Question

Please can Any One help!
A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. In 5000 draws, the chance that the digit 0 appears fewer than 495 times is closest to: 25%, 30%, 35%, 40%, 45% or 50%?
My solution
P(0 appears fewer than 495)=1/10=0.1

n=5000

k<495; K=494

5000C494*(0.1^494) * (0.9^4506)=?

Answer 1

makes sense to be... as a binomial probability solution...

Answer 2

This can be solved by the Normal approximation to the binomial distribution.
\[\mu=np=5000\times 0.1=500\]
\[\sigma=\sqrt{np(1-p)}=\sqrt{5000\times 0.1\times 0.9}=21.2132\]
The z-score for 494 can be calculated, with the continuity correction, as follows:
\[z=\frac{494-500+0.5}{21.2132}=-0.25927\]
Reference to a standard Normal distribution table shows that the chance that the digit 0 appears fewer than 495 times is closest to 40%.

Answer 3

Thanks Kropot72

Answer 4

You're welcome :)