Problem 1:
# Graph a binomial distribution in R.
# First we compute all probabilities.
# X is binomial(10,0.5), therefore, X=0,1,2,..., 10.
x <- seq(0,10)
px <- dbinom(x, size = 10, prob = 0.5)
barplot(px, space = 0, xlab = "X", ylab = "P(X)", ylim = c(0, .25),
main = "Binomial probability distribution: n=10, p=0.5", names=seq(0,10,1))
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Problem 2:
a. X ~ B(20, 0.3).
1. pbinom(4, 20, 0.3)
2. pbinom(7, 20, 0.3, lower.tail=FALSE)
3. pbinom(5, 20, 0.3, lower.tail=FALSE)
4. pbinom(2, 20, 0.3)
5. pbinom(6, 20, 0.3) - pbinom(2, 20, 0.3)
b. X ~ N(13, 2.5)
1. pnorm(16.1, mean=13, sd=2.5)
2. 1-pnorm(16.5, mean=13, sd=2.5) OR pnorm(16.5, mean=13, sd=2.5, lower.tail=FALSE)
3. pnorm(13.5, mean=13, sd=2.5)
4. pnorm(15, mean=13, sd=2.5) - pnorm(12, mean=13, sd=2.5)
5. qnorm(0.61, mean=13, sd=2.5)
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Problem 3:
We have X ~ b(b, 0.58). By trial and error, you will find that n=568. We want the probability to be at least 95% that at least 310 student will be enrolled. So we want, P(X >= 310) >= 95%.
Let's begin with n=500:
pbinom(309, 400, .58, lower.tail=FALSE)
0.03813603 (very small!).
Let's increase to 530:
pbinom(309, 530, .58, lower.tail=FALSE)
0.4275875 (still not close to 95%).
Let's increase to 560:
pbinom(309, 560, .58, lower.tail=FALSE)
0.9046458 (much closer to 95%).
Try n=570:
pbinom(309, 570, .58, lower.tail=FALSE)
0.9629718 (this is above 95%).
Try now, n=567
pbinom(309, 567, .58, lower.tail=FALSE)
0.9498989 (a little bit below 95%).
Finally, use n=568:
pbinom(309, 568, .58, lower.tail=FALSE)
0.9546223 (this satisfies "at least 95 %").
Important: Ask the students to think about it, by giving them the following hint:
Another way to answer this question is to set up an equation using normal approximation to binomial. How?
-1.645 = (309.5 - n*0.58) / sqrt(n*0.58*0.42), and solve for n.
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Problem 4:
It is given that X ~ N(3, 0.3).
a. pnorm(2.5, mean=3, sd=0.3)
b. pnorm(3.2, mean=3, sd=0.3) - pnorm(2.5, mean=3, sd=0.3)
c. pnorm(3.2, mean=3, sd=0.3, lower.tail=FALSE)
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Problem 5:
This is binomial with n=275, p=0.95.
Exact probability:
P(X >265) = sum(x=265 to 275) (275 choose x) * 0.95^x * 0.05^(275-x)
Approximate using normal:
P(X > 265) = P(Z>265.5 - n*p) / (n*p*(1-p))^0.5
where np=275*0.95=261.25 and
(n*p*(1-p))^0.5 = (275*0.95*0.05)^0.5=3.614208.
Therefore, P(X > 265) = P(Z > 265.5-261.25 / 3.614208) = P(Z>1.175915)
Using R - exact:
pbinom(265, 275, 0.95, lower.tail=FALSE)
Answer: 0.1155464
Using R - approximate:
pnorm(265.5, mean= 261.25, sd= 3.614208, lower.tail=FALSE)
Answer: 0.1198145
Exact and approximate are very close.