Question

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 $290. 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 70 one-bedroom apartments and finding the mean to be at least $2,100 per month? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Solution:

Given that,

mean = = 2200

standard deviation = = 290

n = 70

So,

   =    2200

=  ( /n) = (290 / 70 ) = 34.6616

   2100 )

= 1 - p (    2100 )

= 1 - p ( - /)   (2100 - 2200 /34.6616 )

= 1 - p( z   - 100 / 34.6616 )

= 1 - p ( z -2.89 )   

Using z table

= 1 - 0.0019

= 0.9981

Probability = 0.9981