Question

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.

Answer 1

Answer:

The maximum price that the dealer will sell is $7471 and the minimum is $5513.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6492, \sigma = 1025$

He decides to sell boats that will appeal to the middle 66% of the market in terms of price.

50 - (66/2)  = 17th percentile

50 + (66/2) = 83rd percentile

17th percentile

X when Z has a pvalue of 0.17. So X when Z = -0.955.

$Z = \frac{X - \mu}{\sigma}$

$-0.955 = \frac{X - 6492}{1025}$

$X - 6492 = -0.955*1025$

$X = 5513$

83rd percentile

X when Z has a pvalue of 0.83. So X when Z = 0.955.

$Z = \frac{X - \mu}{\sigma}$

$0.955 = \frac{X - 6492}{1025}$

$X - 6492 = 0.955*1025$

$X = 7471$

The maximum price that the dealer will sell is $7471 and the minimum is $5513.