# Thread: [SOLVED] Mean and Variance Question...

1. ## [SOLVED] Mean and Variance Question...

Question:
The random variable Y has the probability distribution given in the following table.

Given that $\displaystyle E(Y) = 4.9$, show that $\displaystyle a = b$, and find the standard deviation of Y.

Attempt:

Need help continuing...

2. $\displaystyle \begin{gathered} \sum\limits_i {y_i P(Y = y_i ) = 4.9 \Rightarrow 3.55 + 4a + 5b = 4.9} \hfill \\ \sum\limits_i {P\left( {Y = y_i } \right) = 1 \Rightarrow 0.7 + a + b = 1} \hfill \\ \end{gathered}$

2 equations in 2 unknowns...

3. Originally Posted by Peritus
$\displaystyle \begin{gathered} \sum\limits_i {y_i P(Y = y_i ) = 4.9 \Rightarrow 3.55 + 4a + 5b = 4.9} \hfill \\ \sum\limits_i {P\left( {Y = y_i } \right) = 1 \Rightarrow 0.7 + a + b = 1} \hfill \\ \end{gathered}$

2 equations in 2 unknowns...
But in the textbook, the answer is $\displaystyle a = b = 0.15$

4. If you plug in a = 0.15 and b = 0.15 into both equations Peritus provided, don't they both work out? Hence a = b = 0.15 is the solution.