Question

Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 employees; the Hawaii plant has 20. A random sample of 10 employees is to be asked to fill out a benefits questionnaire.
a. What is the probability that none of the employees in the sample work at the plant in Hawaii (to 4 decimals)?b. What is the probability that 1 of the employees in the sample works at the plant in Hawaii (to 4 decimals)?c. What is the probability that 2 or more of the employees in the sample work at the plant in Hawaii (to 2 decimals)?d. What is the probability that 9 of the employees in the sample work at the plant in Texas (to 2 decimals)?

Answer 1

Answer:

a) There is an 1.74% probability that none of the employees in the sample work at the plant in Hawaii.

b) There is an 8.70% probability that 1 of the employees in the sample works at the plant in Hawaii.

c) There is an 89.56% probability that 2 or more of the employees in the sample work at the plant in Hawaii.

d) There is an 8.70% probability that 9 of the employees in the sample work at the plant in Texas.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they work at the Texas plant, or they work at the Hawaii plant. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

Our sample has 10 employees, so $n = 10$.

a. What is the probability that none of the employees in the sample work at the plant in Hawaii (to 4 decimals)?

Of the 60 employees in total, 20 work in Hawaii, so:

$p = \frac{20}{60} = 0.333$

We want P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.333)^{0}.(0.667)^{10} = 0.0174$

There is an 1.74% probability that none of the employees in the sample work at the plant in Hawaii.

b. What is the probability that 1 of the employees in the sample works at the plant in Hawaii?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.333)^{1}.(0.667)^{9} = 0.0870$

There is an 8.70% probability that 1 of the employees in the sample works at the plant in Hawaii.

c. What is the probability that 2 or more of the employees in the sample work at the plant in Hawaii?

There are two possible outcomes. Either there are less than 2 employees in the sample working at the plant in Hawaii, or there are more than two. The sum of the probabilities of these events is decimal 1. So:

$P(X < 2) + P(X \geq 2) = 1$

We want to find $P(X \geq 2)$.

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

From a. and b., we have both these probabilities.

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0174 + 0.0870 = 0.1044$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1044 = 0.8956$

There is an 89.56% probability that 2 or more of the employees in the sample work at the plant in Hawaii.

d. What is the probability that 9 of the employees in the sample work at the plant in Texas?

This is the same as the probability that 1 of the employees in the sample work at Hawaii, that we found in b.

So, there is an 8.70% probability that 9 of the employees in the sample work at the plant in Texas.