Question

In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,175 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 70 one-bedroom apartments and finding the mean to be at least $2,115 per month? (Round your z values to 2 decimal places and final answer to 4 decimal places.) Probability

Answer 1

Solution :

Given that ,

mean = $\mu$ = 2175

standard deviation = $\sigma$ = 250

n = 70

$\mu$_{$\bar x$} = $\mu$ = 2175 and

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 250 / $\sqrt$ 70 = 29.8807

P($\bar x$$\geq$ 2115) = 1 - P($\bar x$$\leq$ 2115)

= 1 - P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$}$\leq$ (2115 -2175) / 29.8807)

= 1 - P(z $\leq$ -2.01)

= 1 - 0.0222 Using standard normal table.

= 0.9778

Probability = 0.9778