Question

Binomial Distribution Problem 1: (a) An urn contains 1000 balls, 100 are green and 900 are white. One ball is chosen from the urn 100 times with replacement. Use Excel (binom.dist) to find the probability that six or seven green balls are selected. (b) An urn contains 1000 balls, 100 are green and 900 are white. One ball is chosen from the urn 1000 times. Use Excel (binom.dist) to find the probability that between 110 and 120 of the balls, inclusive, are green. (c) Redo (a) and (b) again using Excel but use the normal approximation (normal.dist). How do the answers compare with the above? Are there any discrepancies? If so, please explain why they happened.

Answer 1

	Total	1000
	Green	100
	White	900
a)	p	0.1	\=Green/Total
	n	100
	P ( X = 6)	0.059579	\=binom.dist(6, n, p, False)
	P( X = 7)	0.088895	\=binom.dist(7, n, p, False)
	P( 6 or 7 balls are selected)	0.148474	\=P(X = 6) + P(X = 7)
b)
	p	0.1
	n	1000
	P( 110 <= X <= 120 )	0.141	\=binom.dist(120, n, p, True) - binom.dist(109, n, p, True)
c)
	p	0.1
	n	100
	Mean	10	\=n*p
	Variance	9	\=n*p*(1-p)
	Standard Deviation (SD)	3	\=sqrt(Variance)
	P ( 6 <= X <= 7)
	P( 5.5 <= X <= 7.5)		continuity correction
	P( 5.5 <= X <= 7.5)	0.135521	\=norm.dist(7.5, mean, sd, true) - norm.dist(5.5, mean, sd, True)
d)
	p	0.1
	n	1000
	Mean	100
	variance	90
	sd	9.486833
	P( 110 <= X <= 120 )
	P( 109.5 <= X <= 120.5 )		continuity correction
	P( 109.5 <= X <= 120.5 )	0.142968

We can see that there is a difference between a and c but the approximation for b is good by d. Because we have a rule that if n*p and n*(1-p) is greater than 10 and n is large, then normal gives a good approximation to the binomial distribution. So, in case (a) np = 10. That's why it is not a good approximation and also n is not that large in comparison to (b) part