Exam 2 (Chapters 5, 6, 7) Summer 2018 instructions I help Question 1 (of 12) Sav

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Exam 2 (Chapters 5, 6, 7) Summer 2018 instructions I help Question 1 (of 12) Save & Exit Submit Time remaining: 2:28:27 16.00 points According to industry sources, the average U.S. customer spends $1,350 for a do-it-yourself bathroom remodeling project (vanity, sink, faucet, mirror, and toilet). The total sale amount has a normal distribution with a standard deviation of $125 What is the probability that a randomly selected customer will have a total sale of more than $1,455? ANSWER: table in your textbook to answer this question. Report your answer to 4 decimal places, using conventional rounding rules) (Use only the appropriate formula and/or statistical % (Use only the appropriate formula and/or statistical table in your What percent of the total sales amounts will be between $1,380 and $1,410? ANSWER: textbook to answer this question. Report your answer as a percentage to 2 decimal places, using conventional rounding rules) What is the probability that a randomly selected customer will have a total sale of less than $1,280? ANSWER: table in your textbook to answer this question. Report your answer to 4 decimal places, using conventional rounding rules) (Use only the appropriate formula and/or statistical Thirty-three percent of the total sales amounts should be below what dollar value? ANSWER: S textbook to answer this question. Report your answer as a whole dollar amount (i.e., no decimal places), using conventional rounding rules) (Use only the appropriate formula and/or statistical table in your

Explanation / Answer

Sol:

mean=1350

sd=125

P(X>1455)

z=x-mean/sd=1455-1350/125=0.84

P(Z>0.84)

=1-P(Z<0.84)

=1-0.7995

=0.2005

ANSWER:0.2005

Solutionb:

P(1380<X<1410)

P(1380-1350/125<Z<1410-1350/125)

P(0.24<Z<0.48)

=0.0896

=0.0896*100

=8.96%

ANSWER:8.96%

SOLUTIONC:

P(X<1280)

P(Z<1280-1350/125)

P(Z<-0.56)

=1-P(Z<0.56)

=1-0.7123

=0.2877

ANSWER:0.2877

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