Question

Sports Drink Consumption The average number of gallons of sports drinks consumed by the football team during a game is 20, with a standard deviation of 3 gallons. Assume the variable is normally distributed. When a game is played, find the probability of using

Answer 1

a)
`P(0<z<1.67)= P(z<1.67) -P(Z<0) = 0.953 -0.5=0.453`

b)
`P(Z<-0.33) = 0.371`

c)
`P(Z>0.33) =1-P(Z<0.33) = 1-0.629= 0.371`

d)
`P(2<z<2.67)= P(z<2.67) -P(Z<2) = 0.996 -0.977=0.019`

Explanation:

Assuming the following questions:

a. Between 20 and 25 gallons

Let X the random variable that represent the sports drink consumption of a population, and for this case we know the distribution for X is given by:

`X \sim N(20,3)`

Where
`\mu=20` and
`\sigma=3`

We are interested on this probability

`P(20<X<25)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

And if we find the z score for each limit we got:

`z = (20-20)/(3)=0`

`z = (25-20)/(3)=1.67`

And we can use the normal standard distirbution table and we got:

`P(0<z<1.67)= P(z<1.67) -P(Z<0) = 0.953 -0.5=0.453`

b. Less than 19 gallons

`z = (19-20)/(3)=-0.33`

And using the normal table we got:

`P(Z<-0.33) = 0.371`

c. More than 21 gallons

`z = (21-20)/(3)=0.33`

And using the normal table and the complement rule we got:

`P(Z>0.33) =1-P(Z<0.33) = 1-0.629= 0.371`

d. Between 26 and 28 gallons

`z = (26-20)/(3)=2`