EXERCISE 14.1

CLASS 10 MATHEMATICS (CH) 14 STATISTICS EXERCISE 14.1
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MEAN OF GROUPED DATA : (With Class-Interval)
When the data is grouped in the form of class interval then the mean can be calculated by three methods.
1. Direct Method
In this method, we use a midpoint which represents the whole class. It is called the class mark. It is the average of the upper limit and the lower limit.
Class Mark = (Upper Class Limit −Lower Class Limit)/2
x = (∑f1 x_i )/(∑f1 )
Example:
A teacher marks the test result of the class of 55 students for mathematics. Find the mean for the given group.

 

Marks of Students
0 – 10
10–20
20–30
30–40
40–50
50–60
Frequency
27
10
7
5
4
2

 

To find the mean we need to find the mid-point or class mark for each class interval which will
be the x and then by multiplying frequency and midpoint we get fx.

Marks of Students
Frequency(f)
Mid point(x)
fx
0–10
27
5
135
10–20
10
15
150
20–30
7
25
175
30–40
5
35
175
40–50
4
45
180
50–60
2
55
110
∑f = 55
 
 
∑fx = 925

 

 

x = (f1 x1 + f2 x2 + f3 x3 ……fnx1)/(f1 + f2 + f3 ……fn )
x = 925/55 = 16.8 Marks

2. Deviation or Assumed Mean Method:
If we have to calculate the large numbers then we can use this method to make our calculations easy. In this method, we choose one of the x’s as assumed mean and let it be “a”. Then we find the deviation which is the difference of assumed mean and each of the x. The rest of the method is the same as the direct method.
x = a + (∑f1 d1 )/(∑f1)

Example : If we have the table of the expenditure of the company’s workers in the household, then what will be the mean of their expenses in Rs?

Expenses
100-150
150-200
200-250
250–300
300–350
350–400
400–450
Frequency
24
40
33
28
30
22
33



Solution: As we can see that there are big values of x to calculate so we will use the assumed
mean method .Here we take 275 as the assumed mean.

Expenses (RS.)
Frequency (f)
Mid value(x)
d = x – 275
fd
100-150
24
125
-150
-3600
150-200
40
175
-100
-4000
200-250
36
225
-50
-1650
250-300
28
275
0
0
300-350
20
325
50
1500
350-400
22
375
100
2200
 
∑ f =180
 
 
∑ f d= −5550

 

x = 275 + (−5550)/180
= 275 − 30.83
= 244.17
3. Step Deviation Method:
In this method, we divide the values of d with a number “h” to make our calculations easier.
x = a + ((∑f1 u1 )/(∑f1 )) × h
Example : The wages of the workers are given in the table. Find the mean by step deviation method.

Wages
10-20
20–30
30–40
40–50
50–60
No. of workers
8
9
12
11
6

 

Solution:
Wages No. of workers (f)

Class interval
f1
Midpoint (xi)
d = xi – a
u = (x – a)/h
fiui
10-20
8
15
-20
-2
-16
20-30
9
25
-10
-1
-9
30-40
12
35
0
0
0
40-50
11
45
10
1
11
50-60
6
55
20
2
12
 
∑f= 46
 
 
 
∑fu = -2

 

x = a + (∑f1 u1)/(∑f1 )) × h
x = 35 + (−2)/46) × 10 = 34.57

EXERCISE 14.1
1.A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants
0-2
2-4
4-6
6-8
8-10
10-12
12-14
Number of Houses
1
2
1
5
6
2
3

 

Which method did you use for finding the mean, and why?
Solution: In order to find the mean value, we will use direct method because the numerical value of fi and xi are small.
Find the midpoint of the given interval using the formula.
Midpoint xi = (Upper Limit + Lower Limit)/2

Class Interval
Frequency(fi)
Mid point(xi)
fixi
0–2
1
1
1
2-4
2
3
6
4-6
1
5
5
6-8
5
7
35
8-10
6
9
54
10-12
2
11
22
12-14
3
13
39
 
