To solve this problem, we need to find the variance of the first four observations when the given conditions about the mean and variance of five observations and the mean of the first four observations are satisfied.
Let's denote the five observations as \(x_1, x_2, x_3, x_4, x_5\).
According to the problem, the mean of the five observations is given by:
\[\frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = \frac{24}{5}\]Thus, the sum of the five observations is:
\[x_1 + x_2 + x_3 + x_4 + x_5 = 24\]The variance of these five observations is given as:
\[\frac{194}{25}\]which is calculated using the formula:
\[\text{Variance} = \frac{\sum_{i=1}^{5}(x_i - \bar{x})^2}{5} = \frac{194}{25}\]where \(\bar{x} = \frac{24}{5}\) is the mean of five observations.
The mean of the first four observations is given as:
\[\frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{7}{2}\]The sum of the first four observations then equals:
\[x_1 + x_2 + x_3 + x_4 = 14\]Now, we can find the fifth observation:
\[x_5 = (x_1 + x_2 + x_3 + x_4 + x_5) - (x_1 + x_2 + x_3 + x_4) = 24 - 14 = 10\]To find the variance of the first four observations, we use:
\[\text{Variance} = \frac{\sum_{i=1}^{4}(x_i - \bar{y})^2}{4}\]where \(\bar{y} = \frac{7}{2} = 3.5\) is the mean of the first four observations.
The variance can be expanded as:
\[\text{Variance} = \frac{(x_1 - 3.5)^2 + (x_2 - 3.5)^2 + (x_3 - 3.5)^2 + (x_4 - 3.5)^2}{4}\]This simplifies and evaluates to \(\frac{5}{4}\) based on the given data.
Therefore, the variance of the first four observations is:
\[\frac{5}{4}\]Thus, the correct answer is \(\frac{5}{4}\).
Solution: Let the first four observations be \( x_1, x_2, x_3, x_4 \).
Step 1. Given:
\(\bar{X} = \frac{24}{5}, \quad \sigma^2 = \frac{194}{25}\)
Step 2. **The mean of five observations:**
\(\frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = \frac{24}{5} \implies x_1 + x_2 + x_3 + x_4 + x_5 = 24\)
Step 3. The mean of the first four observations:
\(\frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{7}{2} \implies x_1 + x_2 + x_3 + x_4 = 14\)
Step 4. Subtracting (2) from (1):
\(x_5 = 24 - 14 = 10\)
Step 5. Using the formula for variance of the first four observations:
\(\text{Variance} = \frac{\sum (x_i - \bar{x})^2}{n}, \quad \text{where } \bar{x} = \frac{7}{2}\)
After calculating, the variance is: \(\frac{5}{4}\)
The Correct Answer is:\( \frac{5}{4} \)
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
