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 Toronto is 1,750. Suppose that the standard deviation of the cost of renting a one-bedroom apartment in Toronto is 180 and that one-bedroom apartment rents in Toronto are normally distributed. If a Toronto 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) 1,950 or more?
(b) Between 1,500 and 1,900?
(c) Between 1,475 and 1,675?
(d) Less than 1,600?

Answer:

Given:
The population mean $(\mu)=1,750$

The population standard deviation $(\sigma)= 180$

Solution:

(a) The probability that the price is 1,950 or more:

$P(x > 1,950) = P\left( \frac{x – \mu}{\sigma} > \frac{1,950 – 1,750}{180} \right) = P(z > 1.11) = 0.1335 $

(b) The probability that the price is between 1,500 and 1,900:

$$ P(1,500 < x < 1,900) = P\left(\frac{1,500 – 1,750}{180} < \frac{x – \mu}{\sigma} < \frac{1,900 – 1,750}{180}\right) $$ $$ = P(-1.39 < z < 0.83) $$ $$ = P(z < 0.83) – P(z < -1.39) $$ $$ = 0.7967 – 0.0823 $$ $$ = 0.7144$$

(c) The probability that the price is between 1,475 and 1,675:

$$ P(1,475 < x < 1,675) = P\left(\frac{1,475 – 1,750}{180} < \frac{x – \mu}{\sigma} < \frac{1,675 – 1,750}{180}\right) $$ $$ = P(-1.53 < z < -0.42) $$ $$ = P(z < -0.42) – P(z < -1.53) $$ $$ = 0.3372 – 0.0630 $$ $$ = 0.2742$$

(d) The probability that the price is less than 1,600:

$P(x < 1,600) = P\left(\frac{x – \mu}{\sigma} < \frac{1,600 – 1,750}{180}\right) = P(z < -0.83) = 0.2033 $

Final answer:

(a) The probability that the price is 1,950 or more = 0.1335

(b) The probability that the price is between 1,500 and 1,900 = 0.7144

(c) The probability that the price is between 1,475 and 1,675 = 0.2742

(d) The probability that the price is less than 1,600 = 0.2033