Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 emp

Question

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Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 emp

loyees; 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)?

Mathematics

1 answer:

Zina [12.3K] · Answer 1 · 2026-03-05T22:20:40+03:00

Answer:

a) The likelihood that none of the sampled employees are from the Hawaii plant is 1.74%.

b) The chance that exactly 1 employee from the sample is found working in the Hawaii plant is 8.70%.

c) There is an 89.56% chance that 2 or more employees in the sample are from the Hawaii plant.

d) The probability that 9 employees from the sample are working at the Texas plant is 8.70%.

Step-by-step explanation:

Each employee has two potential employment locations: either Texas or Hawaii. Thus, the binomial probability distribution can be utilized to solve this scenario.

Binomial probability distribution

This distribution defines the probability of achieving exactly x successes in n trials where there are only two outcomes.

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

Here, $C_{n,x}$ denotes the number of ways to choose x objects from a set of n, represented by the subsequent formula.

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

And p is the probability of success occurring.

In this context, we know:

The sample comprises 10 employees, therefore $n = 10$ .

a. Calculate the probability that none of the sampled employees are from the Hawaii plant (to 4 decimals)?

Given that 20 out of 60 employees are based in Hawaii:

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

We aim to find P(X = 0).

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

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

Thus, the likelihood that none in the sample are from Hawaii stands at 1.74%.

b. Calculate the probability that 1 employee from the sample is from the Hawaii plant?

This is represented as P(X = 1).

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

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

Therefore, there is an 8.70% possibility that 1 employee in the sample comes from Hawaii.

c. Calculate the probability that 2 or more employees in the sample are from the Hawaii plant?

We can observe two scenarios: either fewer than 2 employees are from Hawaii or 2 and beyond. The combined probabilities equal decimal 1. So:

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

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

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

From problems a and b, we possess values for both probabilities.

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

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

Accordingly, the chance that 2 or more employees in this sample operate at the Hawaii plant is 89.56%.

d. Calculate the likelihood that 9 employees in the sample are working at the Texas plant?

This corresponds to the probability found in part b for 1 employee working in Hawaii.

Consequently, there is an 8.70% chance that 9 employees belong to the Texas plant.