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प्रश्न
Consider the following frequency distributions
| Class | 65 - 85 | 85 - 105 | 105 - 125 | 125 - 145 | 145 - 165 | 165 - 185 | 185-205 |
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
The difference of the upper limit of the median class and the lower limit of the modal class is?
विकल्प
0
19
20
38
उत्तर
20
Explanation:-
| Class | Frequency | Cumulative frequency |
| 65 - 85 | 4 | 4 |
| 85 - 105 | 5 | 9 |
| 105 - 125 | 13 | 22 |
| 125 - 145 | 20 | 42 |
| 145 - 165 | 14 | 56 |
| 165 - 185 | 7 | 63 |
| 185 - 205 | 4 | 67 |
Here, N = 67.
`therefore N/2=33.5,` which lies in the interval 125 - 145.
Therefore, the lower limit of the median class is 125.
The highest frequency is 20, which lies in the interval 125 - 145.
Therefore, the upper limit of modal class is 145.
So, the required difference is 145 - 125 = 20.
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संबंधित प्रश्न
The heights of 50 girls of Class X of a school are recorded as follows:
| Height (in cm) | 135 - 140 | 140 – 145 | 145 – 150 | 150 – 155 | 155 – 160 | 160 – 165 |
| No of Students | 5 | 8 | 9 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.
From the following frequency, prepare the ‘more than’ ogive.
| Score | Number of candidates |
| 400 – 450 | 20 |
| 450 – 500 | 35 |
| 500 – 550 | 40 |
| 550 – 600 | 32 |
| 600 – 650 | 24 |
| 650 – 700 | 27 |
| 700 – 750 | 18 |
| 750 – 800 | 34 |
| Total | 230 |
Also, find the median.
The following frequency distribution gives the monthly consumption of electricity of 64 consumers of locality.
| Monthly consumption (in units) | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 |
| Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 |
Form a ‘ more than type’ cumulative frequency distribution.
Calculate the missing frequency form the following distribution, it being given that the median of the distribution is 24
| Age (in years) | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Number of persons |
5 | 25 | ? | 18 | 7 |
If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
The arithmetic mean of the following frequency distribution is 53. Find the value of k.
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 12 | 15 | 32 | k | 13 |
If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
| Frequency | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
Calculate the mean of the following frequency distribution :
| Class: | 10-30 | 30-50 | 50-70 | 70-90 | 90-110 | 110-130 |
| Frequency: | 5 | 8 | 12 | 20 | 3 | 2 |
If the sum of all the frequencies is 24, then the value of z is:
| Variable (x) | 1 | 2 | 3 | 4 | 5 |
| Frequency | z | 5 | 6 | 1 | 2 |
The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
| Marks | Number of students |
| Below 10 | 10 |
| Below 20 | 50 |
| Below 30 | 130 |
| Below 40 | 270 |
| Below 50 | 440 |
| Below 60 | 570 |
| Below 70 | 670 |
| Below 80 | 740 |
| Below 90 | 780 |
| Below 100 | 800 |
Construct a frequency distribution table for the data above.
