Advertisements
Advertisements
प्रश्न
For a certain frequency distribution, the values of mean and median are 72 and 78 respectively. Find the value of mode.
उत्तर
Mean = 72
Median = 78
Mean – Mode = 3(Mean – Median)
72 – Mode = 3(72 – 78)
Mode = 72 + 18 = 90
संबंधित प्रश्न
For a certain frequency distribution, the value of mean is 20 and mode is 11. Find the value of median.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kg) | 40−45 | 45−50 | 50−55 | 55−60 | 60−65 | 65−70 | 70−75 |
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
For a certain frequency distribution, the values of Assumed mean (A) = 1300, `sumf_id_i` = 900 and `sumfi` = 100. Find the value of mean (`barx`) .
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.
Find:
- Median
- Lower quartile
- Upper quartile
- Inter-quartile range
Calculate the median from the following frequency distribution table:
| Class | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 |
| Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 2 | 2 |
The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.
| Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |
| Frequency | 12 | a | 12 | 15 | b | 6 | 6 | 4 |
In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.
| Runs scored | 2500 – 3500 | 3500 – 4500 | 4500 – 5500 | 5500 – 6500 | 6500 – 7500 | 7500 - 8500 |
| Number of batsman | 5 | x | y | 12 | 6 | 2 |
The following frequency distribution table shows the number of mango trees in a grove and their yield of mangoes, and also the cumulative frequencies. Find the median of the data.
| Class (No. of mangoes) |
Frequency (No. of trees) |
Cumulative frequency (less than) |
| 50-100 | 33 | 33 |
| 100-150 | 30 | 63 |
| 150-200 | 90 | 153 |
| 200-250 | 80 | 233 |
| 250-300 | 17 | 250 |
In the graphical representation of a frequency distribution, if the distance between mode and mean is ktimes the distance between median and mean, then write the value of k.
If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by
Obtain the median for the following frequency distribution:
| x : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| f : | 8 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |
For the following distribution
| Marks | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
| No. of Students | 3 | 9 | 13 | 10 | 5 |
the number of students who got marks less than 30 is?
If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by ______.
In a hospital, weights of new born babies were recorded, for one month. Data is as shown:
| Weight of new born baby (in kg) | 1.4 - 1.8 | 1.8 - 2.2 | 2.2 - 2.6 | 2.6 - 3.0 |
| No of babies | 3 | 15 | 6 | 1 |
Then the median weight is?
Consider the data:
| 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:
Will the median class and modal class of grouped data always be different? Justify your answer.
Find the median of the following distribution:
| Marks | 0 – 10 | 10 –20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Number of students | 5 | 8 | 20 | 15 | 7 | 5 |
Consider the following frequency distribution:
| Class | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |
| Frequency | 12 | 10 | 15 | 8 | 11 |
The median class is:
A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
| Family size | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 | 9 – 11 |
| Numbers of Families | 7 | 8 | 2 | 2 | 1 |
Find the median of this data.
The following table shows classification of number of workers and number of hours they work in software company. Prepare less than upper limit type cumulative frequency distribution table:
| Number of hours daily | Number of workers |
| 8 - 10 | 150 |
| 10 - 12 | 500 |
| 12 - 14 | 300 |
| 14 - 16 | 50 |
