Access Answers for SAT Maths Chapter 14 – Statistics
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 f i and xi are
small.
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
No. of plants No. of houses Mid-point (xi) fixi
Frequency (fi)
(Class interval)
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
Sum fi = 20 Sum fixi = 162
The formula to find the mean is:
Mean = x̄ = ∑fi xi /∑fi
= 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 (in Rs.) 100-120 120-140 140-160 160-180 180-200
Number 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 (xi) is very large, so let us assume the mean value, A = 150 and
class interval is h = 20.
So, ui = (xi – A)/h = ui = (xi – 150)/20
Substitute and find the values as follows:
Daily wages Number of workers Mid-point (xi) ui = (xi – 150)/20 fiui
frequency (fi)
(Class interval)
100-120 12 110 -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
Total Sum fi = 50 Sum fiui = -12
So, the formula to find out the mean is:
Mean = x̄ = A + h∑fiui /∑fi =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 23-35
, 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 (xi) mean x̄ = 18
Class interval Number of children (fi) Mid-point (xi) fixi
11-13 7 12 84
13-15 6 14 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
Total fi = 44+f Sum fixi = 752+20f
The mean formula is
Mean = x̄ = ∑fixi /∑fi = (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+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 – 752 = 20f – 18f
, ⇒ 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.
Number of heart beats per minute 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
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the ui and fiui as follows:
Class Interval Number of women (fi) Mid-point (xi) ui = (xi – 75.5)/h fiu i
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
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 f i and xi are
small.
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
No. of plants No. of houses Mid-point (xi) fixi
Frequency (fi)
(Class interval)
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
Sum fi = 20 Sum fixi = 162
The formula to find the mean is:
Mean = x̄ = ∑fi xi /∑fi
= 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 (in Rs.) 100-120 120-140 140-160 160-180 180-200
Number 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 (xi) is very large, so let us assume the mean value, A = 150 and
class interval is h = 20.
So, ui = (xi – A)/h = ui = (xi – 150)/20
Substitute and find the values as follows:
Daily wages Number of workers Mid-point (xi) ui = (xi – 150)/20 fiui
frequency (fi)
(Class interval)
100-120 12 110 -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
Total Sum fi = 50 Sum fiui = -12
So, the formula to find out the mean is:
Mean = x̄ = A + h∑fiui /∑fi =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 23-35
, 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 (xi) mean x̄ = 18
Class interval Number of children (fi) Mid-point (xi) fixi
11-13 7 12 84
13-15 6 14 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
Total fi = 44+f Sum fixi = 752+20f
The mean formula is
Mean = x̄ = ∑fixi /∑fi = (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+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 – 752 = 20f – 18f
, ⇒ 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.
Number of heart beats per minute 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
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the ui and fiui as follows:
Class Interval Number of women (fi) Mid-point (xi) ui = (xi – 75.5)/h fiu i
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
