NCERT Solutions for Class 10th: Ch 14 Statistics Maths
Page No: 270Exercise 14.1
1. A survey was conducted by a group of students as a part of their environment awareness programme, 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 |
Answer
| No. of plants (Class interval) | No. of houses (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 |
| Sum fi = 20 | Sum fixi = 162 |
Mean = x̄ = ∑fixi/∑fi = 162/20 = 8.1
We would use direct method because the numerical value of fi and xi are small.
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 |
Answer
Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui = (xi - A)/h = ui = (xi - 150)/20
| Daily wages (Class interval) | Number of workers frequency (fi) | Mid-point (xi) | ui = (xi - 150)/20 | fiui |
| 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 |
Thus, mean daily wage = 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.
Answer
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 |
Mean = x̄ = ∑fixi/∑fi = (752+20f)/(44+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
Page No: 271
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Answer
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5
| Class Interval | Number of women (fi) | Mid-point (xi) | ui = (xi - 75.5)/h | fiui |
| 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 |
| Sum fi= 30 | Sum fiui = 4 |
Mean = x̄ = A + h∑fiui/∑fi = 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
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.
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Answer
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
| Class Interval | Number of boxes (fi) | Mid-point (xi) | di = xi - A | fidi |
| 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 |
| Sum fi = 400 | Sum fidi = 75 |
Mean = x̄ = A + ∑fidi/∑fi = 57 + (75/400) = 57 + 0.1875 = 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.
Here, assumed mean (A) = 225
| Class Interval | Number of households (fi) | Mid-point (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 |
| Sum fi = 25 | Sum fidi = -350 |
Mean = x̄ = A + ∑fidi/∑fi = 225 + (-350/25) = 225 - 14 = 211
The mean daily expenditure on food is 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:
Answer
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