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Old 06-12-2012, 07:51 PM   #1
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statistics class problem

A circus performer who gets shot from a cannon is supposed to land in a safety net
positioned at the other end of the arena. The distance he travels is normally distributed
with a mean of 175 feet and a standard deviation of 15 feet. His landing net is 50 feet
long and the mid-point of the net is positioned 175 feet from the cannon. What is the
probability that the performer will hit the net on a given night? (Round your answer(s) to
3 decimal places.)

is it P(160<X<190)?
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Old 06-12-2012, 08:06 PM   #2
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Re: statistics class problem

the only way he'll hit a 50 feet net is if he lands in a 30 feet interval?
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Old 06-12-2012, 08:14 PM   #3
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Re: statistics class problem

i'm not really sure how to do the problem.

do i do 25/15 = 1.67 std devation.

so use the z score for 1.67?
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Old 06-12-2012, 08:33 PM   #4
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Re: statistics class problem

okay i figured it out.

net is centered 175. 25 to the left, 25 to the right for the total of 50 of the net.
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Old 06-12-2012, 11:48 PM   #5
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Re: statistics class problem

Go on... (seriously, I want to see you get this)
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Old 06-13-2012, 12:03 AM   #6
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Re: statistics class problem

The OP should have been posted in the SPM official homework question thread. Make sure to do that next time. Thanks.
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Old 06-13-2012, 12:42 AM   #7
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Re: statistics class problem

Quote:
Originally Posted by kingofsurvivorbb View Post
i'm not really sure how to do the problem.

do i do 25/15 = 1.67 std devation.

so use the z score for 1.67?
This looks right to me. Comes out to 90.5% (0.905 is the requested form).

As an aside, the net is too damn small. An interesting followup question would be for you to calculate the needed size to ensure, say, 99.99% chance of hitting the net, so the guy only dies once in 10,000 tries. I think that would be a reasonable standard.

Last edited by NewOldGuy; 06-13-2012 at 12:56 AM.
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Old 06-13-2012, 03:02 AM   #8
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Re: statistics class problem

He has to land between 150 and 200 with avg 175 which is +-1.667sd from center. This means 1-2*P(x>1.667)=1-2*(1-P(x<1.667))=1-2*(1-0.952)=0.904

90.4% chance.
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