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 (xáµ¢) and frequencies (fáµ¢) are small, making the direct calculation of fáµ¢xáµ¢ straightforward.
Calculation Table:
| Number of plants | Number of houses (fáµ¢) | Class mark (xáµ¢) | fáµ¢xáµ¢ |
|---|---|---|---|
| 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 (fáµ¢) | Class mark (xáµ¢) | dáµ¢ = xáµ¢ – A | uáµ¢ = dáµ¢/h | fáµ¢uáµ¢ |
|---|---|---|---|---|---|
| 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 (fáµ¢) | Class mark (xáµ¢) | fáµ¢xáµ¢ |
|---|---|---|---|
| 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 (fáµ¢) | Class mark (xáµ¢) | dáµ¢ = xáµ¢ – A | fáµ¢dáµ¢ |
|---|---|---|---|---|
| 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 (fáµ¢) | Class mark (xáµ¢) | dáµ¢ = xáµ¢ – A | uáµ¢ = dáµ¢/h | fáµ¢uáµ¢ |
|---|---|---|---|---|---|
| 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 (fáµ¢) | Class mark (xáµ¢) | dáµ¢ = xáµ¢ – A | fáµ¢dáµ¢ |
|---|---|---|---|---|
| 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 (xÌ„) = A + (Σfáµ¢dáµ¢ / Σfáµ¢) = 225 + (-350 / 25) = 225 – 14 = 211.
Answer: The mean daily expenditure is ₹211.
7. To find out the concentration of SOâ‚‚ in the air… Find the mean concentration of SOâ‚‚ in the air.
| Concentration of SOâ‚‚ (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 (fáµ¢) | Class mark (xáµ¢) | fáµ¢xáµ¢ |
|---|---|---|---|
| 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 SOâ‚‚ 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 (fáµ¢) | Class mark (xáµ¢) | dáµ¢ = xáµ¢ – A | fáµ¢dáµ¢ |
|---|---|---|---|---|
| 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 (xÌ„) = A + (Σfáµ¢dáµ¢ / Σfáµ¢) = 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 (fáµ¢) | Class mark (xáµ¢) | dáµ¢ = xáµ¢ – A | uáµ¢ = dáµ¢/h | fáµ¢uáµ¢ |
|---|---|---|---|---|---|
| 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 (xÌ„) = A + ( (Σfáµ¢uáµ¢ / Σfáµ¢) × h ) = 70 + ((-2 / 35) × 10) = 70 – 20/35 = 70 – 4/7 ≈ 70 – 0.57 = 69.43.
Answer: The mean literacy rate is 69.43%.
