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Snowcat [4.5K]
3 years ago
6

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.
Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

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

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 = \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.

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