The Canada Mortgage and Housing Corporation lists average monthly apartment rents for many cities in Canada. According to its report, the average cost of renting a one-bedroom apartment in Vancouver is 1,850. Suppose that the standard deviation of the cost of renting a one-bedroom apartment in Vancouver is 200 and that one-bedroom apartment rents in Vancouver are normally distributed. If a Vancouver one-bedroom apartment is randomly selected, what is the probability that the price is: (Round the values of z to 2 decimal places, e.g., 0.75. Round your answers to 4 decimal places, e.g., 0.7578.)

(a) 2,100 or more?
(b) Between 1,600 and 2,000?
(c) Between 1,575 and 1,825?
(d) Less than 1,700?


Answer:

Given: The population mean $(\mu)=1,850$

The population standard deviation $(\sigma)= 200$

Solution:

(a) The probability that the price is 2,100 or more:

$P(x > 2,100) = P\left( \frac{x – \mu}{\sigma} > \frac{2,100 – 1,850}{200} \right) = P(z > 1.25) = 0.1056 $

(b) The probability that the price is between 1,600 and 2,000:

$$ P(1,600 < x < 2,000) = P\left(\frac{1,600 – 1,850}{200} < \frac{x – \mu}{\sigma} < \frac{2,000 – 1,850}{200}\right) $$ $$ = P(-1.25 < z < 0.75) $ $ = P(z < 0.75) – P(z < -1.25) $$ $$ = 0.7734 – 0.1056 $$ $$ = 0.6678$$

(c) The probability that the price is between 1,575 and 1,825:

$$ P(1,575 < x < 1,825) = P\left(\frac{1,575 – 1,850}{200} < \frac{x – \mu}{\sigma} < \frac{1,825 – 1,850}{200}\right) $$ $$ = P(-1.38 < z < -0.12) $$ $$ = P(z < -0.12) – P(z < -1.38) $$ $$ = 0.4522 – 0.0838 $$ $$ = 0.3684 $$

(d) The probability that the price is less than 1,700:

$P(x < 1,700) = P\left(\frac{x – \mu}{\sigma} < \frac{1,700 – 1,850}{200}\right) = P(z < -0.75) = 0.2266 $

Final answer:

(a) The probability that the price is 2,100 or more = 0.1056

(b) The probability that the price is between 1,600 and 2,000 = 0.6678

(c) The probability that the price is between 1,575 and 1,825 = 0.3684

(d) The probability that the price is less than 1,700 = 0.2266