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:

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

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

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

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

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

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

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

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