Stuck at a binomial and Poisson distribution assignment
$begingroup$
In one service center every day a number of devices for repair arrive. (The probability that two device arrive at the same time is $0$.) Let $X$ be the random variable that gives the number of devices that arrived for repair in one day.
What is the assumption for the distribution of $X$?
The manager of the service center has gathered data for the number of devices that have arrived in one day:
$$ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5$$
Does the data that manager has gathered match with the assumed distribution?
Let the average number of the arrived devices be 8. If the total number of arrived devices in 40 days exceeds 350 then the manager will make a decision for the service to work for only one shift. Find the probability for that to happen.
Now for the first one X obviously has a Binomial distribution of $ X sim B(n, p)$
For the second one we can calculate n and p by finding the average value and standard deviation for the data since $E(x)=np$ and $D(x)=np(1-p)$;
However for the third one I have no idea. I assume this is a Poisson distribution with $λ=8 times 40=320$ and I need to find $Plbrace X<=350 rbrace $.
However I have no idea how to calculate that!
Any ideas?
probability statistics
edited Jan 10 at 19:42
Bernard
119k740113
asked Jan 10 at 18:55
David MathersDavid Mathers
143
$endgroup$
add a comment |
$begingroup$
In one service center every day a number of devices for repair arrive. (The probability that two device arrive at the same time is $0$.) Let $X$ be the random variable that gives the number of devices that arrived for repair in one day.
What is the assumption for the distribution of $X$?
The manager of the service center has gathered data for the number of devices that have arrived in one day:
$$ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5$$
Does the data that manager has gathered match with the assumed distribution?
Let the average number of the arrived devices be 8. If the total number of arrived devices in 40 days exceeds 350 then the manager will make a decision for the service to work for only one shift. Find the probability for that to happen.
Now for the first one X obviously has a Binomial distribution of $ X sim B(n, p)$
For the second one we can calculate n and p by finding the average value and standard deviation for the data since $E(x)=np$ and $D(x)=np(1-p)$;
However for the third one I have no idea. I assume this is a Poisson distribution with $λ=8 times 40=320$ and I need to find $Plbrace X<=350 rbrace $.
However I have no idea how to calculate that!
Any ideas?
probability statistics
edited Jan 10 at 19:42
Bernard
119k740113
asked Jan 10 at 18:55
David MathersDavid Mathers
143
$endgroup$
$begingroup$
hi, please use LaTeX or MathJax to edit your questions as I did.
$endgroup$
– Ahmad Bazzi
Jan 10 at 18:59
$begingroup$
Okey I'll keep that in mind next time
$endgroup$
– David Mathers
Jan 10 at 19:01
$begingroup$
Th meaning of the line with 1's, 2's, ..., 5's is unclear. If these are 34 daily counts sorted in order, then they do not fit either a binomial or a Poisson distribution. For one thing values in this range would need to show some 0's to match either distribution. // Later on, perhaps you meant to say "does not exceed 350." In any case, if $X sim mathsf{Pois}(lambda=320),$ then $P(X le 350)=0.9543,$ computed in R statistical software with codeppois(350, 320), which returns 0.9543056; you can probably come "close" to this answer using a normal aprx with $mu = 320, sigma=sqrt{320}.$
$endgroup$
– BruceET
Jan 11 at 7:32
$begingroup$
Thanks for the answer. Anyway you're right for the second one a binomial distribution doesn't fit and a Poisson distribution is more logical(We can estimate λ using the data given). However can the binomial distribution also be applied?
$endgroup$
– David Mathers
Jan 12 at 15:58
add a comment |
$begingroup$
In one service center every day a number of devices for repair arrive. (The probability that two device arrive at the same time is $0$.) Let $X$ be the random variable that gives the number of devices that arrived for repair in one day.
What is the assumption for the distribution of $X$?
The manager of the service center has gathered data for the number of devices that have arrived in one day:
$$ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5$$
Does the data that manager has gathered match with the assumed distribution?
Let the average number of the arrived devices be 8. If the total number of arrived devices in 40 days exceeds 350 then the manager will make a decision for the service to work for only one shift. Find the probability for that to happen.
