A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Find the chance that the digit 5 appears on more than 11% of the draws, if:1000 draws are made

Question

anonymous · Answer

did u get the ans for 100 draws

anonymous · Answer

the answer is 0.2969668992

anonymous · Answer

how did u get that

anonymous · Answer

0.1348v is the ans 4 this ques

anonymous · Answer

ok I used the graphic calculator and used the binopdf function

perl · Answer

let X = # fives in 100 random numbers drawn
X is a bnomial random variable , p = 1/10 ,

perl · Answer

it says "more than 11%"

anonymous · Answer

yes pearl but when I use 1000 I get the answer in calculator is 1 but correct answer is 1.348. Why is it so? will you explain it?

perl · Answer

wait, what is the answer for the first one?

perl · Answer

for 1000 draws the answer is .134776

anonymous · Answer

the answer is 0.2969668992

anonymous · Answer

yes you are write but when I used the same logic and inset in Ti-84 I got 1.

perl · Answer

wait,

perl · Answer

how do you know you got the first part right?

anonymous · Answer

Great! Pearl I got it.
Thanks a lot

perl · Answer

we want probability picking 5 for more than 11% of the sample (11% of 100 )
so we want P( X > 11 ) 
P( X > 11 ) = 1 - P(X<=11) = 1 - binomcdf (100, .1, 100*.11)

perl · Answer

similiarly for 1000 draws we have
P( X > 110 ) = 1 - P( X<=110) = 1 - binomcdf ( 1000, .1, 110)

anonymous · Answer

Yes Bro I got it and understood where I was wrong. Thanks

perl · Answer

because 11% of 1000 = 110

perl · Answer

is this online?

perl · Answer

please tell us if we are right :)

anonymous · Answer

Yes you are 100% right

perl · Answer

correct answer is 
a).2969...
b) .1348

anonymous · Answer

b

perl · Answer

yeah i was doing the other one from yesterday . any more questions>

anonymous · Answer

Thanks bro. Not right now