Probability in Process Control Limit Charts
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I am currently taking an Operations Managment class where we are discussing control limits for various processes/tasks. For example, we discuss a machine that produces memory cards of a specific width $x_n$. However, because the machine is not perfect, the produced memory cards width's follows a normal distribution.
What is the probability that the next 5 (or however many) chips it produces would have widths of increasing order $(x_1<x_2<x_3...x_N)$
If the question is reduced to asking for the probability that the next two are in increasing order, $P(x_1>x_2)=1/2$, since $EV(x_1)=μ$. Would this logic work for the rest?
probability operations-research
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Gabe Hoffman
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I am currently taking an Operations Managment class where we are discussing control limits for various processes/tasks. For example, we discuss a machine that produces memory cards of a specific width $x_n$. However, because the machine is not perfect, the produced memory cards width's follows a normal distribution.
What is the probability that the next 5 (or however many) chips it produces would have widths of increasing order $(x_1<x_2<x_3...x_N)$
If the question is reduced to asking for the probability that the next two are in increasing order, $P(x_1>x_2)=1/2$, since $EV(x_1)=μ$. Would this logic work for the rest?
probability operations-research
asked yesterday
Gabe Hoffman
12
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am currently taking an Operations Managment class where we are discussing control limits for various processes/tasks. For example, we discuss a machine that produces memory cards of a specific width $x_n$. However, because the machine is not perfect, the produced memory cards width's follows a normal distribution.
What is the probability that the next 5 (or however many) chips it produces would have widths of increasing order $(x_1<x_2<x_3...x_N)$
If the question is reduced to asking for the probability that the next two are in increasing order, $P(x_1>x_2)=1/2$, since $EV(x_1)=μ$. Would this logic work for the rest?
probability operations-research
asked yesterday
Gabe Hoffman
12
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am currently taking an Operations Managment class where we are discussing control limits for various processes/tasks. For example, we discuss a machine that produces memory cards of a specific width $x_n$. However, because the machine is not perfect, the produced memory cards width's follows a normal distribution.
What is the probability that the next 5 (or however many) chips it produces would have widths of increasing order $(x_1<x_2<x_3...x_N)$
If the question is reduced to asking for the probability that the next two are in increasing order, $P(x_1>x_2)=1/2$, since $EV(x_1)=μ$. Would this logic work for the rest?
probability operations-research
probability operations-research
asked yesterday
Gabe Hoffman
12
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Gabe Hoffman
12
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Gabe Hoffman
12
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Gabe Hoffman
12
asked yesterday
Gabe Hoffman
12
12
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gabe Hoffman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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If the widths of the chips are $x_1,x_2,x_3,x_4,x_5$, then all $5!$ possible orderings of these variables are equally likely, due to symmetry and the fact that the probability that any two widths have exactly the same value is zero. (This is true of any continuous distribution, not necessarily the normal.) So the probability that $x_1 <x_2<x_3<x_4<x_5$ is
$$frac{1}{5!}$$
answered yesterday
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5,54611021
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If the widths of the chips are $x_1,x_2,x_3,x_4,x_5$, then all $5!$ possible orderings of these variables are equally likely, due to symmetry and the fact that the probability that any two widths have exactly the same value is zero. (This is true of any continuous distribution, not necessarily the normal.) So the probability that $x_1 <x_2<x_3<x_4<x_5$ is
$$frac{1}{5!}$$
answered yesterday
awkward
5,54611021
add a comment |
up vote
0
down vote
If the widths of the chips are $x_1,x_2,x_3,x_4,x_5$, then all $5!$ possible orderings of these variables are equally likely, due to symmetry and the fact that the probability that any two widths have exactly the same value is zero. (This is true of any continuous distribution, not necessarily the normal.) So the probability that $x_1 <x_2<x_3<x_4<x_5$ is
$$frac{1}{5!}$$
answered yesterday
awkward
5,54611021
add a comment |
up vote
0
down vote
up vote
0
down vote
If the widths of the chips are $x_1,x_2,x_3,x_4,x_5$, then all $5!$ possible orderings of these variables are equally likely, due to symmetry and the fact that the probability that any two widths have exactly the same value is zero. (This is true of any continuous distribution, not necessarily the normal.) So the probability that $x_1 <x_2<x_3<x_4<x_5$ is
$$frac{1}{5!}$$
answered yesterday
awkward
5,54611021
If the widths of the chips are $x_1,x_2,x_3,x_4,x_5$, then all $5!$ possible orderings of these variables are equally likely, due to symmetry and the fact that the probability that any two widths have exactly the same value is zero. (This is true of any continuous distribution, not necessarily the normal.) So the probability that $x_1 <x_2<x_3<x_4<x_5$ is
$$frac{1}{5!}$$
answered yesterday
awkward
5,54611021
answered yesterday
awkward
5,54611021
answered yesterday
awkward
5,54611021
answered yesterday
awkward
5,54611021
5,54611021
add a comment |
add a comment |
Gabe Hoffman is a new contributor. Be nice, and check out our Code of Conduct.
Gabe Hoffman is a new contributor. Be nice, and check out our Code of Conduct.
Gabe Hoffman is a new contributor. Be nice, and check out our Code of Conduct.
Gabe Hoffman is a new contributor. Be nice, and check out our Code of Conduct.
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