The American Automobile Association (AAA) reported that families planning to travel over the Labor Day weekend would spend an average of $749 (The Associated Press, August 12, 2012). Assume that the amount spent is normally distributed with a standard deviation of $225

Pavukol

Pavukol

Answered question

2022-09-06

The American Automobile Association (AAA) reported that families planning to travel over the Labor Day weekend would spend an average of $749 (The Associated Press, August 12, 2012). Assume that the amount spent is normally distributed with a standard deviation of $225
a) What is the probability that family expenses for the weekend will be less than $400?
b) What is the probability that family expenses for the weekend will be $800 or more?
c) What is the probability that family expenses for the weekend will be between $500 and $1000?
d) What are Labor Day weekend expenses for 5% of the families with the most expensive travel plans?

Answer & Explanation

coccusk7

coccusk7

Beginner2022-09-07Added 14 answers

Step 1
Mean = 749
Standard Deviation = 225
a) Z-Score for 400 400 749 225 = 1.5511 P ( Z < 1.5511 ) = 0.0604
b) Z-Score for 800 800 749 225 = 0.22667 P ( Z > 0.22667 ) = 0.4103
c) Z-Score for 500 500 749 225 = 1.10667
Z-Score for 1000 1000 749 225 = 1.1155 P ( 1.10667 < Z < 1.1155 ) = 0.7335
d) Z-Score for Top 5 % 1.645 ( From Z-Score Table)
X 749 225 = 1.645 X 1119.125
Therefore , expenses should be greater than $1119.125

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