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

ok think of this as a binomial problem. you are drawing 100 numbers randomly , the success if picking 5 , a failure is picking a different number.
probability of success (picking a 5) is 1/10. probability of failure is 9/10
so we have a binomial distribution.
now i think you can approximate it with a normal distribution

Answer 2

the mean of a binomial distribution is
mean = n*p
std dev = sqrt ( n p q )

Answer 3

now you want probability ( X > = 11% of the 100 draws) = P ( X > = 11)

Answer 4

P( X > = 11) = 1 - P ( X < 11) =.4168

Answer 5

and you can solve this using a normal distribution approximation to the binomial distribution
the normal distribution will have mean = n * p , and st dev = sqrt(n*p*q)

Answer 6

Unfortunately, I haven't got correct results.

Answer 7

0.312

Answer 8

i THINK , ANSWER 1

Answer 9

sorry
, I dont understand

Answer 10

pls any one can ans above askd question clearly