Question

answer: d

g

the distribution of monthly rent for one - bedroom apartments in a city is approximately normal with mean $936 and standard deviation $61. steven is looking for a one - bedroom apartment and wants to pay no more than $800 in monthly rent. what is the percentage of one bedroom apartments that are available to steven?
(g) 1.3%
(h) 2.5%
(i) 50%
(j) 95%
(k) 97.5%

Explanation:

Step1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value we are interested in, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $x = 800$, $\mu=936$, and $\sigma = 61$.
So, $z=\frac{800 - 936}{61}=\frac{- 136}{61}\approx - 2.23$ (rounded to two decimal places).

Step2: Find the area to the left of the z - score

We can use a standard normal distribution table (z - table) to find the area to the left of $z=-2.23$. Looking up $z = - 2.23$ in the z - table, the area to the left (which represents the percentage of data values less than $x = 800$) is approximately 0.0129 or 1.29%, which is approximately 1.3%.

Answer:

G. 1.3%