In 2008, the per capita consumption of soft drinks in Country A was reported to

Question

In 2008, the per capita consumption of soft drinks in Country A was reported to be

19.27

gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normallydistributed, with a mean of

19.27

gallons and a standard deviation of

44

gallons. Complete parts (a) through (d) below.

a. What is the probability that someone in Country A consumed more than

13

gallons of soft drinks in 2008?The probability is

nothing.

(Round to four decimal places as needed.)

b. What is the probability that someone in Country A consumed between 7 and 10 gallons of soft drinks in 2008?

The probability is

(Round to four decimal places as needed)

c. What is the probability that someone in Country A consumed between 5 and 8 gallons of soft drinks in 2008?

The probability is

(Round up to four decimal places as needed)

d. What is the probability that someone in Country A consumed less than 10 gallons of soft drinks in 2008?

The probability is

(Round up to four decimal places as needed)

e. What is the probability that someone in Country A consumed less than 8 gallons of soft drinks in 2008?

The probability is

(Round up to four decimals as needed)

f. 97% of the people in Country A consumed less than how many gallons soft drinks?

(Round to two decimal places as needed)

Explanation / Answer

Mean = 19.27
Stdev 44

a. P(X>13) = P(Z> (13-19.27)/44) = P(Z>-.1425) = 1-.4443 = .5567
b. P(7<X<10)= P( -.28 <Z< -.21) = .4168 .3897 = .0271
c. P(5<X<8) = P(-.324<Z<-.256) = .3974-.3745 = .0229
d. P(X<10) = P(Z<-.21) = .3897
e. P(X<8) = P(Z<-.256) = .3974
f. P(X<x) = .97
So, x = 19.27 +44*1.88 = 102.99

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