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
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 |
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
| Age in years | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
| No. of persons | 5 | 25 | ? | 18 | 7 |
The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:
| Class interval | Frequency |
| 0 - 100 | 2 |
| 100 - 200 | 5 |
| 200 - 300 | f1 |
| 300 - 400 | 12 |
| 400 - 500 | 17 |
| 500 - 600 | 20 |
| 600 - 700 | f2 |
| 700 - 800 | 9 |
| 800 - 900 | 7 |
| 900 - 1000 | 4 |
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
| Height in cm | Number of Girls |
| Less than 140 | 4 |
| Less than 145 | 11 |
| Less than 150 | 29 |
| Less than 155 | 40 |
| Less than 160 | 46 |
| Less than 165 | 51 |
Find the median height.
From the following data, find:
Median
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
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 |
Calculate the median for the following data:
| Class | 19 – 25 | 26 – 32 | 33 – 39 | 40 – 46 | 47 – 53 | 54 - 60 |
| Frequency | 35 | 96 | 68 | 102 | 35 | 4 |
In the following table, Σf = 200 and mean = 73. Find the missing frequencies f1, and f2.
| x | 0 | 50 | 100 | 150 | 200 | 250 |
| f | 46 | f1 | f2 | 25 | 10 | 5 |
The maximum speeds, in km per hour, of 35 cars in a race are given as follows:
| Speed (km/h) | 85 – 100 | 100 – 115 | 115 – 130 | 130 – 145 |
| Number of cars | 5 | 8 | 13 | 9 |
Calculate the median speed.
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 |
