probability homework

#1
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!!:confused:
 

vinux

Dark Knight
#2
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
 

vinux

Dark Knight
#6
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.
 
#9
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.
 
#12
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?
 
#14
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!
 

vinux

Dark Knight
#15
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)
 
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