Answer to Question #185804 in Statistics and Probability for Betty

Question

1. A discrete random variable 𝑋 has the following probability mass function

𝑃(𝑋=𝑥)={2𝑘𝑥 𝑥=2,4,6

{𝑘(𝑥+2) 𝑥=8

{ 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

where 𝑘 is a constant

a. Show that 𝑘=1/34

b. Find the exact value of 𝑃(4<𝑥≤8)

c. Find the exact value of 𝑃(2<𝑥<4)

d. What is the expected value of the random variable 𝑋?

e. What is the variance of the random variable 𝑋?

f. Determine 𝑉𝑎𝑟(5−3𝑋)

Answer 1

$P(X = x) = \left\{ {\begin{matrix} {2kx,\,\,x = 2,4,6}\\ {k(x + 2),\,\,x = 8}\\ {0,\,\,otherwise} \end{matrix}} \right.$

Find the values ​​of the probabilities

$P(x = 2) = 2k \cdot 2 = 4k$

$P(x = 4) = 2k \cdot 4 = 8k$

$P(x = 6) = 2k \cdot 6 = 12k$

$P(x = 8) = k \cdot (8 + 2) = 10k$

a) We find the value of k from the condition

$\sum {{P_i}} = 1$

Then

$4k + 8k + 12k + 10k = 1 \Rightarrow 34k = 1 \Rightarrow k = \frac{1}{{34}}$

Q.E.D

b) construct a series of probability distributions

$\begin{matrix} X&2&4&6&8\\ P&{\frac{4}{{34}}}&{\frac{8}{{34}}}&{\frac{{12}}{{34}}}&{\frac{{10}}{{34}}} \end{matrix}$

Let's find

$P(4 < x \le 8) = P(6) + P(8) = \frac{{12 + 10}}{{34}} = \frac{{11}}{{17}}$

Answer: $P(4 < x \le 8) = \frac{{11}}{{17}}$

c) Let's find

$P(2 < X < 4) = 0$ (since, by condition, $P(X = x) = 0$ for all $x \in \left( {2;4} \right)$ )

Answer: $P(2 < X < 4) = 0$

d) Let's find he expected value:

$M(x) = \sum {{x_i}} {p_i} = \frac{{2 \cdot 4 + 4 \cdot 8 + 6 \cdot 12 + 8 \cdot 10}}{{34}} = \frac{{96}}{{17}}$

Answer: $M(x) = \frac{{96}}{{17}}$

e) Let's find the variance:

$Var(x) = M({x^2}) - {M^2}(x) = \frac{{4 \cdot 4 + 16 \cdot 8 + 36 \cdot 12 + 64 \cdot 10}}{{34}} - {\left( {\frac{{96}}{{17}}} \right)^2} = \frac{{1120}}{{289}}$

Answer: $Var(x) = \frac{{1120}}{{289}}$

f) Let's find

$Var(5 - 3x) = Var(5) + Var( - 3x) = 0 + {( - 3)^2}Var(x) = 9 \cdot \frac{{1120}}{{289}} = \frac{{{\rm{10080}}}}{{289}}$