If \( \overline{x_1} \) and \( \overline{x_2} \) are the means of two distributions such that \( \overline{x_1} < \overline{x_2} \) and \( \overline{x} \) is the mean of the combined distribution, then
If \( \overline{x_1} \) and \( \overline{x_2} \) are the means of two distributions such that \( \overline{x_1} < \overline{x_2} \) and \( \overline{x} \) is the mean of the combined distribution, then
Show Hint
- $\overline{x}<\overline{x}_{1}$
- $\overline{x}>\overline{x}_{2}$
- $\overline{x}=\frac{\overline{x}_{1}+\overline{x}_{2}}{2}$
- $\overline{x}_{1}<\overline{x}<\overline{x}_{2}$
The Correct Option is D
Solution and Explanation
Top Questions on Mean and Variance of Random variables
For a statistical data \( x_1, x_2, \dots, x_{10} \) of 10 values, a student obtained the mean as 5.5 and \[ \sum_{i=1}^{10} x_i^2 = 371. \] He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively.
The variance of the corrected data is:- JEE Main - 2025
- Mathematics
- Mean and Variance of Random variables
Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that
\[\sum_{i=1}^{10} (x_i - 2) = 30, \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 98, \quad \beta > 2\]
and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of
\[ 2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, \ldots, 2(x_{10} - 1) + 4\beta\]
then $\frac{\beta \mu}{\sigma^2}$ is equal to:- JEE Main - 2025
- Mathematics
- Mean and Variance of Random variables
- Let \( a, b \in \mathbb{R} \). Let the mean and the variance of 6 observations \(-3, 4, 7, -6, a, b\) be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:
- JEE Main - 2024
- Mathematics
- Mean and Variance of Random variables
- The variance \(\sigma^2\) of the data
Is _______.
- JEE Main - 2024
- Mathematics
- Mean and Variance of Random variables
- Want to practice more? Try solving extra ecology questions todayView All Questions