In a certain section of Southern California the distribution
In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,200 and a standard deviation of $250. The distribution of the monthly rent does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 50 one-bedroom apartments and finding the mean to be at least $1,950 per month? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Solution
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
           
 x = critical value =    1950      
 u = mean =    2200      
 n = sample size =    50      
 s = standard deviation =    250      
           
 Thus,          
           
 z = (x - u) * sqrt(n) / s =    -7.07      
           
 Thus, using a table/technology, the right tailed area of this is          
           
 P(z >   -7.07   ) =    1.0000 [ANSWER, it is very close to 1]

