Advertisements
Advertisements
प्रश्न
The monthly pocket money of 50 students of a class are given in the following distribution
| Monthly pocket money (in Rs) | 0 - 50 | 50 – 100 | 100 – 150 | 150 -200 | 200 – 250 | 250 - 300 |
| Number of Students | 2 | 7 | 8 | 30 | 12 | 1 |
Find the modal class and give class mark of the modal class.
उत्तर
Here the maximum class frequency is 30, and the class corresponding to the frequency is 150-200. So, the modal class is 150-200.
Also, class mark of the modal class is `(150+200)/2=175.`
APPEARS IN
संबंधित प्रश्न
During the medical check-up of 35 students of a class, their weights were recorded as follows:
| Weight (in kg | Number of students |
| Less than 38 | 0 |
| Less than 40 | 3 |
| Less than 42 | 5 |
| Less than 44 | 9 |
| Less than 46 | 14 |
| Less than 48 | 28 |
| Less than 50 | 32 |
| Less than 52 | 35 |
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
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 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.
In the formula `barx=a+h((sumf_iu_i)/(sumf_i))`, for finding the mean of grouped frequency distribution ui = ______.
Find the mode of the following frequency distribution.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 8 | 10 | 10 | 16 | 12 | 6 | 7 |
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.
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 |
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 |
For the following distribution:
| C.I. | 0 - 5 | 6 - 11 | 12 - 17 | 18 - 23 | 24 - 29 |
| f | 13 | 10 | 15 | 8 | 11 |
the upper limit of the median class is?
Consider the following distribution:
| Marks obtained | Number of students |
| More than or equal to 0 | 63 |
| More than or equal to 10 | 58 |
| More than or equal to 20 | 55 |
| More than or equal to 30 | 51 |
| More than or equal to 40 | 48 |
| More than or equal to 50 | 42 |
The frequency of the class 30 – 40 is:
