This exercise helps students to find the arithmetic mean of a continuous frequency distribution. The RD Sharma Solutions Class 10 provides a stepwise solution to problems so that students attain strong conceptual knowledge. All the solutions are consistent with the latest CBSE pattern. Further, students can access the RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.3, PDF provided below.
RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.3
Access RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.3
1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.
| Expenditure (in rupees) (x) | Frequency (fi) | Expenditure (in rupees) (xi) | Frequency (fi) |
| 100 – 150 | 24 | 300 – 350 | 30 |
| 150 – 200 | 40 | 350 – 400 | 22 |
| 200 – 250 | 33 | 400 – 450 | 16 |
| 250 – 300 | 28 | 450 – 500 | 7 |
Find the average expenditure (in rupees) per household.
Solution:
Let the assumed mean (A) = 275
| Class interval | Mid value (xi) | di = xi – 275 | ui = (xi – 275)/50 | Frequency fi | fiui |
| 100 – 150 | 125 | -150 | -3 | 24 | -72 |
| 150 – 200 | 175 | -100 | -2 | 40 | -80 |
| 200 – 250 | 225 | -50 | -1 | 33 | -33 |
| 250 – 300 | 275 | 0 | 0 | 28 | 0 |
| 300 – 350 | 325 | 50 | 1 | 30 | 30 |
| 350 – 400 | 375 | 100 | 2 | 22 | 44 |
| 400 – 450 | 425 | 150 | 3 | 16 | 48 |
| 450 – 500 | 475 | 200 | 4 | 7 | 28 |
| N = 200 | Σ fiui = -35 |
It’s seen that A = 275 and h = 50
So,
Mean = A + h x (Σfi ui/N)
= 275 + 50 (-35/200)
= 275 – 8.75
= 266.25
2. A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 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 house: | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Which method did you use for finding the mean, and why?
Solution:
From the given data,
To find the class interval, we know that,
Class marks (xi) = (upper class limit + lower class limit)/2
Now, let’s compute xi and fixi by the following
| Number of plants | Number of house (fi) | 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 |
| Total | N = 20 | Σ fiui = 162 |
Here,
Mean = Σ fiui / N
= 162/ 20
= 8.1
Thus, the mean number of plants in a house is 8.1
We have used the direct method as the values of class mark xi and fi are very small.
3. Consider the following distribution of daily wages of workers of a factory
| Daily wages (in ₹) | 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:
Let the assume mean (A) = 150
| Class interval | Mid value xi | di = xi – 150 | ui = (xi – 150)/20 | Frequency fi | fiui |
| 100 – 120 | 110 | -40 | -2 | 12 | -24 |
| 120 – 140 | 130 | -20 | -1 | 14 | -14 |
| 140 – 160 | 150 | 0 | 0 | 8 | 0 |
| 160 – 180 | 170 | 20 | 1 | 6 | 6 |
| 180 – 200 | 190 | 40 | 2 | 10 | 20 |
| N= 50 | Σ fiui = -12 |
It’s seen that,
A = 150 and h = 20
So,
Mean = A + h x (Σfi ui/N)
= 150 + 20 x (-12/50)
= 150 – 24/5
= 150 = 4.8
= 145.20
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute 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:
Using the relation (xi) = (upper class limit + lower class limit)/ 2
And, class size of this data = 3
Let the assumed mean (A) = 75.5
So, let’s calculate di, ui, fiui as follows:
| Number of heart beats per minute | Number of women (fi) | xi | di = xi – 75.5 | ui = (xi – 755)/h | fiui |
| 65 – 68 | 2 | 66.5 | -9 | -3 | -6 |
| 68 – 71 | 4 | 69.5 | -6 | -2 | -8 |
| 71 – 74 | 3 | 72.5 | -3 | -1 | -3 |
| 74 – 77 | 8 | 75.5 | 0 | 0 | 0 |
| 77 – 80 | 7 | 78.5 | 3 | 1 | 7 |
| 80 – 83 | 4 | 81.5 | 6 | 2 | 8 |
| 83 – 86 | 2 | 84.5 | 9 | 3 | 6 |
| N = 30 | Σ fiui = 4 |
From the table, it’s seen that
N = 30 and h = 3
So, the mean = A + h x (Σfi ui/N)
= 75.5 + 3 x (4/30
= 75.5 + 2/5
= 75.9
Therefore, the mean heartbeats per minute for those women are 75.9 beats per minute.
