Let the mean and variance of 12 observations be \( \frac{9}{2} \) and 4, respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is \( \frac{m}{n} \), where \( m \) and \( n \) are coprime, then \( m + n \) is equal to:
For corrections in statistics, update both the sum and the sum of squares carefully, and recompute the variance using the corrected values.
The mean of the 12 observations is given as:
\[ \frac{\Sigma x}{12} = \frac{9}{2} \implies \Sigma x = 54. \]
The variance of the 12 observations is given as:
\[ \frac{\Sigma x^2}{12} - \left(\frac{\Sigma x}{12}\right)^2 = 4. \]
Substitute \( \Sigma x = 54 \):
\[ \frac{\Sigma x^2}{12} - \left(\frac{9}{2}\right)^2 = 4 \implies \frac{\Sigma x^2}{12} - \frac{81}{4} = 4. \]
Simplify:
\[ \Sigma x^2 = 12 \left(4 + \frac{81}{4}\right) = 12 \cdot \frac{97}{4} = 291. \]
After correction, the observations 9 and 10 are replaced with 7 and 14.
The corrected sum of the observations is:
\[ \Sigma x_{\text{new}} = 54 - (9 + 10) + (7 + 14) = 56. \]
The corrected sum of squares is:
\[ \Sigma x^2_{\text{new}} = 291 - (81 + 100) + (49 + 196) = 355. \]
The corrected variance is:
\[ \sigma^2_{\text{new}} = \frac{\Sigma x^2_{\text{new}}}{12} - \left(\frac{\Sigma x_{\text{new}}}{12}\right)^2. \]
Substitute \( \Sigma x^2_{\text{new}} = 355 \) and \( \Sigma x_{\text{new}} = 56 \):
\[ \sigma^2_{\text{new}} = \frac{355}{12} - \left(\frac{56}{12}\right)^2. \]
Simplify each term:
Thus:
\[ \sigma^2_{\text{new}} = \frac{355}{12} - \frac{49}{36}. \]
Take the LCM of 12 and 36:
\[ \sigma^2_{\text{new}} = \frac{1065}{36} - \frac{49}{36} = \frac{1016}{36}. \]
Simplify the fraction:
\[ \sigma^2_{\text{new}} = \frac{254}{9}. \]
Here, \( m = 254 \) and \( n = 9 \), which are coprime.
Finally, \( m + n = 254 + 9 = 317 \).
Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \]
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \]
An object of uniform density rolls up the curved path with the initial velocity $v_o$ as shown in the figure. If the maximum height attained by an object is $\frac{7v_o^2}{10 g}$ ($g=$ acceleration due to gravity), the object is a _______
