Mathematics, 17.02.2020 18:33 painespeach1849

A manufacturing company has 5 identical machines that procuce nails. The probability that a machine will break down on any given day is 0.1. Define a random variable x to be the number of machines that will break down in a day.

a. What is the appropriate probability distribution for x? Explain how x satisfies the properties of the distribution.
b. Compute the probability that 4 machines will break down.
c. Compute the probability that at least 4 machines will break down.
d. What is expected number of machines that will break down in a day?
e. What is the variance of the number of machines that will break down in a day?

Answers: 2

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Answer from: sneakypanther19

a) Binomial probability distribution. Only two outcomes possible for each machine, with independent probabilities.

b) P(X = 4) = 0.00045

c) $P(X \geq 4) = 0.00046$

d) E(X) = 0.5

e) V(X) = 0.45

Step-by-step explanation:

For each machine, there is only two possibilities. On a given day, either they will break down, or they will not. The probabilities for each machine breaking down are independent. So we use the binomial probability distribution to solve this problem.

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 combinations 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.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

In this problem we have that:

n = 5, p = 0.1

a. What is the appropriate probability distribution for x? Explain how x satisfies the properties of the distribution.

Binomial probability distribution. Only two outcomes possible for each machine, with independent probabilities.

b. Compute the probability that 4 machines will break down.

This is P(X = 4) .

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

$P(X = 4) = C_{5,4}.(0.1)^{4}.(0.9)^{1} = 0.00045$

c. Compute the probability that at least 4 machines will break down.

This is

$P(X \geq 4) = P(X = 4) + P(X = 5)$

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

$P(X = 4) = C_{5,4}.(0.1)^{4}.(0.9)^{1} = 0.00045$

$P(X = 5) = C_{5,5}.(0.1)^{5}.(0.9)^{0} = 0.00001$

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.00045 + 0.00001 = 0.00046$

d. What is expected number of machines that will break down in a day?

E(X) = np = 5*0.1 = 0.5

e. What is the variance of the number of machines that will break down in a day?

V(X) = np(1-p) = 5*0.1*0.9 = 0.45

Answer from: Quest

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(continued on page 79)

step-by-step explanation:

Answer from: Quest

What are the following algebraic equations. you need to post those if you want the answer

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