Find the mean of each of the following frequency distributions: (5 – 14)
5.
| Class interval: | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |
| Frequency: | 6 | 8 | 10 | 9 | 7 |
Solution:
Let’s consider the assumed mean (A) = 15
| Class interval | Mid – value xi | di = xi – 15 | ui = (xi – 15)/6 | fi | fiui |
| 0 – 6 | 3 | -12 | -2 | 6 | -12 |
| 6 – 12 | 9 | -6 | -1 | 8 | -8 |
| 12 – 18 | 15 | 0 | 0 | 10 | 0 |
| 18 – 24 | 21 | 6 | 1 | 9 | 9 |
| 24 – 30 | 27 | 12 | 2 | 7 | 14 |
| N = 40 | Σ fiui = 3 |
From the table, it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σfi ui/N)
= 15 + 6 x (3/40)
= 15 + 0.45
= 15.45
6.
| Class interval: | 50 – 70 | 70 – 90 | 90 – 110 | 110 – 130 | 130 – 150 | 150 – 170 |
| Frequency: | 18 | 12 | 13 | 27 | 8 | 22 |
Solution:
Let’s consider the assumed mean (A) = 100
| Class interval | Mid – value xi | di = xi – 100 | ui = (xi – 100)/20 | fi | fiui |
| 50 – 70 | 60 | -40 | -2 | 18 | -36 |
| 70 – 90 | 80 | -20 | -1 | 12 | -12 |
| 90 – 110 | 100 | 0 | 0 | 13 | 0 |
| 110 – 130 | 120 | 20 | 1 | 27 | 27 |
| 130 – 150 | 140 | 40 | 2 | 8 | 16 |
| 150 – 170 | 160 | 60 | 3 | 22 | 66 |
| N= 100 | Σ fiui = 61 |
From the table, it’s seen that,
A = 100 and h = 20
Mean = A + h x (Σfi ui/N)
= 100 + 20 x (61/100)
= 100 + 12.2
= 112.2
7.
| Class interval: | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |
| Frequency: | 6 | 7 | 10 | 8 | 9 |
Solution:
Let’s consider the assumed mean (A) = 20
| Class interval | Mid – value xi | di = xi – 20 | ui = (xi – 20)/8 | fi | fiui |
| 0 – 8 | 4 | -16 | -2 | 6 | -12 |
| 8 – 16 | 12 | -8 | -1 | 7 | -7 |
| 16 – 24 | 20 | 0 | 0 | 10 | 0 |
| 24 – 32 | 28 | 8 | 1 | 8 | 8 |
| 32 – 40 | 36 | 16 | 2 | 9 | 18 |
| N = 40 | Σ fiui = 7 |
From the table, it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 8 x (7/40)
= 20 + 1.4
= 21.4
8.
| Class interval: | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |
| Frequency: | 7 | 5 | 10 | 12 | 6 |
Solution:
Let’s consider the assumed mean (A) = 15
| Class interval | Mid – value xi | di = xi – 15 | ui = (xi – 15)/6 | fi | fiui |
| 0 – 6 | 3 | -12 | -2 | 7 | -14 |
| 6 – 12 | 9 | -6 | -1 | 5 | -5 |
| 12 – 18 | 15 | 0 | 0 | 10 | 0 |
| 18 – 24 | 21 | 6 | 1 | 12 | 12 |
| 24 – 30 | 27 | 12 | 2 | 6 | 12 |
| N = 40 | Σ fiui = 5 |
From the table, it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σfi ui/N)
= 15 + 6 x (5/40)
= 15 + 0.75
= 15.75
9.
| Class interval: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency: | 9 | 12 | 15 | 10 | 14 |
Solution:
Let’s consider the assumed mean (A) = 25
| Class interval | Mid – value xi | di = xi – 25 | ui = (xi – 25)/10 | fi | fiui |
| 0 – 10 | 5 | -20 | -2 | 9 | -18 |
| 10 – 20 | 15 | -10 | -1 | 12 | -12 |
| 20 – 30 | 25 | 0 | 0 | 15 | 0 |
| 30 – 40 | 35 | 10 | 1 | 10 | 10 |
| 40 – 50 | 45 | 20 | 2 | 14 | 28 |
| N = 60 | Σ fiui = 8 |
From the table, it’s seen that,
A = 25 and h = 10
Mean = A + h x (Σfi ui/N)
= 25 + 10 x (8/60)
= 25 + 4/3
= 79/3 = 26.333
10.
