Chapter 12, Problem 1CACQ

(A)

To determine

The expected value of each option is to be calculated.

Explanation of Solution

Expected value is calculated as

  $E(x)=q1x1+q2x2+\mathrm{.........}+qnxn$

Consider an option 1,

  $E(x)=\frac{1}{16}\times 150+\frac{4}{16}\times 300+\frac{6}{16}\times 750+\frac{4}{16}\times 300+\frac{1}{16}\times 150$

  $E(x)=\frac{150}{16}+\frac{1200}{16}+\frac{4500}{16}+\frac{1200}{16}+\frac{150}{16}$

  = $\frac{7200}{16}$

  $=450.$

Consider an option 2,

  $E(x)=\frac{1}{5}\times 120+\frac{1}{5}\times 255+\frac{1}{5}\times 1500+\frac{1}{5}\times 255+\frac{1}{5}\times 120$

  $=24+51+300+51+24$

  $=450.$

Therefore, in both the options, expected value is 450.

(B)

To determine

The variance and standard deviation of each option is to be calculated.

Explanation of Solution

Variance is calculated as

σ = [ q₁(x₁ - (E(x))2 + q₂(x₂ - (E(x))2 + ................+ q_n(x_n - (E(x))2 ]

Standard deviation is the square root of variance.

SD = $\sqrt{\sigma }$

Consider an option 1,

σ = $\frac{1}{16}\times {\left(150-450\right)}^2$$+$$\frac{4}{16}\times {\left(300-450\right)}^2$$+$$\frac{6}{16}\times {\left(\text{750 }-450\right)}^2$$+$$\frac{4}{16}\times {\left(300-450\right)}^2$$+$$\frac{1}{16}\times {\left(150-450\right)}^2$

  = $\frac{1}{16}\times {\left(300\right)}^2$$+$$\frac{4}{16}\times {\left(150\right)}^2$$+$$\frac{6}{16}\times {\left(300\right)}^2$$+$$\frac{4}{16}\times {\left(150\right)}^2$$+$$\frac{1}{16}\times {\left(300\right)}^2$

  = $\frac{1}{16}\times 90000$$+$$\frac{4}{16}\times 22500$$+$$\frac{6}{16}\times 90000$$+$$\frac{4}{16}\times 22500$$+$$\frac{1}{16}\times 90000$

  = $\frac{1}{16}$ $\left[90000+900000+540000+900000+90000\right]$

  = $\frac{2520000}{16}$

  $=157500.$

Consider an option 2,

σ = $\frac{1}{5}\times {(120-450)}^2$$+$$\frac{1}{5}\times {(255-450)}^2$ 2 $+$$\frac{1}{5}\times {(1500-450)}^2$$+$$\frac{1}{5}\times {(255-450)}^2$$+$$\frac{1}{5}\times {(120-450)}^2$

  = $\frac{1}{5}\times {(330)}^2$$+$$\frac{1}{5}\times {(195)}^2$$+$$\frac{1}{5}\times {(1050)}^2$$+$$\frac{1}{5}\times {(195)}^2$$+$$\frac{1}{5}\times {(330)}^2$

  = $\frac{1}{5}\times 108900$$+$$\frac{1}{5}\times 38025$$+$$\frac{1}{5}\times 1102500$$+$$\frac{1}{5}\times 38025$$+$$\frac{1}{5}\times 108900$

  = $\frac{1}{5}$ $\left[108900+38025+1102500+38025+108900\right]$

  = $\frac{1396350}{5}$

  $=279270.$

Variance in option 1 is 157500 and in option 2 is 279270.

Standard deviation is the square root of variance.

SD = v σ

Consider an option 1,

  $SD=\surd 157500$

  $=396.86$

Consider an option 2,

  $SD=\surd 279270$

  $=528.46$

Variance of option 1 is 157500 and variance of option 2 is 279270.

Standard deviation of option 1 is 396.86 and standard deviation of option 2 is 528.46.

(C)

To determine

The option that is riskier to be ascertained.

Explanation of Solution

Basically, the option with high variability is more risky option.

Higher variability means high variance or standard deviation.In this question, Option 2 has more variance and standard deviation. Thus, it is riskier.

The riskier option is Option 2.