Question

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
1. 100 draws are made
2. 1000 draws are made

Answer 1

Let X be the number of times the digit 5 appears out of n draws. X ~ Binomial(n,p). As n gets large, X ~ N(np,np(1-p)). Find P(X>11) for 1(use n=100) and P(X>110) for 2 (use n=1000).

Answer 2

Thank you for your response drawar, I am however still confused on how to calculate it right now. Would you mind explaining it a bit more ? Thanks in advance

Answer 3

Which part escapes you?

Answer 4

@drawar Would you explaining it a bit more(part 1)???

Answer 5

1. X~Bin(100,1/10) => X~N(10,9) => P(X>11)=P(Z>0.333)=.... (look up the answer in any normal distribution table).
2. Just replace n=100 by n=1000, so X~Bin(1000,1/10) =>....

Answer 6

@drawar 0.3707 is correct????(part1)

Answer 7

Don't know cause I don't have a table with me, just follow the hint and you SHOULD get the correct answer.

Answer 8

What's the (given) correct answer?

Answer 9

hello , i got a different answer

Answer 10

@perl your ans is correct??

Answer 11

hello

Answer 12

I got
1- binomcdf( 100, .10, 11 ) = .296966899

Answer 13

if you use normal approximation you should get after continuity correct
normalcdf( 11.5, 10^99, 10, 3 )

Answer 14

or using z score
(11.5 - 10) / 3 =.5
=.30853

Answer 15

or using z score
P( Z > (11.5 - 10) / 3 =.5)
=.30853

Answer 16

@perl n wht about
2. 1000 draws are made

Answer 17

@abhi_abhi
mate if u got 1st one than plzzz tell me

Answer 18

1)0.297
2)0.135