| Class interval: | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |
| Frequency: | 5 | 9 | 10 | 8 | 8 |
Solution:
Let’s consider the assumed mean (A) = 20
| Class interval | Mid – value xi | di = xi – 20 | ui = (xi – 20)/8 | fi | fiui |
| 0 – 8 | 4 | -16 | -2 | 5 | -10 |
| 8 – 16 | 12 | -4 | -1 | 9 | -9 |
| 16 – 24 | 20 | 0 | 0 | 10 | 0 |
| 24 – 32 | 28 | 4 | 1 | 8 | 8 |
| 32 – 40 | 36 | 16 | 2 | 8 | 16 |
| N = 40 | Σ fiui = 5 |
From the table, it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 8 x (5/40)
= 20 + 1
= 21
11.
| Class interval: | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |
| Frequency: | 5 | 6 | 4 | 3 | 2 |
Solution:
Let’s consider the assumed mean (A) = 20
| Class interval | Mid – value xi | di = xi – 20 | ui = (xi – 20)/8 | fi | fiui |
| 0 – 8 | 4 | -16 | -2 | 5 | -12 |
| 8 – 16 | 12 | -8 | -1 | 6 | -8 |
| 16 – 24 | 20 | 0 | 0 | 4 | 0 |
| 24 – 32 | 28 | 8 | 1 | 3 | 9 |
| 32 – 40 | 36 | 16 | 2 | 2 | 14 |
| N = 20 | Σ fiui = -9 |
From the table, it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 6 x (-9/20)
= 20 – 72/20
= 20 – 3.6
= 16.4
12.
| Class interval: | 10 – 30 | 30 – 50 | 50 – 70 | 70 – 90 | 90 – 110 | 110 – 130 |
| Frequency: | 5 | 8 | 12 | 20 | 3 | 2 |
Solution:
Let’s consider the assumed mean (A) = 60
| Class interval | Mid – value xi | di = xi –60 | ui = (xi – 60)/20 | fi | fiui |
| 10 – 30 | 20 | -40 | -2 | 5 | -10 |
| 30 – 50 | 40 | -20 | -1 | 8 | -8 |
| 50 – 70 | 60 | 0 | 0 | 12 | 0 |
| 70 – 90 | 80 | 20 | 1 | 20 | 20 |
| 90 – 110 | 100 | 40 | 2 | 3 | 6 |
| 110 – 130 | 120 | 60 | 3 | 2 | 6 |
| N = 50 | Σ fiui = 14 |
From the table, it’s seen that,
A = 60 and h = 20
Mean = A + h x (Σfi ui/N)
= 60 + 20 x (14/50)
= 60 + 28/5
= 60 + 5.6
= 65.6
13.
| Class interval: | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 |
| Frequency: | 6 | 10 | 8 | 12 | 4 |
Solution:
Let’s consider the assumed mean (A) = 50
| Class interval | Mid – value xi | di = xi – 50 | ui = (xi – 50)/10 | fi | fiui |
| 25 – 35 | 30 | -20 | -2 | 6 | -12 |
| 35 – 45 | 40 | -10 | -1 | 10 | -10 |
| 45 – 55 | 50 | 0 | 0 | 8 | 0 |
| 55 – 65 | 60 | 10 | 1 | 12 | 12 |
| 65 – 75 | 70 | 20 | 2 | 4 | 8 |
| N = 40 | Σ fiui = -2 |
From the table, it’s seen that,
A = 50 and h = 10
Mean = A + h x (Σfi ui/N)
= 50 + 10 x (-2/40)
= 50 – 0.5
= 49.5
14.
| Class interval: | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |
| Frequency: | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
Solution:
Let’s consider the assumed mean (A) = 42
| Class interval | Mid – value xi | di = xi – 42 | ui = (xi – 42)/5 | fi | fiui |
| 25 – 29 | 27 | -15 | -3 | 14 | -42 |
| 30 – 34 | 32 | -10 | -2 | 22 | -44 |
| 35 – 39 | 37 | -5 | -1 | 16 | -16 |
| 40 – 44 | 42 | 0 | 0 | 6 | 0 |
| 45 – 49 | 47 | 5 | 1 | 5 | 5 |
| 50 – 54 | 52 | 10 | 2 | 3 | 6 |
| 55 – 59 | 57 | 15 | 3 | 4 | 12 |
| N = 70 | Σ fiui = -79 |
From the table, it’s seen that,
A = 42 and h = 5
Mean = A + h x (Σfi ui/N)
= 42 + 5 x (-79/70)
= 42 – 79/14
= 42 – 5.643
= 36.357
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