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प्रश्न
For a certain frequency distribution, the value of mean is 20 and mode is 11. Find the value of median.
उत्तर
The inter-relation between the measures of central tendency is given by
Mean – Mode = 3(Mean-Median)
20 – 11 = 3 (20-Median)
9 = 3 (20-Median)
9/3= 20-Median
3=20 -Median
median =20-3
Median = 17
संबंधित प्रश्न
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 |
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters | Number of surnames |
1 - 4 | 6 |
4 − 7 | 30 |
7 - 10 | 40 |
10 - 13 | 6 |
13 - 16 | 4 |
16 − 19 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
An incomplete distribution is given as follows:
Variable: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 |
Frequency: | 10 | 20 | ? | 40 | ? | 25 | 15 |
You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.
Compute the median for the following data:
Marks | No. of students |
Less than 10 | 0 |
Less than 30 | 10 |
Less than 50 | 25 |
Less than 70 | 43 |
Less than 90 | 65 |
Less than 110 | 87 |
Less than 130 | 96 |
Less than 150 | 100 |
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
The ages of 37 students in a class are given in the following table:
Age (in years) | 11 | 12 | 13 | 14 | 15 | 16 |
Frequency | 2 | 4 | 6 | 10 | 8 | 7 |
The weight of 60 boys are given in the following distribution table:
Weight (kg) | 37 | 38 | 39 | 40 | 41 |
No. of boys | 10 | 14 | 18 | 12 | 6 |
Find:
- Median
- Lower quartile
- Upper quartile
- Inter-quartile range
If the median of the following frequency distribution is 32.5, find the values of `f_1 and f_2`.
Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 -40 | 40 – 50 | 50 – 60 | 60 – 70 | Total |
Frequency | `f_1` |
5 |
9 | 12 | `f_2` | 3 | 2 | 40 |
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Milk (Litre) | 1 - 3 | 3 - 5 | 5 - 7 | 7 - 9 | 9 - 11 | 11 - 13 |
No. of hotels | 7 | 5 | 15 | 20 | 35 | 18 |
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Age (years) | Less than 5 | 5 - 9 | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 |
No. of patients | 38 | 32 | 50 | 36 | 24 | 20 |
If the difference of Mode and Median of a data is 24, then the difference of median and mean is ______.
Pocket expenses of a class in a college are shown in the following frequency distribution:
Pocket expenses |
0 - 200 |
200 - 400 |
400 - 600 |
600 - 800 |
800 - 1000 |
1000 - 1200 |
1200 - 1400 |
Number of students | 33 | 74 | 170 | 88 | 76 | 44 | 25 |
Then the median for the above data is?
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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:
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Weekly income (in Rs) |
Number of families |
0 – 1000 | 250 |
1000 – 2000 | 190 |
2000 – 3000 | 100 |
3000 – 4000 | 40 |
4000 – 5000 | 15 |
5000 – 6000 | 5 |
Total | 600 |
Compute the median income.
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
Class interval | Frequency |
0 – 100 | 2 |
100 – 200 | 5 |
200 – 300 | x |
300 – 400 | 12 |
400 – 500 | 17 |
500 – 600 | 20 |
600 – 700 | y |
700 – 800 | 9 |
800 – 900 | 7 |
900 – 1000 | 4 |
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Class: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Frequency: | 6 | 3 | x | 12 | 19 |
Find the median of the following frequency distribution:
Class: | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
Frequency: | 6 | 8 | 5 | 9 | 7 |
The median of first seven prime numbers is ______.