Question

Consider the two options in the following table, both of which have random outcomes:

a. Determine the expected value of each option.

b. Determine the variance and standard deviation of each option.

c. Which option is most risky?

Short answer

a. The expected value for option $1$and $2$is the same, i.e.$\text{\$450}$,.

b. The variance and standard deviation of each option is –

Option$1$ – Variance is $\text{56,250}$and Standard Deviation is$\text{237.17}$ .

Option$2$– Variance is$\text{279,270}$ and Standard Deviation is$\text{528.46}$.

c. Due to high variance and standard deviation Option$2$ is risky.

Step-by-step solution

Concept Introduction

Variance – The expectation of a random variable's squared deviation from its population mean or sample mean is known as variance in probability theory and statistics. Variance is a measure of dispersion, or how far a set of numbers deviates from its average value.

Standard Deviation – The standard deviation is a statistic that measures the amount of variation or dispersion in a set of numbers. A low standard deviation implies that the values are close to the set's mean, whereas a high standard deviation suggests that the values are dispersed over a larger range.

Information Provided

The following table shows the possibilities of occurrence of the different outcome scenarios together with the possible outcomes for each scenario of the two options –

Expected Value for Each Option

a.

The expected value of the Option $1$can be calculated as the sum of the probabilities that different outcomes will occur times the resulting payoffs –

$\begin{array}{c}\text{E}\left(\text{option1}\right)=\frac{1}{16}\times 150+\frac{4}{16}\times 300+\frac{6}{16}\times 750+\frac{4}{16}\times 300+\frac{1}{16}\times 150\\ \mathrm{E}\left(\mathrm{option}1\right)=9.375+75+281.25+75+9.375\\ =450\end{array}$

The expected value of the Option$2$ can be calculated as follows –

$\begin{array}{c}\text{E}\left(\text{option}2\right)=\frac{1}{5}\times 120+\frac{1}{5}\times 255+\frac{1}{5}\times 1,500+\frac{1}{5}\times 255+\frac{1}{5}\times 120\\ \mathrm{E}\left(\mathrm{option}2\right)=24+51+300+51+24\\ =450\end{array}$

Therefore, the value for option$1$ and$2$ are$\text{\$450}$ .

Variance and Standard Deviation

b.

Through the variance, the risk associated with random outcomes can be measured. The variance of a variable can be determined as the sum of the probabilities that different outcomes will occur times the squared deviations from the mean of the random variable. It can be formulated as follows –

${\text{σ}}^{\text{2}}{\text{=q}}_{\text{1}}{\left({\text{x}}_{\text{1}}\text{-E(x)}\right)}^{\text{2}}{\text{+q}}_{\text{2}}{\left({\text{x}}_{\text{2}}\text{-E(x)}\right)}^{\text{2}}{\text{+…+q}}_{\text{n}}{\left({\text{x}}_{\text{n}}\text{-E(x)}\right)}^{\text{2}}\mathrm{......}\text{(1)}$

Substituting with the values from Option $1$from the table in equation $1$it remains –

$\begin{array}{c}{\text{σ}}^{\text{2}}\text{=}\frac{\text{1}}{\text{16}}{\text{(150-450)}}^{\text{2}}\text{+}\frac{\text{4}}{\text{16}}{\text{(300-450)}}^{\text{2}}\text{+}\frac{\text{6}}{\text{16}}{\text{(750-450)}}^{\text{2}}\text{+}\frac{\text{4}}{\text{16}}{\text{(300-450)}}^{\text{2}}\text{+}\frac{\text{1}}{\text{16}}{\text{(150-450)}}^{\text{2}}\\ {\text{σ}}^{\text{2}}\text{=5,625+5,625+33,750+5,625+5,625}\\ \text{=56,250}\end{array}$

Substituting with the values from Option $2$from the table in equation $1$it remains –

$\begin{array}{c}{\text{σ}}^{\text{2}}\text{=}\frac{\text{1}}{\text{5}}{\text{(120-450)}}^{\text{2}}\text{+}\frac{\text{1}}{\text{5}}{\text{(255-450)}}^{\text{2}}\text{+}\frac{\text{1}}{\text{5}}{\text{(1,500-450)}}^{\text{2}}\text{+}\frac{\text{1}}{\text{5}}{\text{(255-450)}}^{\text{2}}\text{+}\frac{\text{1}}{\text{5}}{\text{(120-450)}}^{\text{2}}\\ {\text{σ}}^{\text{2}}\text{=21,780+7,605+220,500+7,605+21,780}\\ \text{=279,270}\end{array}$

The standard deviationis can be calculated as the square root of the variance. The formula is –

$\begin{array}{c}\text{σ=}\sqrt{{\text{σ}}^{\text{2}}}\\ \text{=}\sqrt{{\text{q}}_{\text{1}}{\left({\text{x}}_{\text{1}}\text{-E(x)}\right)}^{\text{2}}{\text{+q}}_{\text{2}}{\left({\text{x}}_{\text{2}}\text{-E(x)}\right)}^{\text{2}}{\text{+…+q}}_{\text{n}}{\left({\text{x}}_{\text{n}}\text{-E(x)}\right)}^{\text{2}}}\end{array}$

The standard deviation for option and can be calculated, substituting the previous formula –

For Option$1$ –

$\begin{array}{c}\text{σ=}\sqrt{\text{56,250}}\\ \text{=237.17}\end{array}$

For Option $2$–

$\begin{array}{c}\text{σ=}\sqrt{\text{279,270}}\\ \text{=528.46}\end{array}$

Therefore, the value for variance and standard deviation for option$1$ and$2$ are$\text{56,250}$ and $\text{237.17}$and $\text{279,270}$and$\text{528.46}$ respectively.

Riskiest Option

(c)

Comparing both options, Option$2$ will be riskier as it has a higher variance and since the deviations from the mean aremuch larger under option$2$ than under option $1$.

Therefore, Option$2$ is riskier.