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प्रश्न
The median class for the given distribution is:
Class Interval | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 |
Cumulative Frequency | 2 | 6 | 11 | 18 |
पर्याय
1 – 5
6 – 10
11 – 15
11 – 20
उत्तर
11 – 15
Explanation:
n = 18
`n/2 = 18/2 = 9`
9 lies between 6 to 11.
so, median class is 11 -15.
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संबंधित प्रश्न
The marks of 200 students in a test were recorded as follows:
Marks | No. of students |
10-19 | 7 |
20-29 | 11 |
30-39 | 20 |
40-49 | 46 |
50-59 | 57 |
60-69 | 37 |
70-79 | 15 |
80-89 | 7 |
Construct the cumulative frequency table. Drew the ogive and use it too find:
(1) the median and
(2) the number of student who score more than 35% marks.
Find the mean of the following frequency distribution :
Class | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
Frequency | 8 | 6 | 12 | 11 | 13 |
Find the mean of the following frequency distribution by the short cut method :
Class | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 |
Frequency | 7 | 10 | 14 | 17 | 15 | 11 | 6 |
Find the mean of the following frequency distribution by the step deviation method :
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | 120-140 |
Frequency | 12 | 24 | 52 | 88 | 66 | 42 | 16 |
Find the mode of the following frequency distribution:
Hrs. Spent daily in studies | 3 | 3.5 | 4 | 4.5 | 5 | 5.5 | 6 | 6.5 |
No. of students | 8 | 7 | 3 | 5 | 10 | 6 | 3 | 4 |
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Heights (in cm) | 138 | 139 | 140 | 141 | 142 |
Frequency | 6 | 11 | 16 | 10 | 7 |
Find the median, the upper quartile and the lower quartile of the heights.
Find the mean of: first five odd natural numbers
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25, 11, 15, 10, 17, 6, 5, 12.
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