SOLUTION: A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of m

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Question 1110339: A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 35 months and a standard deviation of 5 months. Using the empirical rule (as presented in the book), what is the approximate percentage of cars that remain in service between 20 and 30 months?
Do not enter the percent symbol.
ans as a % =

AND

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 7. Using the empirical rule (as presented in the book), what is the approximate percentage of lightbulb replacement requests numbering between 48 and 55?
Do not enter the percent symbol.
ans =

LASTLY

The combined SAT scores for the students at a local high school are normally distributed with a mean of 1481 and a standard deviation of 293. The local college includes a minimum score of 1217 in its admission requirements.
What percentage of students from this school earn scores that fail to satisfy the admission requirement?
P(X < 1217) =
Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
mean 35 sd 5
20 is 3 sd s to the left of the mean
30 is 1 sd to the left.
The empirical rule has 68% within 1 sd or 34% on one side
It has 95% within 2 sd or 47.5% on one side
It has 99.7% within 3 sd or 49.85% on one side
Therefore between 3 sd and 1 sd on one side is 49.85-34=15.85% or enter 15.9
mean of 48 sd 7
between 48 and 55 is between -1 and 0 sd or 34% enter 34. The last one doesn't seem to post easily:z=(x-mean)/sd or z< (1217-1481)/293 or z <-264/293 or -0.901,
That probability is 0.1838 or 18.4% or enter 18.4