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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.
उत्तर
The frequency distribution table of more than type is as follows:
| Height (in cm) (lower class limit | Cumulative frequency (cf) |
| More than 135 | 5 + 45 = 50 |
| More than 140 | 8 + 37 = 45 |
| More than 145 | 9 + 28 = 37 |
| More than 150 | 12 + 16 = 28 |
| More than 155 | 14 + 2 = 16 |
| More than 160 | 2 |
Taking lower class limits of on x-axis and their respective cumulative frequencies on y-axis,its ogive can be drawn as follows:
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संबंधित प्रश्न
Write the median class of the following distribution:
| Class | 0 – 10 | 10 -20 | 20- 30 | 30- 40 | 40-50 | 50- 60 | 60- 70 |
| Frequency | 4 | 4 | 8 | 10 | 12 | 8 | 4 |
What is the lower limit of the modal class of the following frequency distribution?
| Age (in years) | 0 - 10 | 10- 20 | 20 -30 | 30 – 40 | 40 –50 | 50 – 60 |
| Number of patients | 16 | 13 | 6 | 11 | 27 | 18 |
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
| Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 -70 |
| Number of patients | 6 | 42 | 55 | 70 | 53 | 20 |
Form a ‘less than type’ cumulative frequency distribution.
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.
The following table, construct the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
| Marks obtained (in percent) | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 |
| Number of Students | 141 | 221 | 439 | 529 | 495 | 322 | 153 |
(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark.
(c) Find the modal class and write its cumulative frequency.
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 |
Consider the following frequency distribution :
| Class: | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
| Frequency: | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is
Change the following distribution to a 'more than type' distribution. Hence draw the 'more than type' ogive for this distribution.
| Class interval: | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 |
| Frequency: | 10 | 8 | 12 | 24 | 6 | 25 | 15 |
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 are the ages of 300 patients getting medical treatment in a hospital on a particular day:
| Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Form: Less than type cumulative frequency distribution.