Now for the first one X obviously has a Binomial distribution of $ X sim B(n, p)$
For the second one we can calculate n and p by finding the average value and standard deviation for the data since $E(x)=np$ and $D(x)=np(1-p)$;
However for the third one I have no idea. I assume this is a Poisson distribution with $λ=8 times 40=320$ and I need to find $Plbrace X<=350 rbrace $.
However I have no idea how to calculate that!
Any ideas?
probability statistics
edited Jan 10 at 19:42
Bernard
119k740113
asked Jan 10 at 18:55
David MathersDavid Mathers
143
$endgroup$
In one service center every day a number of devices for repair arrive. (The probability that two device arrive at the same time is $0$.) Let $X$ be the random variable that gives the number of devices that arrived for repair in one day.
What is the assumption for the distribution of $X$?
The manager of the service center has gathered data for the number of devices that have arrived in one day:
$$ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5$$
Does the data that manager has gathered match with the assumed distribution?
Let the average number of the arrived devices be 8. If the total number of arrived devices in 40 days exceeds 350 then the manager will make a decision for the service to work for only one shift. Find the probability for that to happen.
Now for the first one X obviously has a Binomial distribution of $ X sim B(n, p)$
For the second one we can calculate n and p by finding the average value and standard deviation for the data since $E(x)=np$ and $D(x)=np(1-p)$;
However for the third one I have no idea. I assume this is a Poisson distribution with $λ=8 times 40=320$ and I need to find $Plbrace X<=350 rbrace $.
However I have no idea how to calculate that!
Any ideas?
probability statistics
probability statistics
edited Jan 10 at 19:42
Bernard
119k740113
asked Jan 10 at 18:55
David MathersDavid Mathers
143
edited Jan 10 at 19:42
Bernard
119k740113
asked Jan 10 at 18:55
David MathersDavid Mathers
143
edited Jan 10 at 19:42
Bernard
119k740113
edited Jan 10 at 19:42
Bernard
119k740113
edited Jan 10 at 19:42
Bernard
119k740113
119k740113
asked Jan 10 at 18:55
David MathersDavid Mathers
143
asked Jan 10 at 18:55
David MathersDavid Mathers
143
asked Jan 10 at 18:55
David MathersDavid Mathers
143
143
$begingroup$
hi, please use LaTeX or MathJax to edit your questions as I did.
$endgroup$
– Ahmad Bazzi
Jan 10 at 18:59
$begingroup$
Okey I'll keep that in mind next time
$endgroup$
– David Mathers
Jan 10 at 19:01
$begingroup$
Th meaning of the line with 1's, 2's, ..., 5's is unclear. If these are 34 daily counts sorted in order, then they do not fit either a binomial or a Poisson distribution. For one thing values in this range would need to show some 0's to match either distribution. // Later on, perhaps you meant to say "does not exceed 350." In any case, if $X sim mathsf{Pois}(lambda=320),$ then $P(X le 350)=0.9543,$ computed in R statistical software with codeppois(350, 320), which returns 0.9543056; you can probably come "close" to this answer using a normal aprx with $mu = 320, sigma=sqrt{320}.$
$endgroup$
– BruceET
Jan 11 at 7:32
$begingroup$
Thanks for the answer. Anyway you're right for the second one a binomial distribution doesn't fit and a Poisson distribution is more logical(We can estimate λ using the data given). However can the binomial distribution also be applied?
$endgroup$
– David Mathers
Jan 12 at 15:58
add a comment |
$begingroup$
hi, please use LaTeX or MathJax to edit your questions as I did.
$endgroup$
– Ahmad Bazzi
Jan 10 at 18:59
$begingroup$
Okey I'll keep that in mind next time
$endgroup$
– David Mathers
Jan 10 at 19:01
$begingroup$
Th meaning of the line with 1's, 2's, ..., 5's is unclear. If these are 34 daily counts sorted in order, then they do not fit either a binomial or a Poisson distribution. For one thing values in this range would need to show some 0's to match either distribution. // Later on, perhaps you meant to say "does not exceed 350." In any case, if $X sim mathsf{Pois}(lambda=320),$ then $P(X le 350)=0.9543,$ computed in R statistical software with codeppois(350, 320), which returns 0.9543056; you can probably come "close" to this answer using a normal aprx with $mu = 320, sigma=sqrt{320}.$
$endgroup$
– BruceET
Jan 11 at 7:32
$begingroup$
Thanks for the answer. Anyway you're right for the second one a binomial distribution doesn't fit and a Poisson distribution is more logical(We can estimate λ using the data given). However can the binomial distribution also be applied?
