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प्रश्न
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | Number of consumers |
| 65 - 85 | 4 |
| 85 - 105 | 5 |
| 105 - 125 | 13 |
| 125 - 145 | 20 |
| 145 - 165 | 14 |
| 165 - 185 | 8 |
| 185 - 205 | 4 |
उत्तर
The given data is shown below.
| Monthly Consumption (in units) | No. of consumers (fi) | xi | fixi | C.f. |
| 65−85 | 4 | 75 | 300 | 4 |
| 85−105 | 5 | 95 | 475 | 9 |
| 105−125 | 13 | 115 | 1495 | 22 |
| 125−145 | 20 | 135 | 2700 | 42 |
| 145−165 | 14 | 155 | 2170 | 56 |
| 165−185 | 8 | 175 | 1400 | 64 |
| 185−205 | 4 | 195 | 780 | 68 |
| Total | `sumf=68` | `sumf_1x_1=9320` |
Here, the maximum frequency is 20 so the modal class is 125 − 145.
Therefore,
l = 125, h = 20, f = 20, f1 = 13 and f2 = 14
Mode `=l+(f-f_1)/(2f-f_1-f_2)xxh`
`=125+7/13xx20`
`=125+140/13`
= 125 + 10.76
= 135.76
Mode = 135.76 units
Thus, the mode of the monthly consumption of electricity is 135.76 units.
Mean `=(sumf_1x_1)/(sumf)`
`=9320/68=137.05`
Mean = 137.05
Thus, the mean of the monthly consumption of electricity is 137.05 units.
Here,
Total number of consumers, N = 68 (even)
Then, `N/2=34`
Median `=l+(N/2_f)/fxxh`
`=125+(68/2_22)/20xx20`
`=125+(34-22)/20xx20`
`=125+12/20xx20`
= 125 + 12
= 137
Thus, the median of the monthly consumption of electricity is 137 units.
संबंधित प्रश्न
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| Lifetimes (in hours) | 0 − 20 | 20 − 40 | 40 − 60 | 60 − 80 | 80 − 100 | 100− 120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
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| Number of cars | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 | 60 − 70 | 70 − 80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Find the mode of the following data:
3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.
| Marks: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 | 80 - 90 | 90 - 100 |
| Frequency: | 3 | 5 | 16 | 12 | 13 | 20 | 5 | 4 | 1 | 1 |
The following is the distribution of height of students of a certain class in a certain city:
| Height (in cm): | 160 - 162 | 163 - 165 | 166 - 168 | 169 - 171 | 172 - 174 |
| No. of students: | 15 | 118 | 142 | 127 | 18 |
Find the average height of maximum number of students.
Compute the mode of the following data:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 25 | 16 | 28 | 20 | 5 |
The frequency distribution for agriculture holdings in a village is given below:
| Area of land (in hectares) | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 | 9 – 11 | 11 – 13 |
| Number of families | 20 | 45 | 80 | 55 | 40 | 12 |
Find the modal agriculture holding per family.
If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
If mode of a series exceeds its mean by 12, then mode exceeds the median by
Find the mode of the following frequency distribution:
| x | 10 | 11 | 12 | 13 | 14 | 15 |
| f | 1 | 4 | 7 | 5 | 9 | 3 |
The monthly salary of 10 employees in a factory are given below:
₹ 5000, ₹ 7000, ₹ 5000, ₹ 7000, ₹ 8000, ₹ 7000, ₹ 7000, ₹ 8000, ₹ 7000, ₹ 5000
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Find the mode of the following distribution:
| Weight (in kgs) | 25 − 34 | 35 − 44 | 45 − 54 | 55 − 64 | 65 − 74 | 75 − 84 |
| Number of students | 4 | 8 | 10 | 14 | 8 | 6 |
Construction of a cumulative frequency table is useful in determining the ______.
If L = 10, f1 = 70, f0 = 58, f2 = 42, h = 2, then find the mode by using formula.
If mode of the following frequency distribution is 55, then find the value of x.
| Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
| Frequency | 10 | 7 | x | 15 | 10 | 12 |
The mode of a grouped frequency distribution is 75 and the modal class is 65-80. The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8. Find the frequency of the modal class.
The mode of the numbers 2, 3, 3, 4, 5, 4, 4, 5, 3, 4, 2, 6, 7 is ______.
The following frequency distribution table shows the classification of the number of vehicles and the volume of petrol filled in them. To find the mode of the volume of petrol filled, complete the following activity:
| Class (Petrol filled in Liters) |
Frequency (Number of Vehicles) |
| 0.5 - 3.5 | 33 |
| 3.5 - 6.5 | 40 |
| 6.5 - 9.5 | 27 |
| 9.5 - 12.5 | 18 |
| 12.5 - 15.5 | 12 |
Activity:
From the given table,
Modal class = `square`
∴ Mode = `square + [(f_1 - f_0)/(2f_1 -f_0 - square)] xx h`
∴ Mode = `3.5 + [(40 - 33)/(2(40) - 33 - 27)] xx square`
∴ Mode = `3.5 +[7/(80 - 60)] xx 3`
∴ Mode = `square`
∴ The mode of the volume of petrol filled is `square`.
