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.

Respuesta :

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 6492, \sigma = 1025[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.955 = \frac{X - 6492}{1025}[/tex]

[tex]X - 6492 = -0.955*1025[/tex]

[tex]X = 5513[/tex]

83rd percentile

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.955 = \frac{X - 6492}{1025}[/tex]

[tex]X - 6492 = 0.955*1025[/tex]

[tex]X = 7471[/tex]

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