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प्रश्न
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.
उत्तर
| Height(exclusive) | Height(inclusive) | No. of students |
| 160 - 162 | 159.5-162.5 | 15 |
| 163 - 165 | 162.5-165.5 | 118 |
| 166 - 168 | 165.5-168.5 | 142 |
| 169 - 171 | 168.5-171-.5 | 127 |
| 172 - 174 | 171.5-174.5 | 18 |
Here the maximum frequency is 142, then the corresponding class 165.5 – 168.5 is modal class
L = 165.5, h = 168.5 – 165.5 = 3, f = 142, f1 = 118, f2 = 127
Mode `rArr=l+(f-f1)/(2f-f1-f2)xxh`
`=165.5+(142-118)/(2xx142-118-127)xx3`
`=165.5+(24xx3)/39`
`= 165.5 + 72/39`
= 165.5 + 1.85
= 167.35 cm
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संबंधित प्रश्न
Find the mode of the following distribution.
| Class-interval: | 25 - 30 | 30 - 35 | 35 - 40 | 40 - 45 | 45 - 50 | 50 - 55 |
| Frequency: | 25 | 34 | 50 | 42 | 38 | 14 |
Find the mode of the given data:
| Class Interval | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 |
| Frequency | 15 | 6 | 18 | 10 |
If mode of a series exceeds its mean by 12, then mode exceeds the median by
Find the mode from the following information:
L = 10, h = 2, f0 = 58, f1 = 70, f2 = 42.
Find the mode of the following data:
| Marks | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 |
| Number of students | 22 | 38 | 46 | 34 | 20 |
Mrs. Garg recorded the marks obtained by her students in the following table. She calculated the modal marks of the students of the class as 45. While printing the data, a blank was left. Find the missing frequency in the table given below.
| Marks Obtained |
0 − 20 | 20 − 40 | 40 − 60 | 60 − 80 | 80 − 100 |
| Number of Students |
5 | 10 | − | 6 | 3 |
From one footwear shop, 12 pairs of chappals were sold. The sizes of these chappals are given below.
7, 8, 6, 7, 7, 5, 9, 7, 6, 7, 8, 7
Find their mode.
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 frequency distribution of daily working expenditure of families in a locality is as follows:
| Expenditure in ₹ (x): |
0 – 50 | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 |
| No. of families (f): |
24 | 33 | 37 | b | 25 |
If the mode of the distribution is ₹ 140 then the value of b 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`.
