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
asked Nov 14 at 5:53
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 Nov 14 at 5:53
Gabe Hoffman
12
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 Nov 14 at 5:53
Gabe Hoffman
12
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 Nov 14 at 5:53
Gabe Hoffman
12
asked Nov 14 at 5:53
Gabe Hoffman
12
asked Nov 14 at 5:53
Gabe Hoffman
12
asked Nov 14 at 5:53
Gabe Hoffman
12
asked Nov 14 at 5:53
Gabe Hoffman
12
12
add a comment |
add a comment |
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 Nov 14 at 15:10
awkward
5,59111021
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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 Nov 14 at 15:10
awkward
5,59111021
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 Nov 14 at 15:10
awkward
5,59111021
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 Nov 14 at 15:10
awkward
5,59111021
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 Nov 14 at 15:10
awkward
5,59111021
answered Nov 14 at 15:10
awkward
5,59111021
answered Nov 14 at 15:10
awkward
5,59111021
answered Nov 14 at 15:10
awkward
5,59111021
5,59111021
add a comment |
add a comment |
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