How can I get N trials from binomial distribution (Edited)
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$$
sum_k=0^17_NC_ktimes 0.1^ktimes 0.9^N-k<0.004
$$
How can I get a $N$ from above inequality?
binomial-distribution
asked Sep 7 at 11:32
baeharam
396
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up vote
-1
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$$
sum_k=0^17_NC_ktimes 0.1^ktimes 0.9^N-k<0.004
$$
How can I get a $N$ from above inequality?
binomial-distribution
asked Sep 7 at 11:32
baeharam
396
If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39
$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42
This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16
I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$$
sum_k=0^17_NC_ktimes 0.1^ktimes 0.9^N-k<0.004
$$
How can I get a $N$ from above inequality?
binomial-distribution
asked Sep 7 at 11:32
baeharam
396
$$
sum_k=0^17_NC_ktimes 0.1^ktimes 0.9^N-k<0.004
$$
How can I get a $N$ from above inequality?
binomial-distribution
binomial-distribution
asked Sep 7 at 11:32
baeharam
396
asked Sep 7 at 11:32
baeharam
396
asked Sep 7 at 11:32
baeharam
396
asked Sep 7 at 11:32
baeharam
396
asked Sep 7 at 11:32
baeharam
396
396
If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39
$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42
This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16
I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22
add a comment |Â
If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39
$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42
This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16
I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22
If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39
If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39
$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42
$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42
This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16
This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16
I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22
I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
As commented by @joriki, in general for this summation a numerical approach is the way to go.
Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.
The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.
Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2500$ with $p=0.1 = frac110$) is beyond me, and I'd like to here from anyone who has an idea.
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
As commented by @joriki, in general for this summation a numerical approach is the way to go.
Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.
The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.
Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2500$ with $p=0.1 = frac110$) is beyond me, and I'd like to here from anyone who has an idea.
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
add a comment |Â
up vote
0
down vote
accepted
As commented by @joriki, in general for this summation a numerical approach is the way to go.
Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.
The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.
Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2500$ with $p=0.1 = frac110$) is beyond me, and I'd like to here from anyone who has an idea.
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
As commented by @joriki, in general for this summation a numerical approach is the way to go.
Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.
The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.
Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2500$ with $p=0.1 = frac110$) is beyond me, and I'd like to here from anyone who has an idea.
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
As commented by @joriki, in general for this summation a numerical approach is the way to go.
Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.
The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.
Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2500$ with $p=0.1 = frac110$) is beyond me, and I'd like to here from anyone who has an idea.
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
edited Sep 7 at 13:22
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
answered Sep 7 at 13:13
Lee David Chung Lin
1,9363926
1,9363926
add a comment |Â
add a comment |Â
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If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39
$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42
This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16
I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22