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 Halifax is 1,650. Suppose that the standard deviation of the cost of renting a one-bedroom apartment in Halifax is 170 and that one-bedroom apartment rents in Halifax are normally distributed. If a Halifax 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) $1,850 or more?
(b) Between $1,450 and $1,800?
(c) Between $1,425 and $1,595?
(d) Less than $1,500?

Answer:

Given:
The population mean $(μ)=1,650$

The population standard deviation $(\sigma)= 170$

Solution:
(A) The probability that the price is 1,850 or more:
$P(x > 1850) = P\left( \frac{x – \mu}{\sigma} > \frac{1850 – 1650}{170} \right) = P(z > 1.18) = 0.1190 $

(B) The probability that the price is between 1,450 and 1,800:
$$P(1,450 < x < 1,800) = P\left(\frac{1,450 – 1,650}{170} < \frac{x – \mu}{\sigma} < \frac{1,800 – 1,650}{170}\right) $$ $$ = P(-1.18 < z < 0.88) $$ $$ = P(z < 0.88) – P(z < -1.18) $$ $$ = 0.8106 – 0.119 $$ $$ = 0.6916$$

(C) The probability that the price is between 1,425 and 1,595:
$$P(1,425 < x < 1,595) = P\left(\frac{1,425 – 1,650}{170} < \frac{x – \mu}{\sigma} < \frac{1,595 – 1,650}{170}\right) $$ $$ = P(-1.32 < z < -0.32) $$ $$ = P(z < -0.32) – P(z < -1.32) $$ $$ = 0.3745 – 0.0934 $$ $$ = 0.2811$$

(D) The probability that the price is less than 1500:
$P(x < 1,500) = P\left(\frac{x – \mu}{\sigma} < \frac{1,500 – 1,650}{170}\right) = P(z < -0.88) = 0.1894 $

Final answer:
(A) The probability that the price is 1,850 or more = 0.1190

(B) The probability that the price is between 1,450 and 1,800 = 0.6916

(C) The probability that the price is between 1,425 and 1,595 = 0.2811

(D) The probability that the price is less than 1500 = 0.1894