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

Answer:

a) $P(0$

b) $P(Z$

c) $P(Z>0.33) =1-P(Z$

d) $P(2$

Step-by-step 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

$P(20$

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

$z=\frac{x-\mu}{\sigma}$

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

$z = \frac{20-20}{3}=0$

$z = \frac{25-20}{3}=1.67$

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

$P(0$

b. Less than 19 gallons

$z = \frac{19-20}{3}=-0.33$

And using the normal table we got:

$P(Z$

c. More than 21 gallons

$z = \frac{21-20}{3}=0.33$

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

$P(Z>0.33) =1-P(Z$

d. Between 26 and 28 gallons

$z = \frac{26-20}{3}=2$

$z = \frac{28-20}{3}=2.67/tex]And we can use the normal standard distirbution table and we got:[tex] P(2$