Mathematics, 08.02.2021 19:10 samsmith666

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, 3% of the components are identified as defective, and the components can be assumed to be independent. Required:
a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?
b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?
c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

Answers: 1

Show answers

Answers

Answer from: rjmarin592ovzbn9

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

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.

3% of the components are identified as defective

This means that p = 0.03

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is P(X = 0) when n = 90 . So

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

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when n = 102

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when n = 105

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

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

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Answer from: Quest

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64. therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

Answer from: Quest

what is

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 18:00

Clara schumann is buying bagels for her coworkers, she buys a dozen bagels priced at $5.49 a dozen. the bakery's cost for making the bagels is $2.25 per dozen. what is the markup rate based on selling price on a dozen bagels?

Answers: 1

Answer

Mathematics, 21.06.2019 19:30

Combine the information in the problem and the chart using the average balance method to solve the problem.

Answers: 2

Answer

Mathematics, 21.06.2019 21:00

The sum of the first eight terms of an arithmetic progression is 9m + 14. the fifth term of the06 progression is 2m - 6, where m is a constant.given that the first term of the progression is 26,find the value of the common difference, d and thevalue of m.

Answers: 1

Answer

Mathematics, 21.06.2019 23:30

Zach spent 2/3 hour reading on friday a 11/3 hlurs reading on saturday. how much more time did he read on saturday tha on friday

Answers: 1

Answer

You know the right answer?

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process...

Questions