Approx 10.3% drop out before graduation. Find prob. that 6

MRS.FREE

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I got some help with this type of problem and am trying to do one on my own now, can someone look this over really good and see if im doing this correctly please. im sorry, i don't know how to add powers and all that so i just typed it :)

APPROXIMATLY 10.3% OF AMERICAN HIGH SCHOOL STUDENTS DROP OUT OF SCHOOL BEFORE GRADUATION. CHOOSE 10 STUDENTS ENTERING HIGH SCHOOL AT RANDOM. FIND THE PROBABILITY THAT AT LEAST 6 GRADUATE.

n=10
p=.103
q=.897
x= 6,7,8,9,10

p(6)= 10!/(10-6)!6! (.6)to the 6th power (.4) to the fourth power=.251
p(7)= 10!/(10-7)!7! (.6)to the seventh power (.4) to the third power = .215
p(8)= 10!/(10-8)!8! (.6) to the eighth power (.4) squared= .121
p(9)= 10!/(10-9)!9! (.6) to the ninth power (.4) to the first power= .040
p(10)= 10!/(10-10)!10! (.6) to the tenth power (.4) to the 0 power= .006

(.251)(.215)(.121)(.006)= .633

one little mistake and it messes EVERYTHING up...am i on the right track? thak you :)
 
Re: please look to see if i did it right and ty!

Your final answer comes from adding p(6) through p(10).
 
Re: please look to see if i did it right and ty!

ok, some how i added wrong...the answer is .593...what im asking is did i do the work in p(6)-p(10) correctly, can someone please tell me :)
 
Re: please look to see if i did it right and ty!

See Here
 

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for the same word roblem:

no more than 2 drop out=
p(1)= 10!/(10-1)!1! (.103) to the first power (.897) to the ninth= .387
p(2)= 10!/(10-2)!2! (.103) to the 2nd power (.897) to the eighth= .200

.387 + .200= .587


all 10 graduate=
i came up with all the answers from p(1)-p(10)
p(1)=.387
p(2)=.200
p(3)=.061
p(4)=.012
p(5)=.001
here's where i started secon guessing myself
p(6)=.000162335
p(7)=.000010652
p(8)=.000000459
p(9)=.000011704
p(10)= 1.343 to the negative tenth power does that mean.00000000001343 ???

so when i added them all up i got .661

am i correct?

i really want to make sure im not learning this wrong and applying it incorrectly :)
 
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