$endgroup$
– David Mathers
Jan 12 at 15:58
$begingroup$
hi, please use LaTeX or MathJax to edit your questions as I did.
$endgroup$
– Ahmad Bazzi
Jan 10 at 18:59
$begingroup$
hi, please use LaTeX or MathJax to edit your questions as I did.
$endgroup$
– Ahmad Bazzi
Jan 10 at 18:59
$begingroup$
Okey I'll keep that in mind next time
$endgroup$
– David Mathers
Jan 10 at 19:01
$begingroup$
Okey I'll keep that in mind next time
$endgroup$
– David Mathers
Jan 10 at 19:01
$begingroup$
Th meaning of the line with 1's, 2's, ..., 5's is unclear. If these are 34 daily counts sorted in order, then they do not fit either a binomial or a Poisson distribution. For one thing values in this range would need to show some 0's to match either distribution. // Later on, perhaps you meant to say "does not exceed 350." In any case, if $X sim mathsf{Pois}(lambda=320),$ then $P(X le 350)=0.9543,$ computed in R statistical software with code
ppois(350, 320), which returns 0.9543056; you can probably come "close" to this answer using a normal aprx with $mu = 320, sigma=sqrt{320}.$$endgroup$
– BruceET
Jan 11 at 7:32
$begingroup$
Th meaning of the line with 1's, 2's, ..., 5's is unclear. If these are 34 daily counts sorted in order, then they do not fit either a binomial or a Poisson distribution. For one thing values in this range would need to show some 0's to match either distribution. // Later on, perhaps you meant to say "does not exceed 350." In any case, if $X sim mathsf{Pois}(lambda=320),$ then $P(X le 350)=0.9543,$ computed in R statistical software with code
ppois(350, 320), which returns 0.9543056; you can probably come "close" to this answer using a normal aprx with $mu = 320, sigma=sqrt{320}.$$endgroup$
– BruceET
Jan 11 at 7:32
$begingroup$
Thanks for the answer. Anyway you're right for the second one a binomial distribution doesn't fit and a Poisson distribution is more logical(We can estimate λ using the data given). However can the binomial distribution also be applied?
$endgroup$
– David Mathers
Jan 12 at 15:58
$begingroup$
Thanks for the answer. Anyway you're right for the second one a binomial distribution doesn't fit and a Poisson distribution is more logical(We can estimate λ using the data given). However can the binomial distribution also be applied?
$endgroup$
– David Mathers
Jan 12 at 15:58
add a comment |
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$begingroup$
hi, please use LaTeX or MathJax to edit your questions as I did.
$endgroup$
– Ahmad Bazzi
Jan 10 at 18:59
$begingroup$
Okey I'll keep that in mind next time
$endgroup$
– David Mathers
Jan 10 at 19:01
$begingroup$
Th meaning of the line with 1's, 2's, ..., 5's is unclear. If these are 34 daily counts sorted in order, then they do not fit either a binomial or a Poisson distribution. For one thing values in this range would need to show some 0's to match either distribution. // Later on, perhaps you meant to say "does not exceed 350." In any case, if $X sim mathsf{Pois}(lambda=320),$ then $P(X le 350)=0.9543,$ computed in R statistical software with code
ppois(350, 320), which returns 0.9543056; you can probably come "close" to this answer using a normal aprx with $mu = 320, sigma=sqrt{320}.$$endgroup$
– BruceET
Jan 11 at 7:32
$begingroup$
Thanks for the answer. Anyway you're right for the second one a binomial distribution doesn't fit and a Poisson distribution is more logical(We can estimate λ using the data given). However can the binomial distribution also be applied?
$endgroup$
– David Mathers
Jan 12 at 15:58