A marine sales dealer finds that the average price of a previously owned boat is \$6492. He decides to sell boats that will appeal to the middle 66% of the market in terms of price. Find the maximum and minimum prices of the boats the dealer will sell. The standard deviation is \$1025, and the variable is normally distributed.

#### Solution

Given that mean $\mu=6492$ and standard deviation $=\sigma = 1025$.

Let the maximum price be "a" and the minimum price be "b" for the middle 66% of the market

$P(a\leq X\leq b) =1-\alpha=0.66$ means

$P(X< a)= (1-\alpha)/2 = 0.17$

and $P(X> b) = (1-\alpha)/2 = 0.17$.

` $$ \begin{aligned} & P(X< a) =0.17\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{a-6492}{1025}\big)=0.17\\ &\Rightarrow P(Z< \frac{a-6492}{1025}\big)=0.17\\ &\Rightarrow \frac{a-6492}{1025}= -0.954\\ &\quad\quad\text{(From normal table)}\\ &\Rightarrow a = 6492 +( -0.954)* 1025\\ &\Rightarrow a = 5514.15 \end{aligned} $$ `

And

` $$ \begin{aligned} & P(X>b) = 0.17\\ &\Rightarrow P(X< b) =0.83\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{b-6492}{1025}\big)=0.83\\ &\Rightarrow P(Z< \frac{b-6492}{1025}\big)=0.83\\ &\Rightarrow \frac{b-6492}{1025}= 0.954\\ &\quad\quad\text{(From normal table)}\\ &\Rightarrow b = 6492 + 0.954* 1025\\ &\Rightarrow b = 7469.85 \end{aligned} $$ `

The maximum price of the boats the dealer will sell is \$ 7469.85 and the minimum prices of the boats the dealer will sell is \$ 5514.15.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators