Question

Consider the discrete probability distribution shown here.

x	10	12	18	20
p	.2	.3	.1	.4

a. Calculate$\mathit{\mu }\mathbf{,}{\mathit{\sigma }}^{\mathbf{2}}$ and$\mathit{\sigma }$ .

b. What is$\mathit{P}\left(x<15\right)$ ?

c. Calculate $\mathit{\mu }\mathbf{\pm }\mathbf{2}\mathit{\sigma }$ .

d. What is the probability that xis in the interval $\mathit{\mu }\mathbf{\pm }\mathbf{2}\mathit{\sigma }$ ?

Short answer

(a) $\mu =15.4,{\sigma }^2=18.44and\sigma =4.294.$

(b) $P\left(x<15\right)=0.5$

(c) $\mu \pm 2\sigma =(6.812,23.988)$

(d) The required probability is 0.5.

Step-by-step solution

Given information

X is a random variable.

Identifying the type of random variable

a.

Mean=$\mu$

$$\begin{gathered} \mu =\sum _{i=1}^4{x}_{j}P\left(x_i\right) \\ =\left(10\times 0.2\right)+\left(12\times 0.3\right)+\left(18\times 0.1\right)+\left(20\times 0.4\right) \\ =2+3.6+1.8+8 \\ =15.4 \end{gathered}$$

Variance=${\sigma }^2$

$$\begin{gathered} {\sigma }^2=\sum _{i=1}^4\left[{\left(x-\mu \right)}^{2}P\left(x\right)\right] \\ =\left({\left(10\times 15.4\right)}^2\times 0.2\right)+\left({\left(12-15.4\right)}^2\times 0.3\right)+\left({\left(18-15.4\right)}^2\times 0.1\right)+\left({\left(20-15.4\right)}^2\times 0.4\right) \\ =\left(29.16\times 0.2\right)+\left(11.56\times 0.3\right)+\left(6.76\times 0.1\right)+\left(21.16\times 0.4\right) \\ =5.832+3.468+0.676+8.464 \\ =18.44 \end{gathered}$$

Standard Deviation=$\sigma$

$$\begin{gathered} \sigma =\sqrt{18.44} \\ =4.294 \end{gathered}$$

Hence, $\mu =15.4,{\sigma }^2=18.44$ and $\sigma =4.294$ .

When  p(x<15)

b.

$$\begin{gathered} p\left(x<15\right)=p\left(x=10\right)+p\left(x=12\right) \\ =0.2+0.3 \\ =0.5 \end{gathered}$$

Hence, $P\left(x<15\right)=0.5$

Calculating the value when  μ±2σ

c.

The interval is

$$\begin{gathered} \mu \pm 2\sigma =15.4\pm \left(2\times 4.294\right) \\ =15.4\pm 8.588 \\ =\left(6.812,23.988\right) \end{gathered}$$

Calculating P (x) when μ±2σ

d.

$$\begin{gathered} P\left(6.812\le x\le 23.988\right)=P\left(0\le x\le 17.176\right) \\ =1-P\left(x\ge 17.176\right) \\ =1-\left[P\left(x=18\right)+P\left(x=20\right)\right] \\ =1-\left(0.1+0.4\right) \\ =1-0.5 \\ =0.5 \end{gathered}$$

Hence, the required probability is 0.5.