Exercise 13.1 Solutions
Statistics
1. A survey was conducted… 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?
We will use the Direct Method because the numerical values of the class marks (xᵢ) and frequencies (fᵢ) are small, making the direct calculation of fᵢxᵢ straightforward.
Calculation Table:
| Number of plants | Number of houses (fᵢ) | Class mark (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 | Σfᵢ = 20 | Σfᵢxᵢ = 162 |
Calculating the Mean (x̄):
Mean (x̄) = Σfᵢxᵢ / Σfᵢ = 162 / 20 = 8.1
Answer: 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 ₹) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
We will use the Step-Deviation Method as the class size (h=20) is uniform and the class marks are large, which simplifies calculations.
Let’s take the Assumed Mean (A) = 550. Class size (h) = 20.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | uᵢ = dᵢ/h | fᵢuᵢ |
|---|---|---|---|---|---|
| 500 – 520 | 12 | 510 | -40 | -2 | -24 |
| 520 – 540 | 14 | 530 | -20 | -1 | -14 |
| 540 – 560 | 8 | 550 | 0 | 0 | 0 |
| 560 – 580 | 6 | 570 | 20 | 1 | 6 |
| 580 – 600 | 10 | 590 | 40 | 2 | 20 |
| Total | Σfᵢ = 50 | Σfᵢuᵢ = -12 |
Calculating the Mean (x̄):
Mean (x̄) = A + ( (Σfᵢuᵢ / Σfᵢ) × h )
= 550 + ( (-12 / 50) × 20 ) = 550 – (12/5) × 2 = 550 – 4.8 = 545.2
Answer: The mean daily wage is ₹545.20.
3. The following distribution shows the daily pocket allowance of children… The mean pocket allowance is ₹18. Find the missing frequency f.
| Daily pocket allowance (in ₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
|---|---|---|---|---|---|---|---|
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
We can use the Direct Method as the numbers are manageable and we need to solve for f.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | fᵢxᵢ |
|---|---|---|---|
| 11 – 13 | 7 | 12 | 84 |
| 13 – 15 | 6 | 14 | 84 |
| 15 – 17 | 9 | 16 | 144 |
| 17 – 19 | 13 | 18 | 234 |
| 19 – 21 | f | 20 | 20f |
| 21 – 23 | 5 | 22 | 110 |
| 23 – 25 | 4 | 24 | 96 |
| Total | Σfᵢ = 44+f | Σfᵢxᵢ = 752+20f |
Using the Mean Formula:
Mean (x̄) = Σfᵢxᵢ / Σfᵢ
18 = (752 + 20f) / (44 + f)
18(44 + f) = 752 + 20f
792 + 18f = 752 + 20f
792 – 752 = 20f – 18f
40 = 2f => f = 20.
Answer: The missing frequency f is 20.
4. Thirty women were examined… Find the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats 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 |
The Assumed Mean Method is suitable here. Let Assumed Mean (A) = 75.5.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | fᵢdᵢ |
|---|---|---|---|---|
| 65 – 68 | 2 | 66.5 | -9 | -18 |
| 68 – 71 | 4 | 69.5 | -6 | -24 |
| 71 – 74 | 3 | 72.5 | -3 | -9 |
| 74 – 77 | 8 | 75.5 | 0 | 0 |
| 77 – 80 | 7 | 78.5 | 3 | 21 |
| 80 – 83 | 4 | 81.5 | 6 | 24 |
| 83 – 86 | 2 | 84.5 | 9 | 18 |
| Total | Σfᵢ = 30 | Σfᵢdᵢ = 12 |
Calculating the Mean (x̄):
Mean (x̄) = A + (Σfᵢdᵢ / Σfᵢ)
= 75.5 + (12 / 30) = 75.5 + 0.4 = 75.9.
Answer: The mean heartbeats per minute is 75.9.
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes… Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
| Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
|---|---|---|---|---|---|
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
The class intervals are inclusive. We can use the mid-points directly. The difference between mid-points is uniform (h=3). The Step-Deviation Method is a good choice to simplify calculations with large frequencies.
