Question

A marine sales dealer Önds 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

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

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 = (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`

Middle 66%

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 = (X - \mu)/(\sigma)`

`-0.955 = (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 = (X - \mu)/(\sigma)`

`0.955 = (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.