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प्रश्न
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
विकल्प
17
17.5
18
18.5
उत्तर
The given classes in the table are non-continuous. So, we first make the classes continuous by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit in each class.
| Class | Frequency | Cumulative Frequency |
| 0.5–5.5 | 13 | 13 |
| 5.5–11.5 | 10 | 23 |
| 11.5–17.5 | 15 | 38 |
| 17.5–23.5 | 8 | 46 |
| 23.5–29.5 | 11 | 57 |
Now, from the table we see that N = 57.
So,
\[\frac{N}{2} = \frac{57}{2} = 28 . 5\]
28.5 lies in the class 11.5–17.5.
The upper limit of the interval 11.5–17.5 is 17.5.
Hence, the correct answer is option (b).
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संबंधित प्रश्न
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 median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.
| Class interval : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency : | 4 | x | 5 | y | 1 |
Write the modal class for the following frequency distribution:
| Class-interval: | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
| Frequency: | 30 | 35 | 75 | 40 | 30 | 15 |
For a frequency distribution, mean, median and mode are connected by the relation
The median of a given frequency distribution is found graphically with the help of
The marks obtained by 100 students of a class in an examination are given below.
| Mark | No. of Student |
| 0 - 5 | 2 |
| 5 - 10 | 5 |
| 10 - 15 | 6 |
| 15 - 20 | 8 |
| 20 - 25 | 10 |
| 25 - 30 | 25 |
| 30 - 35 | 20 |
| 35 - 40 | 18 |
| 40 - 45 | 4 |
| 45 - 50 | 2 |
Draw 'a less than' type cumulative frequency curves (ogive). Hence find the median.
For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are.
| Number of days | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 | TOTAL |
| Total Number of students | 15 | 16 | x | 8 | y | 8 | 6 | 4 | 70 |
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 |
Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class:
| Height (in cm) |
Frequency | Cumulative frequency |
| 150 – 155 | 12 | a |
| 155 – 160 | b | 25 |
| 160 – 165 | 10 | c |
| 165 – 170 | d | 43 |
| 170 – 175 | e | 48 |
| 175 – 180 | 2 | f |
| Total | 50 |
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: More than type cumulative frequency distribution.
