Statistics-And-Probability Question 628
Question: The marks obtained by 60 students in a certain test are given below:
| Marks | No. of students |
|---|---|
| 10-20 | 2 |
| 20-30 | 3 |
| 30-40 | 4 |
| 40-50 | 5 |
| 50-60 | 6 |
| 60-70 | 12 |
| 70-80 | 14 |
| 80-90 | 10 |
| 90-100 | 4 |
Mean, median and mode of the above data are respectively
Options:
A) $ 64.33,68.33,76.33 $
B) $ 60,70,80 $
C) $ 66.11,71.11,79.11 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We construct the following table taking assumed mean a = 55 (step deviation method).
| Class | $x_{i}$ | $f_{i}$ | $c.f.$ | $u_{i}=\frac{x_{i}-a}{10}f_{i}u_{i}$ |
|---|---|---|---|---|
| 10-20 | 15 | 2 | 2 | -4-8 |
| 20-30 | 25 | 3 | 5 | -3-9 |
| 30-40 | 35 | 4 | 9 | -2-8 |
| 40-50 | 45 | 5 | 14 | -1-5 |
| 50-60 | 55 | 6 | 20 | 0 |
| 60-70 | 65 | 12 | 32 | 1 |
| 70-80 | 75 | 14 | 46 | 2 |
| 80-90 | 85 | 10 | 56 | 3 |
| 90-100 | 95 | 4 | 60 | 4 |
| Total | 60 | 56 |
$ \therefore $ The mean $ =a+\frac{\sum{f_{i}u_{i}}}{\sum{f_{i}}}\times c $ $ =55\times \frac{56}{60}\times 10=55+\frac{56}{6}=64.333 $ Here $ n=60\Rightarrow \frac{n}{2}=30, $ therefore, 60-70 is the median class Using the formula: $ M=l+\frac{\frac{n}{2}-C}{f}\times c=60+\frac{30-20}{12}\times 10=68.333 $ Using Empirical formula, we have Mode = 3 Median -2 Mean $ =(68.333)-2(64.333)=204.999-128.666=76.333 $
BETA
