Chapter 6, Problem 33SE

(a)

To determine

To find: The probability density function of sale price of the house on market.

Answer to Problem 33SE

  $f_X\left(x\right)=\frac{1}{25000}200000<x<225000$

Explanation of Solution

Given information:

Minimum sale price if left on the market $\left(a\right)=200000$

Maximum sale price if left on the market $\left(b\right)=225000$

Offer amount $=\$210000$

Calculation:

Let X be the amount received from market for another month.

As the sale price is distributed uniformly, the density function could be written as shown below

  $\begin{array}{c}{f}_X\left(x\right)=\frac{1}{b-a}a<x<b\\ =\frac{1}{225000-200000}200000<x<225000\\ =\frac{1}{25000}200000<x<225000\end{array}$

The density function is $f\left(x\right)=\frac{1}{25000}200000<x<225000$

(b)

To determine

To find: The probability of amount greater than or equal to $\$215000$

Answer to Problem 33SE

The probability of amount greater than or equal to $\$215000$ is 0.4.

Explanation of Solution

Calculation:

Computingprobability of amount less than $215000.

  $\begin{array}{c}P\left(X<215000\right)={\displaystyle \underset{200000}{\overset{215000}{\int }}f\left(x\right)dx}\\ ={\displaystyle \underset{200000}{\overset{215000}{\int }}\frac{1}{25000}dx}\\ ={\left[\frac{x}{25000}\right]}_{200000}^{215000}\\ =\frac{215000-200000}{25000}\end{array}$

On solving further,

  $\begin{array}{c}P\left(X<215000\right)=\frac{15000}{25000}\\ =0.6\end{array}$

So,

  $\begin{array}{c}P\left(X\ge 215000\right)=1-P\left(X<215000\right)\\ =1-0.6\\ =0.4\end{array}$

The required probability is 0.4

(c)

To determine

To find: The probability of amount less than $\$210000$

Answer to Problem 33SE

The probability of amount less than $\$210000$ is 0.4

Explanation of Solution

Calculation:

Computing probability of amount less than $210000.

  $\begin{array}{c}P\left(X<210000\right)={\displaystyle \underset{200000}{\overset{210000}{\int }}f\left(x\right)dx}\\ ={\displaystyle \underset{200000}{\overset{210000}{\int }}\frac{1}{25000}dx}\\ ={\left[\frac{x}{25000}\right]}_{200000}^{210000}\\ =\frac{210000-200000}{25000}\end{array}$

On solving further,

  $\begin{array}{c}P\left(X<210000\right)=\frac{10000}{25000}\\ =0.4\end{array}$

The required probability is 0.4.

(d)

To determine

To find: The option to leave the house on market should be opted or not.

Answer to Problem 33SE

The option of leaving on the market should be opted.

Explanation of Solution

Calculation:

The offer price by the executive’s employer is $21000. So, if more amount can be earned from the market, then the option of leaving it on market would seem appropriate.

Using the result from subpart (C),

  $\begin{array}{c}P\left(X<210000\right)=0.4\\ P\left(X\ge 210000\right)=1-P\left(X<210000\right)\\ =1-0.4\\ =0.6\end{array}$

It can be seen from the above subpart that the probability of getting amount more the offer price of $210000 is 0.6.

So, the probability associated with earning more amount than quoted by employer is 0.6 which is more than half, so the option of leaving on the market seems appropriate as the probability of earning more than the offer price is higher.