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 19.47 galons. Assume that the per capita consumption ol soft drinks in Country A is approdimately normal distributed, with a mean of 19.47 gallons and a standard deviation of 5 gallons. Complete parts (a) through (d) below a. What is the probability that someone in Country A consumed more than 14 galions of soft drinks in 20087 The probability is (Round to four decimal places as needed.) b. What is the probability that someone in Country A consumed between 8 and 7 gallons of soft drinks in 2008 The probability is Round to four decimal places as needed.) e. What is the probablity that someone in Country A consumed less than 7 galilons of soft drinks in 2008 The probability is (Round to four deoimal places as needed) d.00% ofthe people in Gountry A consumed less than how many gallons of soft drinks? The probabilty is 99% that someone n Country A consumed less than [-] gallons of soft dnks Round to two deomai ciaces as needed Enter your answer in each of the answer boxes

Explanation / Answer

Data given to us is as follows:

Mean m = 19.47

Standard deviation S = 5

(a)

At X = 14, we have:

z = (X-m)/S = (14-19.47)/5 = -1.094

So,

P(X > 14) = P(z > -1.094) = 0.8630

(b)

At X = 6, we have:

z = (X-m)/S = (6-19.47)/5 = -2.694

So,

P(X < 6) = P(z < -2.694) = 0.0035

At X = 7, we have:

z = (X-m)/S = (7-19.47)/5 = -2.494

So,

P(X < 7) = P(z < -2.494) = 0.0063

So,

P(6 < X < 7) = P(X < 7) - P(X < 6) = 0.0063-0.0035 = 0.0028

(c)

As calculated in part (c) above:

P(X < 7) = P(z < -2.494) = 0.0063

(d)

At p = 0.99, we have:

z = 2.33

So,

X = z*S + m = 2.33*5 + 19.47 = 31.12 gallons

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