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:
a. 100 draws are made
b. 1000 draws are made

Answer 1

Let X = # of 5's drawn in 100 number generated draws.
so X can be 0,1,2,...99,100
X is a binomial random variable. each trial is independent, has 1/10 chance of picking a 5.

Answer 2

with me so far?

Answer 3

wrong :(

Answer 4

can you take a screen shot of the question online , how many decimal places round?

Answer 5

sorry i gave you the wrong answer

Answer 6

0.2969668992

Answer 7

let me try

Answer 8

i made a mistake earlier
the correct command is
1 - binomcdf(100,1/10, 11) , you might have to round

Answer 9

are you using stata?

Answer 10

TI 83

Answer 11

ok

Answer 12

we want probability more than 11% of 100 draws have 5 (more than 11 draws),
so we are asking P( X > 11 ) , where X = # of 5's drawn by number generator.
it is easier to look at the complement P (X>11) = 1 - P(X <= 11)
1- binomialcdf(100, .10, 11)

Answer 13

= 0.2969668992

Answer 14

so whats up?

Answer 15

wrong :(

Answer 16

im pretty sure thats correct, can i see this online?

Answer 17

one moment

Answer 18

maybe we have to use stata

Answer 19

are you using a special calculator or something

Answer 20

does it say to use normal approximation

Answer 21

can you tell me what the correct answer should be?

Answer 22

at least i might be able to get the next one correct

Answer 23

Solution
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