MCQ 11 Mark
Consider the following data
Assertion (A): The variance of the data is 45.8.
Reason (R): The standard deviation of the data is 6.77.
| $x _{ i }$ | 4 | 8 | 11 | 17 | 20 | 24 | 32 |
| $f _{ i }$ | 3 | 5 | 9 | 5 | 4 | 3 | 1 |
Reason (R): The standard deviation of the data is 6.77.
- ABoth A and R are true and R is the correct explanation of A.
- BBoth A and R are true but R is not the correct explanation of A.
- CA is true but R is false.
- DA is false but R is true.
Answer
View full question & answer→(b) Both A and R are true but R is not the correct explanation of A.
Explanation: Assertion: Presenting the data in tabular form, we get
$N =30, \sum_{i=1}^7 f_i x_i=420, \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2=1374$
Therefore, $\bar{x}=\frac{\sum_{i=1}^7 f_i x_i}{N}=\frac{1}{30} \times 420=14$
$\therefore$ Variance $\left(\sigma^2\right)=\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2$
$\frac{1}{30} \times 1374=458$
Reason: Standard deviation $(\sigma)=\sqrt{45.8}=6.77$
Explanation: Assertion: Presenting the data in tabular form, we get
| $x _{ i }$ | $f _{ i }$ | $f _{ i } x _{ i }$ | $x _{ i }-\bar{x}$ | $\left( x _{ i }-\bar{x}\right)^2$ | $f _{ i }\left( x _{ i }-\bar{x}\right)^2$ |
| 4 | 3 | 12 | -10 | 100 | 300 |
| 8 | 5 | 40 | -6 | 36 | 180 |
| 11 | 9 | 99 | -3 | 9 | 81 |
| 17 | 5 | 85 | 3 | 9 | 45 |
| 20 | 4 | 80 | 6 | 36 | 144 |
| 24 | 3 | 72 | 10 | 100 | 300 |
| 32 | 1 | 32 | 18 | 324 | 324 |
| 30 | 420 | 1374 |
Therefore, $\bar{x}=\frac{\sum_{i=1}^7 f_i x_i}{N}=\frac{1}{30} \times 420=14$
$\therefore$ Variance $\left(\sigma^2\right)=\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2$
$\frac{1}{30} \times 1374=458$
Reason: Standard deviation $(\sigma)=\sqrt{45.8}=6.77$