Advertisements
Advertisements
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 |
उत्तर
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below
Class Interval | Number of policy holders (f) | Cumulative frequency (cf) |
Below 20 | 2 | 2 |
20 - 25 | 6 - 2 = 4 | 6 |
25 - 30 | 24 - 6 = 18 | 24 |
30 - 35 | 45 - 24 = 21 | 45 |
35 - 40 | 78 - 45 = 33 | 78 |
40 - 45 | 89 - 78 = 11 | 89 |
45 - 50 | 92 - 89 = 3 | 92 |
50 - 55 | 98 - 92 = 6 | 98 |
55 - 60 | 100 - 98 = 2 | 100 |
It is given that n = 100
Cumulative frequency (cf) just greater than `n/2(100/2 = 50)` is 78, belonging to interval 35 - 40.
Therefore, median class = 35 - 40
Lower limit of median class (l) = 35
Class size (h) = 5
Frequency of median class (f) = 33
Cumulative frequency (cf) of class preceding median class = 45
Median = `l + ((n/2-cf)/f)xxh`
= `35 + ((50-45)/33)xx5`
= `35 + 25/33`
= 35 + 0.76
= 35.76
Therefore, the median age is 35.76 years.
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 |
Find the following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours) | Number of lamps |
1500 – 2000 | 14 |
2000 – 2500 | 56 |
2500 – 3000 | 60 |
3000 – 3500 | 86 |
3500 – 4000 | 74 |
4000 – 4500 | 62 |
4500 – 5000 | 48 |
Find the median life time of a lamp.
For a certain frequency distribution, the values of mean and median are 72 and 78 respectively. Find the value of mode.
The numbers 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x
Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.
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
If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by
Find the Median of the following distribution:
x | 3 | 5 | 10 | 12 | 8 | 15 |
f | 2 | 4 | 6 | 10 | 8 | 7 |
The following are the marks scored by the students in the Summative Assessment exam
Class | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 |
No. of Students | 2 | 7 | 15 | 10 | 11 | 5 |
Calculate the median.
The median of first 10 natural numbers is ______.