is 60.

So, the median class is 40−50.
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Question 122 Marks
The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:
Shirt size 37 38 39 40 41 42 43 44
Number of persons 15 25 39 41 36 14 15 12
Find the model shirt size worn by the group.
Answer
Shirt size
37
38
39
40
41
42
43
44
Number of persons
15
25
39
41
36
17
15
12
We see that frequency of 40 is maximum which is 41
$\therefore\ \text{Mode}=40$
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Question 132 Marks
The following table gives production yield per hectare of wheat of 100 farms of a village:
Production yield
50-55
55-60
60-65
65-70
70-75
75-80 in kg per hecctare
Number of farms
2
8
12
24
38
16
Draw ‘less than’ ogive and ‘more than’ ogive.
Answer
Prepare a table for less than type
Production yield
No. of farms
Production yild (less than)
Cumulative frequency
Suitable points
50-55
2
55
2
(55, 2)
55-60
8
60
10
(60, 10)
60-65
12
65
22
(65, 22)
65-70
24
70
46
(70, 46)
70-75
38
75
84
(75, 84)
75-80
16
80
100
(80, 100)
Now, plot the less than ogive using suitable points.

Again, prepare a table for more than type.
Production yield
No. of farms
Production yield (more than)
Cumulative frequency
Suitable points
50-55
2
50
100
(50, 100)
55-60
8
55
98
(55, 98)
60-65
12
60
90
(60, 90)
65-70
24
65
78
(65, 78)
70-75
38
70
54
(70, 54)
75-80
13
75
16
(75, 16)
Now, plot the more than ogive with suitable points.
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2 Marks Questions - Maths STD 10 Questions - Vidyadip