afari File Edit View History Bookmarks Window Hep BUSA 21903 (Spring 2018 Johnat

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afari File Edit View History Bookmarks Window Hep BUSA 21903 (Spring 2018 Johnathon Longstreth&1 4218 404 PM Homework: Homework of Scetion 6.1-6.3 rees 4of6 14 complete) HW Soo 76.67%. 4.6of6ps Score: 0 of 1 pt 6.2.7-T The annual ger capita consumetion of bottied waer was 33.4 galions. Assume that the per capta consumption of botied water is appnolmasely nomaity dselbted with a mean of 33.4 and a standard deviation of 10 galons a. What is the probablity that someone consumed more than 38 galions of bottied water? b. What is the probability that someone consumed between 20 and 30 gallions of botled water e What is the probablity that someone consumed less than 20 gallons of bottled water? d.90% of people cosmod iess than how many g?os of botted wate? a The probability that someone consumed more than 38 galons of boled water is (Round to four decimal places as needed.)

Explanation / Answer

mean= 33.4 and standard deviation= 10 gallons

a) P(X>38)

Since ?=33.4 and ?=10 we have:

P ( X>38 )=P ( X??>38?33.4 )=P ( (X??)/?>(38?33.4)/10)

Since Z=(x??)/? and (38?33.4)/10=0.46 we have:

P ( X>38 )=P ( Z>0.46 )

Use the standard normal table to conclude that:

P (Z>0.46)=0.3228

b) P(20<X<30)

Since ?=33.4 and ?=10 we have:

P ( 20<X<30 )=P ( 20?33.4< X??<30?33.4 )=P ( (20?33.4)/10<(X??)/?<(30?33.4)/10)

Since Z=(x??)/? , (20?33.4)/10=?1.34 and (30?33.4)/10=?0.34 we have:

P ( 20<X<30 )=P ( ?1.34<Z<?0.34 )

Use the standard normal table to conclude that:

P ( ?1.34<Z<?0.34 )=0.2768

c) P(X<20)

Since ?=33.4 and ?=10 we have:

P ( X<20 )=P ( X??<20?33.4 )=P ((X??)/?<(20?33.4)/10)

Since (x??)/?=Z and (20?33.4)/10=?1.34 we have:

P (X<20)=P (Z<?1.34)

Use the standard normal table to conclude that:

P (Z<?1.34)=0.0901

d) Z value corresponding to 90%is 1.28

Z= X-mean/sigma

1.28*10= X-33.4

12.8= X-33.4

X= 33.4+12.8= 46.2 gallons

X= 46.2 gallons

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