gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 31.3 and a standard deviation of 13 gallons.
a. What is the probability that someone consumed more than 41 gallons of bottled water?
b. What is the probability that someone consumed between 30 and 40 gallons of bottled water?
c What is the probability that someone consumed less than 30 gallons of bottled water?
d. 90% of people consumed less than how many gallons of bottled water?
Answer:
Given Data:
The population mean $(\mu)=31.3$
The population standard deviation $(\sigma)=13$
A) The probability that someone consumed more than 41 gallons of bottled water:
$$\therefore P(x > 41) = P\left(\frac{x – \mu}{\sigma} > \frac{41 – 31.3}{13}\right) = P(z > 0.75) = 0.2266 $$
B) The probability that someone consumed between 30 and 40 gallons of bottled water :
$$\therefore P(30 < x < 40) = P\left(\frac{30 – 31.3}{13} < \frac{x – \mu}{\sigma} < \frac{40 – 31.3}{13}\right)$$ $$= P(-0.1 < z < 0.67) $$ $$= P(z < 0.67) – P(z < -0.1)$$ $$ = 0.7486 – 0.4602$$ $$ = 0.2884$$
C) The probability that someone consumed less than 30 gallons of bottled water :
$$\therefore P(x < 30) = P\left(\frac{x – \mu}{\sigma} < \frac{30 – 31.3}{13}\right) = P(z < -0.1) = 0.4602$$
D) 90% of people consumed less than ______ gallons of bottled water :
$$\therefore P(x < z) = 0.9 $$ $$\therefore z = 1.282 $$ $$\text{Now, } z = \frac{x – \mu}{\sigma} $$ $$\therefore 1.282 = \frac{x – 31.3}{13}$$ $$\therefore x = 47.9660$$
