Question

Answer only e) to h) problems a) to d) have the answer included

Problem 1: Relations among Useful Discrete Probability Distributions. A Bernoulli experiment consists of only one trial with two outcomes (success/failure) with probability of success p. The Bernoulli distribution

is

P (X = k) = p^kq^1-k, k=0,1

The sum of n independent Bernoulli trials forms a binomial experiment with parameters n and p. The binomial probability distribution provides a simple, easy-to-compute approximation with reasonable accuracy to hypergeometric distribution with parameters N, M and n when n/N is less than or equal to 0.10. In this case, we can approximate the hypergeometric probabilities by a binomial distribution with parameters n and p = M/N. Further, the Poisson distribution with mean μ = np gives an accurate approximation to binomial probabilities when n is large and p is small.

1. Suppose in a region in Saskatchewan, among a group of 20 adults with cancer, seven were physically abused during their childhood. A random sample of five adult persons is taken from this group. Assume that sampling occurs without replacement, and the random variable X represents the number of adults in the sample who were abused during their childhood period.

(a) Write the formula for p(x), the probability distribution of X. How this distribution is called?

P(X=x)=(5/x)(0.35)^x(1-0.35)^5-x

(b) Using the adequate formulas, find the mean and variance of X?

μ=1.75

variance = 1.1375

(c) Find the probabilities of all the possible values of X. Plot the histogram of X and try the locate the approximative value of the mean μ.

x 0 1 2 3 4 5

P(X=x): 0.116 0.3124 0.3364 0.1811 0.04877 0.005

(d) What is the probability that at least one person was abused during childhood?

0.884

Now suppose another survey in British Columbia reveals that among 180 adults with cancer, only 80 adults were abused in their childhood. Suppose again that a random sample of five adult persons is taken from this group without replacement and let denote by Y the random variable which represents the number of adults abused during their childhood period in the sample.

(e) Find the probabilities of all the possible values of Y and plot the histogram of Y. How do you compare this histogram with the histogram of X.

(f) Find the probabilities of all the possible values of Y using the formula for the binomial distribution with p = 80/180

as an approximation. Plot the histogram and compare it with the histogram obtained using the hypergeometric formula.

(g) Is the precedent approximation close enough? Why or why not?

(h) Calculate the mean and variance using both binomial and hypergeometric distributions, respectively. Provide a comparison and summarize your findings.

Answer 1

1. X=physically abused during their childhood

p(success)=

q(failure)=1-p=1-0.35

n=5

X~binom(5,0.35)

(a) Write the formula for p(x), the probability distribution of X. How this distribution is called?

P(X=x)=

Binomial distribution.

(b) Using the adequate formulas, find the mean and variance of X?

X~binom(5,0.35)

n=5

p=0.35

q=1-0.35=0.65

mean=np=5*0.35=1.75

variance=npq=5*0.35*0.65=1.1375

(c) Find the probabilities of all the possible values of X. Plot the histogram of X and try the locate the approximative value of the mean μ?

possible values of X:

X=0,1,2,3,4,5

P(X=x)=

P(X=0)=

P(X=0)=

P(X=0)=0.116

P(X=1)=

P(X=1)=

P(X=1)=0.3124

P(X=2)=

P(X=2)=

P(X=2)=0.3364

P(X=3)=

P(X=3)=

P(X=3)=0.1811

P(X=4)=

P(X=4)=

P(X=4)=0.04877

P(X=5)=

P(X=5)=

P(X=5)=0.005

SUMMARY:

X	0	1	2	3	4	5
P(X=x)	0.116	0.3124	0.3364	0.1811	0.04877	0.005

Frequency Table
Class	Count
0-0.13999	3
0.14-0.27999	1
0.28-0.41999	2

(d) What is the probability that at least one person was abused during childhood?

P(X1)=1-P(X<1)

P(X1)=1-P(X=0)

P(X1)=1-0.116

P(X1)=0.884

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