The American Automobile Association (AAA) reported that families planning to travel over the Labor Day weekend would spend an average of 749 (The Associated Press, August 12, 2012). Assume that the amount spent is normally distributed with a standard deviation of 225.

a) What is the probability of family expenses for the weekend being less than 400?

b) What is the probability of family expenses for the weekend being 800 or more?

c) What is the probability that family expenses for the weekend will be between 500 and 1000?

d) What would the Labor Day weekend expenses have to be for the 5% of the families with the most expensive travel plans?

Answer :

Given :

The population mean $(\mu )=749$

The population standard deviation $(\sigma )=225$

Solution :

(a) The probability of family expenses for the weekend being less than 400 :

$$P(x<400)=P(\frac{x–\mu }{\sigma }<\frac{400–749}{225})$$ $$=P(z<-1.551)$$ $$=0.0605$$

(b) The probability of family expenses for the weekend being less than 800 or more :

$$P(x\ge 800)=P(\frac{x–\mu }{\sigma }\ge \frac{800–749}{225})$$ $$=P(z\ge 0.227)$$ $$=1–P(z<0.227)$$ $$=1–0.5898$$ $$=0.4102$$

(c) The probability that family expenses for the weekend will be between 500 and 1000 :

$$P(500<x<\mathrm{1,000})=P(\frac{500–749}{225}<\frac{x–\mu }{\sigma }<\frac{\mathrm{1,000}–749}{225})$$ $$=P(-1.107<z<1.116)$$ $$=P(z<1.116)–P(z<-1.107)$$ $$=0.8678–0.1341$$ $$=0.7337$$

(d) Let  $x_0$ would be the minimum Labor Day weekend expenses for the 5% of the families with the most expensive travel plans,

$$\therefore P(x>x_0)=0.05$$ $$\therefore 1–P(x>x_0)=1–0.05$$ $$\therefore P(x<x_0)=0.95$$ $$\therefore P(z<\frac{x_0–\mu }{\sigma })=0.95$$ $$\therefore \frac{x_0–\mu }{\sigma }=1.645$$ $$\therefore \frac{x_0–749}{225}=1.645$$ $$\therefore x_0–749=1.645\times 225$$ $$\therefore x_0=\mathrm{1,119.125}$$