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The mean rent of a 3-bedroom apartment in Orlando is $1250. You randomly select 13 apartments around town. The rents are normally distributed with a standard deviation of $290. What is the probability that the mean rent is more than $1200?Choose one:
A. 0.2676
B. 0.7324
C. 0.2709
D. 0.7291

Respuesta :

pedrocarratore

Answer:

B. 0.7324

Step-by-step explanation:

Population mean (μ) = $1250

Sample size (n) = 13

Sample standard deviation = $290

Assuming a normal distribution, the z-score for any given cost of rent, X, is defined as:

[tex]z = \frac{X-\mu}{\frac{s}{\sqrt{n}}}[/tex]

For X= $1200

[tex]z = \frac{1200-1250}{\frac{290}{\sqrt{13}}}\\z= - 0.62[/tex]

A z-score of -0.62 corresponds to the  26.76-th percentile of a normal distribution. Therefore, the probability that the mean rent is more than $1200 is:

[tex]P(X>1200) = 1 -0.2676 = 0.7324[/tex]

The answer is B. 0.7324

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