Advertisements
Advertisements
प्रश्न
Compute the median for the following data:
| Marks | No. of students |
| More than 150 | 0 |
| More than 140 | 12 |
| More than 130 | 27 |
| More than 120 | 60 |
| More than 110 | 105 |
| More than 100 | 124 |
| More than 90 | 141 |
| More than 80 | 150 |
उत्तर
| Marks | Class Internal | frequency (f.) |
No. of students |
| More than 80 | 80-90 | 9 | 9 |
| More than 90 | 90-100 | 17 | 26 |
| More than 100 | 100-110 | 19 | 45 |
| More than 110 | 110-120 | 45 | 90 |
| More than 120 | 120-130 | 33 | 123 |
| More than 130 | 130-140 | 15 | 138 |
| More than 140 | 140-150 | 12 | 150 |
| More than 150 | 150-160 | 0 | 150 |
| N = 150 |
We have, N = 150
∴ `"N"/2 = 150/2 = 75`
Thus, the cumulative frequency just greater than 75 is 90 and the corresponding class is 110-120.
Therefore, 110-120 is the median class.
l = 110, f = 45, F = 45 and h = 10
∴ Median = `l + {(N/2-F)/f} × h`
∴ Median = `110 + {(75 - 45)/45} × 10`
∴ Median = `110 + (30)/45 × 10`
∴ Median = `110 + (300)/45`
∴ Median = 110 + 6.67
∴ Median = 116.67 (approx)
Hence, the median is 116.67.
APPEARS IN
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 |
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| 20 - 25 | 4 |
| 25 - 30 | 18 |
| 30 - 35 | 21 |
| 35 - 40 | 33 |
| 40 - 45 | 11 |
| 45 - 50 | 3 |
| 50 - 55 | 6 |
| 55 - 60 | 2 |
The mean of following numbers is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence, estimate the median.
The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
| Content of fat (%) | 2 - 3 | 3 - 4 | 4 - 5 | 5 - 6 | 6 - 7 |
| Milk collected (Litre) | 30 | 70 | 80 | 60 | 20 |
The following table shows the number of patients of different age groups admitted to a hospital for treatment on a day. Find the median of ages of the patients.
| Age- group (Yrs.) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| No. of patients | 40 | 32 | 35 | 45 | 33 | 15 |
Find the median of:
66, 98, 54, 92, 87, 63, 72.
Mode and mean of a data are 12k and 15A. Median of the data is ______.
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:
Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median
| Distance (in m) |
0 – 1 | 1 – 2 | 2 – 3 | 3 – 4 | 4 – 5 |
| Number of Students |
40 | 80 | 62 | 38 | 30 |
The median of the following frequency distribution is 25. Find the value of x.
| Class: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency: | 6 | 9 | 10 | 8 | x |
