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प्रश्न
Heights of students of class X are givee in the flowing frequency distribution
| Height (in cm) | 150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 - 175 |
| Number of students | 15 | 8 | 20 | 12 | 5 |
Find the modal height.
Also, find the mean height. Compared and interpret the two measures of central tendency.
उत्तर
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 – 165. So, the modal class is 160 – 165.
Now,
Modal class = 160 – 165, lower limit (l) of modal class = 160, class size (h) = 5,
frequency `(f_1)` of the modal class = 20,
frequency `(f_0)` of class preceding the modal class = 8,
frequency `(f_2)` of class succeeding the modal class = 12
Now, let us substitute these values in the formula:
Mode =` l + ((f_1− f_0 )/(2 f_1− f_0− f_2)) × h`
`= 160 + ((20−8)/(40−8−12)) × 5`
`= 160 + (12/20) × 5`
= 160 + 3
= 163
Hence, the mode is 163.
It represents that the height of maximum number of students is 163cm.
Now, to find the mean let us put the data in the table given below:
| Height (in cm) | Number of students `(f_i)` | Class mark `(x_i)` | `f_i x_i` |
| 150 – 155 | 15 | 152.5 | 2287.5 |
| 155 – 160 | 8 | 157.5 | 1260 |
| 160 – 165 | 20 | 162.5 | 3250 |
| 165 – 170 | 12 | 167.5 | 2010 |
| 170 – 175 | 5 | 172.5 | 862.5 |
| Total | `Ʃ f_i = 60` | `Ʃ f_i x_i` = 9670 |
Mean =`(sum _i f_i x_i)/( sum _i f_i)`
=`9670/60`
=` 161.17
Thus, mean of the given data is 161.17.
It represents that on an average, the height of a student is 161.17cm.
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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.
Compare the modal ages of two groups of students appearing for an entrance test:
| Age (in years): | 16-18 | 18-20 | 20-22 | 22-24 | 24-26 |
| Group A: | 50 | 78 | 46 | 28 | 23 |
| Group B: | 54 | 89 | 40 | 25 | 17 |
Find the mean, median and mode of the following data:
| Classes: | 0-20 | 20-40 | 40-60 | 40-60 | 80-100 | 100-120 | 120-140 |
| Frequency: | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
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.
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.
The relationship between mean, median and mode for a moderately skewed distribution is.
If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
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| Tomatoes per Plant | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 | 21 - 25 |
| Number of Plants | 20 | 50 | 46 | 22 | 12 |
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For the following distribution:
| Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
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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,
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∴ 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`.
