Answer to Question #229286 in Statistics and Probability for cheryl

Question

A discrete random variable 𝑌 has a cumulative probability distribution as shown below.

𝑌 = 𝑦 0 1 2 3

𝑃(𝑌 ≤ 𝑦) 5𝑘 10𝑘 11𝑘 12𝑘

(a) Determine the value of 𝑘.

(b) Calculate expected value of 𝑌.

(c) Calculate variance and standard deviation of 𝑌.

Answer 1

(a) Find the value of k from the condition

$\sum {{p_i}} = 1$

Then

$5k + 10k + 11k + 12k = 1 \Rightarrow 38k = 1 \Rightarrow k = \frac{1}{{38}}$

The distribution series has the form

$\begin{matrix} y&0&1&2&3\\ p&{\frac{5}{{38}}}&{\frac{{10}}{{38}}}&{\frac{{11}}{{38}}}&{\frac{{12}}{{38}}} \end{matrix}$

Answer: $k = \frac{1}{{38}}$

(b) $E(Y) = \sum {{y_i}} {p_i} = \frac{{0 \cdot 5 + 1 \cdot 10 + 2 \cdot 11 + 3 \cdot 12}}{{38}} = \frac{{68}}{{38}} = \frac{{34}}{{19}}$

Answer: $E(Y) = \frac{{34}}{{19}}$

(c) Let's find the variance:

$D(Y) = M({Y^2}) - {M^2}\left( Y \right) = \frac{{0 \cdot 5 + 1 \cdot 10 + 4 \cdot 11 + 9 \cdot 12}}{{38}} - {\left( {\frac{{34}}{{19}}} \right)^2} = \frac{{383}}{{361}}$

Then standard deviation is

$\sigma \left( Y \right) = \sqrt {D(Y)} = \sqrt {\frac{{383}}{{361}}} = \frac{{\sqrt {383} }}{{19}}$

Answer: $D(Y) = \frac{{383}}{{361}}$ ; $\sigma \left( Y \right) = \frac{{\sqrt {383} }}{{19}}$