∑fi = 20
 
∑fixi =162

∑ fi = 20 ∑ fixi = 162

x = (∑f1 x1 )/(∑f1 )
= 162/20
= 8.1
Therefore, the mean number of plants per house is 8.1

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily Wages (Rs)
100-120
120–140
140–160
160–180
180–200
No. of workers
12
14
8
6
10

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution: Find the midpoint of the given interval using the formula.
Midpoint (xi) = (Upper Limit + Lower Limit)/2
In this case, the value of mid-point (x1) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.
So, u1 = (x1−A)/h = u1 = (x1 − 150)/20
Substitute and find the values as follows:

Daily Wages (class interval)
Number of workers Frequency (f1)
Mid point(x1)
d = u1 = (x1 – 150/20)
f1u1
100-120
12
110
d = -2
-24
120-140
14
130
-1
-14
140-160
8
150
0
0
160-180
6
170
1
6
180-200
10
190
2
20
 
∑f1 = 50
 
 
∑ f1u1-12

So, the formula to find out the mean is:

Mean = x = A  = (∑f1 x1 )/(∑f1 )
= 150 + ( 20 × (−12)/50 )
= 150 – 4.8 = 145.20

Thus, mean daily wage of the workers = Rs 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily pocket allowance(in C)
11-13
13-15
15-17
17-19
19-21
21-23
23-25
Number of children
7
6
9
13
f
5
4

Solution: To find out the missing frequency, use the mean formula.
Here, the value of mid-point (x1) mean x̄ = 18

Class interval
Number of children(fi)
Mid point(xi)
fixi
11-13
7
12
84
15-17
9
16
144
17-19
13
18 = A
234
19-21
f
20
20f
21-23
5
22
110
23-25
4
24
96
 
∑fi = 44 + f 
 
∑fixi = 752 + 20f

The mean formula is
Mean = x̄ = (∑f1 x1 )/(∑f1 )
= (752 + 20f))/(44 + f)
Now substitute the values and equate to find the missing frequency (f)
⇒ 18 = (752 + 20f)/(44 + f) 
⇒ 18 (44 + f ) = ( 752 + 20 f )
⇒ 792 + 18 f = 752 + 20 f
⇒ 792 + 18 f = 752 + 20 f
⇒ 792 – 752 = 20 f – 18 f
⇒ 40 = 2f
⇒ f = 20
So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

No.of heartbeats per minutes
65-68
68-71
71-74
74-77
77-80
80-83
83-86
Number of women
2
4
3
8
7
4
2

Solution: From the given data, let us assume the mean as A = 75.5
x1 = (Upper Limit+Lower Limit)/2
Class size (h) = 3
Now, find the u1 and f1 u1 as follows:

class interval
Number of women (f1)
Mid point(x1)
u1 = (x1 – 75.5/h)
f1u1
65-68
2
66.5
-3
-6
68-71
4
69.5
-2
-8
71-74
3
72.5
-1
-3
74-77
8
75.5
0
0
77-80
7
78.5
1
7
80-83
4
81.5
3
8
83-86
2
84.5
3
6
 
∑ f1 =30
 
 
∑ f1u1 = 4

Mean = x = A + (∑f1 u1 )/(∑f1 )
= 75.5 + ( 3 × 4/30 )
= 75.5 + 4/10
= 75.5 + 0.4
= 75.9
Therefore, the mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

No. of Mangoes
50-52
53-55
56-58
59-61
62-64
No. of boxes
15
110
135
115
25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution: Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1
Here, assumed mean (A) = 57
Class size (h) = 3
Here, the step deviation is used because the frequency values are big.

class interval
Number of boxes(f1)
Mid point(x1)
d1 = (x1 – A)
f1d1
49.5-52.5
15
51
-6
90
52.5-55.5
110
54
-3
-330
55.5-58.5
135
57= A
0
0
58.5-61.5
115
60
3
345
61.5-64.5
25
63
6
150
 
