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Question 1182608: Terri Vogel, an amateur motorcycle racer, averages per 129.95 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.28 seconds.
The distribution of her race times is normally distributed.
We are interested in one of her randomly selected laps.
Let X be the number of seconds for a randomly selected lap.
Round to 4 decimals.
a. What is the distribution of X?
=129.95 over 2.28
b. Find the proportion of her laps that are completed between 126.17 and 127.91 seconds.
c. Find the fastest 4% of laps are under ______ seconds.
d. The middle 40% of her laps are from ______ seconds to _______ seconds.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! X~N(129.95, 5.1984) units seconds, seconds^2 (use the variance)
z=(x-mean)/sd
z=-3.78/2.28=-1.658
z=(-2.04/2.28)=-0.895
that probability is 0.1367
can check with 2ndVARS2(normalcdf(126.17,127.91,129.95,2.28)ENTER which is unrounded and 0.1368. I would use the first, unless the course requires no rounding of the z-value until the end.
The fastest 4% of laps: z(0.04), since want the shortest time, z=-1.7506
-1.7506=(x-129.95)/2.28
=125.96 seconds
-
This requires z (0.3) and z(0.7), which are -0.5244 and +0.5244 respectively
-0.5244=(x-129.95)/2.28
x=128.75 sec
and
x=131.15 sec
If this is checked, it is about 0.4013 of the laps, the discrepancy's being that the multiplying z by -0.5244 gives 1.1956 seconds, which was rounded upward to 1.20 seconds.
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