Question

In Exercise 5.129 - 5.132, We have provided the probability distributions of the random variables considered in Exercise 5.109 - 5.112 of section 5.4 For each exercise, do the following tasks.

Part (a) Find the mean of the random variable.

Part (b) Obtain the standard deviation of the random variable by using one of the formulas given in definition 5.10.

Short answer

Part (a) $\mu =2.12$

Part (b)$\sigma \approx 2.4$

Step-by-step solution

Part (a) Step 1. Given information. 

y	0	1	4	6
P(Y=y)	0.36	0.28	0.16	0.20

Part (a) Step 2. A random variable's mean is;

$$\begin{gathered} \mu =\sum _{}x·(X=x) \\ \mu =\left(0\times 0.36\right)+\left(2\times 0.28\right)+\left(4\times 0.16\right)+\left(6\times 0.20\right) \\ \mu =2.12 \end{gathered}$$

Part (b) Step 1.The random variable's standard deviation:

We have subpart (a) $\mu =2.12$

$y$	$P(Y=y)$	$y^2$	$y^{2}P(Y=y)$
0	0.36	0	0
1	0.28	1	0.28
4	0.16	16	2.56
6	0.20	36	7.2
Total			10.04

The standard deviation is defined as

$$\begin{gathered} \sigma =\sqrt{\sum _{}{x}^{2}P(X=x)-{\mu }^2} \\ \sigma =\sqrt{10.04-{\left(2.12\right)}^2} \\ \sigma =\sqrt{5.5456} \\ \sigma =2.3549 \\ \sigma \approx 2.4 \end{gathered}$$