Question #109625

6.7 In 2014, the per capita consumption of bottled water in Spain was recorded to be 32.1 gallons. (Data obtained from http://bit .ly/1pyjHzQ.) Assume that the per capita consumption of bottled water in Spain is approximately normally distributed with a mean of 32.1 gallons and a standard deviation of 8 gallons.

Expert's answer

Let X=X= the number of consumed of bottled water: X∼N(μ,σ2).X\sim N(\mu, \sigma^2).

Then Z=X−μσ∼N(0,1)Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)

Given: μ=32.1,σ=8.\mu=32.1, \sigma=8.

a.The probability that someone consumed more than 32 gallons of bottled water is


P(X>32)=1−P(X≤32)=P(X>32)=1-P(X\leq 32)=

=1−P(Z≤32−32.18)=1−P(Z≤−0.0125)≈=1-P(Z\leq {32-32.1 \over8})=1-P(Z\leq -0.0125)\approx

≈0.504987\approx0.504987

b. The probability that someone consumed between 25 and 35 gallons of bottled water is


P(25<X<35)=P(X<35)−P(X<25)=P(25<X<35)=P(X<35)-P(X<25)=

=P(Z<35−32.18)−P(Z<25−32.18)==P(Z<{35-32.1 \over8})-P(Z< {25-32.1 \over8})=

=P(Z<0.3625)−P(Z<−0.8875)≈=P(Z<0.3625)-P(Z<-0.8875)\approx

≈0.6415108−0.1874049≈0.454106\approx0.6415108-0.1874049\approx0.454106

c. The probability that someone consumed less than 25 gallons of bottled water is


P(X<25)=P(Z<25−32.18)=P(X<25)=P(Z< {25-32.1 \over8})=

P(Z<−0.8875)≈0.187405P(Z<-0.8875)\approx0.187405

d. 99 % of people consumed less than how many gallons of bottled water?


P(X<X∗)=P(Z<X∗−32.18)=0.99P(X<X^*)=P(Z<{X^*-32.1 \over8})=0.99

X∗−32.18≈2.32635{X^*-32.1 \over8}\approx2.32635

X∗≈50.7108≈51X^*\approx50.7108\approx51


LATEST TUTORIALS
APPROVED BY CLIENTS