Statistics And Probability Question 461
Question: Computer the medium form the following table
| Marks obtained | No. of students |
|---|---|
| 0-10 | 2 |
| 10-20 | 18 |
| 20-30 | 30 |
| 30-40 | 45 |
| 40-50 | 35 |
| 50-60 | 20 |
| 60-70 | 6 |
| 70-80 | 3 |
Options:
A) 36.55
B) 35.55
C) 6.85
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a]
| Marks obtained | No. of students | Cumulative frequency |
|---|---|---|
| 0-10 | 2 | 2 |
| 10-20 | 18 | 20 |
| 20-30 | 30 | 50 |
| 30-40 | 45 | 95 |
| 40-50 | 35 | 130 |
| 50-60 | 20 | 150 |
| 60-70 | 6 | 156 |
| 70-80 | 3 | 159 |
$ N=\Sigma f=159 $ (Odd number) Median is $ \frac{1}{2}(n+1)=\frac{1}{2}(159+1)=80th $ value, which lies in the class 30-40 (see the row of cumulative frequency 95, which contains 80). Hence median class is 30-40.
$ \therefore $ We have l = Lower limit of median class =30 f = frequency of median class = 45 C = Total of all frequencies preceding median class = 50 i = width of class interval of median class=10
$ \therefore $ Required median $ =l+\frac{\frac{N}{2}-C}{f}\times i=30+\frac{\frac{159}{2}-50}{45}\times 10 $ $ =3+\frac{295}{45}=36.55 $
BETA
