NCERT Solutions Class 10 Maths
Chapter – 14 (Statistics)
The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 14 Statistics Exercise 14.1 has been provided here to help the students in solving the questions from this exercise.
Chapter : 14 Statistics
- NCERT Class 10 Maths Solution Ex – 14.2
- NCERT Class 10 Maths Solution Ex – 14.3
- NCERT Class 10 Maths Solution Ex – 14.4
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 – To find the mean value, we will use the direct method because the numerical value of fi and xi are small.
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
| No. of plants (Class interval) |
No. of houses Frequency (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 |
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.) | 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.
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 = 550.
Class interval (h) = 20
So, ui = (xi – a)/h
ui = (xi – 550)/20
Substitute and find the values as follows:
| Daily wages (Class interval) |
Number of workers frequency (fi) |
Mid-point (xi) | ui = (xi – 550)/20 | fiui |
| 500-520 | 12 | 510 | -2 | -24 |
| 520-540 | 14 | 530 | -1 | -14 |
| 540-560 | 8 | 550 = a | 0 | 0 |
| 560-580 | 6 | 570 | 1 | 6 |
| 580-600 | 10 | 590 | 2 | 20 |
| Total | Sum fi = 50 | Sum fiui = -12 |
So, the formula to find out the mean is:
Mean = x̄ = a + h(∑fiui /∑fi )
= 550 + [20 × (-12/50)]
= 550 – 4.8
= 545.20
Thus, mean daily wage of the workers = Rs. 545.20
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.
Given, 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 | 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 heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats 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 | 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 = a | 0 | 0 |
| 77-80 | 7 | 78.5 | 1 | 7 |
| 80-83 | 4 | 81.5 | 2 | 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
Therefore, 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.
| Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution – The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.
Here, assumed mean (a) = 57
Class size (h) = 3
Here, the step deviation is used because the frequency values are big.
| Class Interval | Number of boxes (fi) | Mid-point (xi) | ui = (xi – 57)/h | fiui |
| 49.5-52.5 | 15 | 51 | -2 | -30 |
| 52.5-55.5 | 110 | 54 | -1 | -110 |
| 55.5-58.5 | 135 | 57 = a | 0 | 0 |
| 58.5-61.5 | 115 | 60 | 1 | 115 |
| 61.5-64.5 | 25 | 63 | 2 | 50 |
| Sum fi = 400 | Sum fiui = 25 |
The formula to find out the Mean is:
Mean = x̄ = a + h(∑fiui /∑fi )
= 57 + 3(25/400)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19
6. The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure(in c) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Solution – Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
Let us assume the mean (a) = 225
Class size (h) = 50
| Class Interval | Number of households (fi) | Mid-point (xi) | di = xi – A | ui = di/50 | fiui |
| 100-150 | 4 | 125 | -100 | -2 | -8 |
| 150-200 | 5 | 175 | -50 | -1 | -5 |
| 200-250 | 12 | 225 = a | 0 | 0 | 0 |
| 250-300 | 2 | 275 | 50 | 1 | 2 |
| 300-350 | 2 | 325 | 100 | 2 | 4 |
| Sum fi = 25 | Sum fiui = -7 |
Mean = x̄ = a + h(∑fiui /∑fi )
= 225 + 50(-7/25)
= 225 – 14
= 211
Therefore, 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:
| 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 |
Find the mean concentration of SO2 in the air.
Solution – To find out the mean, first find the midpoint of the given frequencies as follows:
| Concentration of SO2 (in ppm) | Frequency (fi) | Mid-point (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 | Sum fi = 30 | Sum (fixi) = 2.96 |
The formula to find out the mean is Mean = x̄ = ∑fixi /∑fi
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 in the air is 0.099 ppm.
8. A class teacher has the following absentee record of 40 students of a class for the whole term. 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 |
Solution – Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
| Class interval | Frequency (fi) | Mid-point (xi) | fixi |
| 0-6 | 11 | 3 | 33 |
| 6-10 | 10 | 8 | 80 |
| 10-14 | 7 | 12 | 84 |
| 14-20 | 4 | 17 | 68 |
| 20-28 | 4 | 24 | 96 |
| 28-38 | 3 | 33 | 99 |
| 38-40 | 1 | 39 | 39 |
| Sum fi = 40 | Sum fixi = 499 |
The mean formula is,
Mean = x̄ = ∑fixi /∑fi
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-98 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
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 = 70.
Class interval (h) = 10
So, ui = (xi – a)/h
ui = (xi – 70)/10
Substitute and find the values as follows:
| Class Interval | Frequency (fi) | (xi) | ui = (xi – 70)/10 | fiui |
| 45-55 | 3 | 50 | -2 | -6 |
| 55-65 | 10 | 60 | -1 | -10 |
| 65-75 | 11 | 70 = a | 0 | 0 |
| 75-85 | 8 | 80 | 1 | 8 |
| 85-95 | 3 | 90 | 2 | 6 |
| Sum fi = 35 | Sum fiui = -2 |
So, Mean = x̄ = a + (∑fiui /∑fi) × h
= 70 + (-2/35) × 10
= 69.43
Therefore, the mean literacy part = 69.43%
