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?

 

Answer:

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:

z = (x-?μ? )/(?σ? /√n)

For X= $1200

z = (1200-1250)/(290/13)

z = -0.62

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:

P(x>1200) = 1-0.2676 = 0.7324