Advertisements
Advertisements
प्रश्न
Find the mode of the following distribution:
| Marks | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Frequency | 12 | 35 | 45 | 25 | 13 |
उत्तर
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 – 40. So, the modal class is 30- 40.
Now,
Modal class = 30 – 40, lower limit (l) of modal class = 30, class size (h) = 10,
frequency `(f_1)` of the modal class = 45,
frequency `(f_0)` of class preceding the modal class = 35,
frequency `(f_2)` of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Mode = `l + ((f_1− f_0)/(2f_1− f_0− f_2)) × h`
`= 30 + ((45−35)/(90−35−25)) × 10`
`= 30 + ((10)/(30)) × 10`
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
APPEARS IN
संबंधित प्रश्न
Find the mode of the following data:
3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
Find the mode of the following distribution.
| Class-interval: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 |
| Frequency: | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
The following table gives the daily income of 50 workers of a factory:
| Daily income (in Rs) | 100 - 120 | 120 - 140 | 140 - 160 | 160 - 180 | 180 - 200 |
| Number of workers: | 12 | 14 | 8 | 6 | 10 |
Find the mean, mode and median of the above data.
Given below is the distribution of total household expenditure of 200 manual workers in a city:
| Expenditure (in Rs) | 1000 – 1500 | 1500 – 2000 | 2000 – 2500 | 2500 – 3000 | 3000 – 3500 | 3500 – 4000 | 4000 – 4500 | 4500 – 5000 |
| Number of manual workers |
24 | 40 | 31 | 28 | 32 | 23 | 17 | 5 |
Find the average expenditure done by maximum number of manual workers.
Calculate the mode from the following data:
| Monthly salary (in Rs) | No of employees |
| 0 – 5000 | 90 |
| 5000 – 10000 | 150 |
| 10000 – 15000 | 100 |
| 15000 – 20000 | 80 |
| 20000 – 25000 | 70 |
| 25000 – 30000 | 10 |
Compute the mode from the following series:
| Size | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 | 95 – 105 | 105 - 115 |
| Frequency | 7 | 12 | 17 | 30 | 32 | 6 | 10 |
The relationship between mean, median and mode for a moderately skewed distribution is.
The weight of coffee in 70 packets are shown in the following table:
| Weight (in g) | Number of packets |
| 200 – 201 | 12 |
| 201 – 202 | 26 |
| 202 – 203 | 20 |
| 203 – 204 | 9 |
| 204 – 205 | 2 |
| 205 – 206 | 1 |
Determine the modal weight.
For the following distribution:
| Class | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 |
| Frequency | 10 | 15 | 12 | 20 | 9 |
the upper limit of the modal class 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`.