Let Assumed Mean (A) = 57. Class size (h) = 3.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | uᵢ = dᵢ/h | fᵢuᵢ |
|---|---|---|---|---|---|
| 50 – 52 | 15 | 51 | -6 | -2 | -30 |
| 53 – 55 | 110 | 54 | -3 | -1 | -110 |
| 56 – 58 | 135 | 57 | 0 | 0 | 0 |
| 59 – 61 | 115 | 60 | 3 | 1 | 115 |
| 62 – 64 | 25 | 63 | 6 | 2 | 50 |
| Total | Σfᵢ = 400 | Σfᵢuᵢ = 25 |
Calculating the Mean (x̄):
Mean (x̄) = A + ( (Σfᵢuᵢ / Σfᵢ) × h )
= 57 + ( (25 / 400) × 3 ) = 57 + (1/16) × 3 = 57 + 0.1875 = 57.1875.
Answer: The mean number of mangoes is approximately 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 ₹) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
The Assumed Mean Method is suitable here. Let Assumed Mean (A) = 225.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | fᵢdᵢ |
|---|---|---|---|---|
| 100-150 | 4 | 125 | -100 | -400 |
| 150-200 | 5 | 175 | -50 | -250 |
| 200-250 | 12 | 225 | 0 | 0 |
| 250-300 | 2 | 275 | 50 | 100 |
| 300-350 | 2 | 325 | 100 | 200 |
| Total | Σfᵢ = 25 | Σfᵢdᵢ = -350 |
Calculating the Mean (x̄):
Mean (x̄) = A + (Σfᵢdᵢ / Σfᵢ) = 225 + (-350 / 25) = 225 – 14 = 211.
Answer: The mean daily expenditure is ₹211.
7. To find out the concentration of SO₂ in the air… Find the mean concentration of SO₂ in the air.
| Concentration of SO₂ (in ppm) | Frequency |
|---|---|
| 0.00 – 0.04 | 4 |
| 0.04 – 0.08 | 9 |
| 0.08 – 0.12 | 9 |
| 0.12 – 0.16 | 2 |
| 0.16 – 0.20 | 4 |
| 0.20 – 0.24 | 2 |
Since the numerical values are very small, the Direct Method is most appropriate.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (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 | 0.22 | 0.44 |
| Total | Σfᵢ = 30 | Σfᵢxᵢ = 2.96 |
Calculating the Mean (x̄):
Mean (x̄) = Σfᵢxᵢ / Σfᵢ = 2.96 / 30 ≈ 0.0987.
Answer: The mean concentration of SO₂ is approximately 0.099 ppm.
8. A class teacher has the following absentee record of 40 students… Find the mean number of days a student was absent.
| Number 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 |
The class intervals are not uniform. Therefore, we cannot use the Step-Deviation method. We will use the Assumed Mean Method.
Let Assumed Mean (A) = 17.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | fᵢdᵢ |
|---|---|---|---|---|
| 0 – 6 | 11 | 3 | -14 | -154 |
| 6 – 10 | 10 | 8 | -9 | -90 |
| 10 – 14 | 7 | 12 | -5 | -35 |
| 14 – 20 | 4 | 17 | 0 | 0 |
| 20 – 28 | 4 | 24 | 7 | 28 |
| 28 – 38 | 3 | 33 | 16 | 48 |
| 38 – 40 | 1 | 39 | 22 | 22 |
| Total | Σfᵢ = 40 | Σfᵢdᵢ = -181 |
Calculating the Mean (x̄):
Mean (x̄) = A + (Σfᵢdᵢ / Σfᵢ) = 17 + (-181 / 40) = 17 – 4.525 = 12.475.
Answer: The mean number of days is 12.48.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (%) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
|---|---|---|---|---|---|
| Number of cities | 3 | 10 | 11 | 8 | 3 |
The Step-Deviation Method is a good choice. Let Assumed Mean (A) = 70. Class size (h) = 10.
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | uᵢ = dᵢ/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 |
| Total | Σfᵢ = 35 | Σfᵢuᵢ = -2 |
Calculating the Mean (x̄):
Mean (x̄) = A + ( (Σfᵢuᵢ / Σfᵢ) × h ) = 70 + ((-2 / 35) × 10) = 70 – 20/35 = 70 – 4/7 ≈ 70 – 0.57 = 69.43.
Answer: The mean literacy rate is 69.43%.
