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

Question

22 days ago

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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 [9.1K] · 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.

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

A. C. E. F.

Which correspond to 2, 3, 5, and 6.