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प्रश्न
The following table gives the frequency distribution of married women by age at marriage:
| Age (in years) | Frequency |
| 15-19 | 53 |
| 20-24 | 140 |
| 25-29 | 98 |
| 30-34 | 32 |
| 35-39 | 12 |
| 40-44 | 9 |
| 45-49 | 5 |
| 50-54 | 3 |
| 55-59 | 3 |
| 60 and above | 2 |
Calculate the median and interpret the results.
उत्तर
| Class interval (exclusive) |
Class interval (inclusive) |
Frequency | Cumulative frequency |
| 15-19 | 14.5 - 19.5 | 53 | 53 |
| 20-24 | 19.5 - 24.5 | 140 | 193 |
| 25-29 | 24.5 - 29.5 | 98 | 291 |
| 30-34 | 29.5 - 34.5 | 32 | 323 |
| 35-39 | 34.5 - 39.5 | 12 | 335 |
| 40-44 | 39.5 - 44.5 | 9 | 344 |
| 45-49 | 44.5 - 49.5 | 5 | 349 |
| 50-54 | 49.5 - 54.5 | 3 | 352 |
| 54-59 | 54.5 - 59.5 | 3 | 355 |
| 60 and above | 59.5 and above | 2 | 357 |
| N = 357 |
N = 357
`N/2=357/2=178.5`
The cumulative frequency just greater than N/2 is 193, then the median class is 19.5-24.5
Such that
l = 19.5, f = 140. F = 53, h = 24.5 - 19.5 = 5
Median `=l+(N/2-F)/fxxh`
`=19.5+(178.5 - 53)/140xx5`
`=19.5+125.5/140xx5`
`=19.5+627.5/140`
= 19.5 + 4.48
= 23.98
Nearly half the a women were married between ages 15 and 25.
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संबंधित प्रश्न
The marks obtained by 30 students in a class assignment of 5 marks are given below.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of Students |
1 | 3 | 6 | 10 | 5 | 5 |
Calculate the mean, median and mode of the above distribution
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.
An incomplete distribution is given below:
| Variable: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency: | 12 | 30 | - | 65 | - | 25 | 18 |
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
From the following data, find:
Upper quartile
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
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 |
Which measure of central tendency can be determine graphically?
Find the median of the scores 7, 10, 5, 8, 9.
Weekly income of 600 families is tabulated below:
| 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.
Find the values of a and b, if the sum of all the frequencies is 120 and the median of the following data is 55.
| Marks | 30 – 40 | 40 – 50 | 50 –60 | 60 – 70 | 70 –80 | 80 – 90 |
| Frequency | a | 40 | 27 | b | 15 | 24 |
Using the empirical relationship between the three measures of central tendency, find the median of a distribution, whose mean is 169 and mode is 175.
