Chapter 15 Statistics Ex 15.1 Solutions
Question - 11 : - Find the mean deviation about median for the following data: | Marks | 0 -10 | 10 -20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Number of girls | 6 | 8 | 14 | 16 | 4 | 2 |
Answer - 11 : -
Let us make the tableof the given data and append other columns after calculations.
| Marks | Number of Girls fi | Cumulative frequency (c.f.) | Mid – points xi | |xi – Med| | fi|xi – Med| |
| 0 – 10 | 6 | 6 | 5 | 22.85 | 137.1 |
| 10 – 20 | 8 | 14 | 15 | 12.85 | 102.8 |
| 20 – 30 | 14 | 28 | 25 | 2.85 | 39.9 |
| 30 – 40 | 16 | 44 | 35 | 7.15 | 114.4 |
| 40 – 50 | 4 | 48 | 45 | 17.15 | 68.6 |
| 50 – 60 | 2 | 50 | 55 | 27.15 | 54.3 |
| 50 | | | | 517.1 |
The class intervalcontaining Nth/2 or 25th item is 20-30
So, 20-30 is themedian class.
Then,
Median = l + (((N/2) –c)/f) × h
Where, l = 20, c = 14,f = 14, h = 10 and n = 50
Median = 20 + (((25 –14))/14) × 10
= 20 + 7.85
= 27.85
Question - 12 : - Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age
| Age (in years) | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 36 – 40 | 41 – 45 | 46 – 50 | 51 – 55 |
| Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]
Answer - 12 : -
The given data isconverted into continuous frequency distribution by subtracting 0.5 from thelower limit and adding the 0.5 to the upper limit of each class intervals andappend other columns after calculations.
| Age | Number fi | Cumulative frequency (c.f.) | Mid – points xi | |xi – Med| | fi|xi – Med| |
| 15.5 – 20.5 | 5 | 5 | 18 | 20 | 100 |
| 20.5 – 25.5 | 6 | 11 | 23 | 15 | 90 |
| 25.5 – 30.5 | 12 | 23 | 28 | 10 | 120 |
| 30.5 – 35.5 | 14 | 37 | 33 | 5 | 70 |
| 35.5 – 40.5 | 26 | 63 | 38 | 0 | 0 |
| 40.5 – 45.5 | 12 | 75 | 43 | 5 | 60 |
| 45.5 – 50.5 | 16 | 91 | 48 | 10 | 160 |
| 50.5 – 55.5 | 9 | 100 | 53 | 15 | 135 |
| 100 | | | | 735 |
The class intervalcontaining Nth/2 or 50th item is 35.5 –40.5
So, 35.5 – 40.5 is themedian class.
Then,
Median = l + (((N/2) –c)/f) × h
Where, l = 35.5, c =37, f = 26, h = 5 and N = 100
Median = 35.5 + (((50– 37))/26) × 5
= 35.5 + 2.5
= 38