∑(f1)= 400
 
 
∑ f1d1= 75

The formula to find out the Mean is:
Mean = x̄ = A + h (∑f1 u1 )/(∑f1 )
= 57 + 3 ( 75/400 )
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Daily Expenditure (in c)
100-150
150–200
200–250
250–300
300–350
No. of households
4
5
12
2
2

 

Solution: Find the midpoint of the given interval using the formula.
Midpoint x1 = (Upper Limit+Lower Limit)/2
Let is assume the mean (A) = 225
Class size (h) = 50 u

Class interval
Number of houses (f1)
Midpoint (x1)
d1 = x1 – A
u1 = d1/50
f1u1
100-150
4
125
-100
-2
-8
150-200
5
175
-50
-1
-5
200-250
12
225
0
0
0
250-300
2
275
50
1
2
300-350
2
325
100
2
4
 
∑f1 = 25
 
 
 
∑ fiui = -7

Mean = x̄ = A + h (∑〖f1 u1 )/(∑f1 )
= 225 + 50 ( (− 7)/25 )
= 225 −14
= 211
Therefore, the mean daily expenditure on food is Rs 211

7. To find out the concentration of SO2 in the air in parts per million, (i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO2 (in ppm)
Frequency
0.00-0.04
4
0.04-0.08
9
0.08-0.12
9
0.16-0.20
4
0.20-0.24
2

Find the mean concentration of SO2 in the air.
Solution : To find out the mean, first find the midpoint of the given frequencies as follows:

conclusion of S04
Frequency(fi)
Mid point(xi)
fixi
0.00-0.04
4
0.02
0.08
0.04-0.08
9
0.06
0.54
0.08-0.12
9
0.10
0.90
0.12-0.16
2
0.14
0.28
0.16-0.20
4
0.18
0.72
0.20-0.24
2
0.20
0.40
 
∑fi= 30
 
∑fixi = 2.96

The formula to find out the mean is
Mean = x̄ = (∑f1 x1 )/(∑f1 )
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 in air is 0.099 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

No.of days
0-6
6-10
10-14
14-20
20-28
28-38
38-40
Number of Students
11
10
7
4
4
3
1

Solution: Find the midpoint of the given interval using the formula.
Midpoint (x1) = (Upper Limit+Lower Limit)/2

Class interval
Frequency(fi)
Mid point(xi)
fixi
0-6
11
3
33
6-10
10
8
80
10-14
7
12
84
14-20
4
17
68
20-28
4
24
96
28-38
3
33
99
38-40
1
39
39
 
∑fi = 40 
 
∑fixi = 499

The mean formula is,
Mean = x̄ = (∑f1 x1)/(∑f1 )
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.

Literacy rate in (%)
45-55
55-65
65-75
75-85
85-95
No. of cities
3
10
11
8
3

Solution: Find the midpoint of the given interval using the formula.
Midpoint x_i = (Upper Limit + Lower Limit)/2
In this case, the value of mid-point ( xi)is very large, so let us assume the mean value, A = 70 and class interval is h = 10.
So, u1 = (x1−A)/h = u1= (x1−70)/10
Substitute and find the values as follows:

Class interval
(fi)
Mid point(xi)
d = (xi – A)
ui = (d – 1/h)
fiui
45-55
3
50
-20
-2
-6
55-65
10
60
-10
-1
-10
65-75
11
70
0
0
0
75-85
8
80
10
1
8
85-95
3
90
20
2
6
 
∑fi = 35
 
 
 
∑fiui = -2

Sum (f1) = 35 Sum
f1 u1 = −2

So, Mean = x̄ = A + h (∑f1 u1)/(∑f1 )
= 70 + 10 ( (− 2)/35 )
= 69.42
Therefore, the mean literacy part = 69.42

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