To solve the given problem, we need to decipher the information about the 10 observations \(x_1, x_2, \ldots, x_{10}\) in terms of their mean and variance, and how they relate to \(\alpha\) and \(\beta\).
Hence, the correct answer is: 2.
We are given:
$\sum_{i=1}^{10} X_i - 10A = 2 \implies \sum_{i=1}^{10} X_i = 10A + 2$.
$\sum_{i=1}^{10} X_i - 10B = 40 \implies \sum_{i=1}^{10} X_i = 10B + 40$.
Equating both expressions for $\sum_{i=1}^{10} X_i$, we get:
$10A + 2 = 10B + 40 \implies 10A - 10B = 38 \implies A - B = 3.8$.
Since A and B are integers, $A = 4$ and $B = 2$.
Thus, $B = 2$.
\(x_i\) | \(f_i\) |
|---|---|
| 0 - 4 | 2 |
| 4 - 8 | 4 |
| 8 - 12 | 7 |
| 12 - 16 | 8 |
| 16 - 20 | 6 |
Find the value of 20M (where M is median of the data)
\(x_i\) | \(f_i\) |
|---|---|
| 0 - 4 | 2 |
| 4 - 8 | 4 |
| 8 - 12 | 7 |
| 12 - 16 | 8 |
| 16 - 20 | 6 |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)