Question

Suppose 5 of 25 Ford subcompact automobiles require adjustment of some kind. Four subcompacts are selected at random. We are interested in the probability that exactly one will require adjustment. a. Solve the problem assuming that of the 25 subcompacts, the samples are drawn without replacement. b. Solve the problem assuming the sampling is done with replacement c. Assuming the replacement, work the problem using the Poisson distribution.

Answer 1

a)

You can use the hypergeometric distribution to find the solution

Let X be the number of subcompacts requiring adjustments. X has the hypergeometric distribution with the following parameters.

K = number of items to be drawn = 4

N = total objects = 25

M = number of objects of a given type = 5

P(X = 1 ) = 0.4505

b)

Let X be the number of subcompacts requiring adjustments. X has the binomial distribution with n = 4 trials and success probability p = 0.2.

In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

the mean of the binomial distribution is n * p = 0.8

the variance of the binomial distribution is n * p * (1 - p) = 0.64

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 0.8

P( X = 1 ) = 0.4096

c)

Let X be the number of subcompacts requiring adjustments. X has the Poisson distribution with parameter λ = np = 0.8

In general you have:

X ~ Poisson( λ )

P(X = x) = ( λ )^x * exp( -λ ) / x! for x = 0, 1, 2, 3, 4, ...

P(X = x) = 0 otherwise

the mean of the Poisson distribution is the parameter, λ

the variance of the Poisson distribution is the parameter, λ

In this problem we have

λ = 4 * 0.2 = 0.8

this results in our random variable X ~ Poisson( 0.8 )

Find P(X = 1 ) = 0.8 ^ 1 * exp( -0.8 ) / 1 ! = 0.3594