a) to c) is look fine.
what s/w you are using? There must be some function for the inverse of the normal probabilities.
d) we have to find value P[time < k seconds] = P[X< k] =.20
e) P[X>k] =.05
Hello, I am having difficulty figuring out my homework. I made attempts and could someone check my answers and see if I'm right or make suggestions? Thank you
A physical-fitness association is including a mile run in its secondary-school fitness test for boys. The time for this event for boys in secondary school is approximately normally distributed with a mean of 450 seconds and a standard deviation of 40 seconds.
a)What is the probability that a randomly selected boy takes between 420 and 540 seconds to run a mile?
solution: normalcdf(420,540,450,40) = .761 = 76.1%
b)What is the probability that a randomly selected boy takes at least 480 seconds to run a mile?
solution: normalcdf(480,E99,450,40) = .227 = 22.7%
c)What is the probability that a randomly selected boy takes at most 390 seconds to run a mile?
solution: normalcdf(E99,390,450,40) = .0668 = 6.68%
d)If the association wants to designate the fastest 20% as "above average, what time should the association set for this criterion?
solution: please give me advice!
e)If the association wants to designate the slowest 5% as "in need of medical evaluation," what time should the association set for this criterion?
solution: please give me advice!
Thank you in advance! I am trying!!
a) to c) is look fine.
what s/w you are using? There must be some function for the inverse of the normal probabilities.
d) we have to find value P[time < k seconds] = P[X< k] =.20
e) P[X>k] =.05
In the long run, we're all dead.
c)
so you are saying to use the invnorm function?
this is what i tried
invNorm(.80,450,40) = 483.66 seconds
is this correct?
it is d) I don't think that value make sense.
Because fastest mean the lowest time.
Try this.
invNorm(.20,450,40)
In the long run, we're all dead.
d)
invNorm(.05,450,40)
=384.206 seconds
could this be it?
No. It is not correct. Do you think slowest and fastest can be identified using same strategy?
Think practically about the values. like "slowest means 384 ? ". Since mean is 450 so it can't be slowest.
This will help to avoid wrong answer.
In the long run, we're all dead.
okay i tried invNorm(.20,450,40) = 416.335, is this the answer?
As an incentive to improve the fitness of the secondary-school boys for a certain town, a rich benefactor offers a monetary prize to every student. Each student will earn P=640-X dollars, where X is their time in the mile run.
a) What is the mean payout to a secondary-school boy (or, the mean of the random variable P)?
solution: P = 640 - 450 = $190
b) What are the variance and the standard deviations of the payout to a secondary-school boy (or, what are the variance and the standard deviations of the random variable P?)
solution: advice needed.
b)
Var(P) = Var(640-X)
hint:
Var( a+bX) = b^2*Var(X)
In the long run, we're all dead.
i have no clue what you are trying to get me to do. thanks for helping
wait, i went back and did the first question and figured it out kind of
e) for the slowest 5%
invnorm(.95,450,40) = 515.79 seconds
d)the fastest 20%
how would i figure this out?
in the variance what would be the "b"
ok i am being confusing i know sorry
these are the answers i have so far
1.
a) 76.1%
b) 22.7%
c) 6.68%
d) 416.34 seconds (fastest 20%, "above average are faster than 416.34)
e) 515.79 seconds (slowest 5%, "in need of medical evaluation")
2.
a)P=640-450= $190
b)i need to find the variance and standard deviation. please help
and also check my answers! thanks!
Answers are correct.
Regarding variance. I don't think hint will work for you.
a and b are constant
In your case b = -1 and a =640
Var(640 - X ) = (-1)^2Var(X) = Var(X).=40^2
I hope you know that standard deviation. = sqrt(Var)
Last edited by vinux; 10-28-2008 at 04:39 PM.
In the long run, we're all dead.
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