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Use the normal model to approximate the binomial to determine the probability of at least 191 passengers...
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Because many passengers who make reservations do not show up, airlines often overbook flights (sell more than there are seats). A certain airplane holds 190 passengers. If the airline believes the rate of passenger no-shows is 9% and sells 208 tickets, is it likely they will not have enough seats and someone will get bumped?
probability statistics normal-distribution
asked Nov 8 '15 at 21:27
ZezimaZezima
61
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$begingroup$
Because many passengers who make reservations do not show up, airlines often overbook flights (sell more than there are seats). A certain airplane holds 190 passengers. If the airline believes the rate of passenger no-shows is 9% and sells 208 tickets, is it likely they will not have enough seats and someone will get bumped?
probability statistics normal-distribution
asked Nov 8 '15 at 21:27
ZezimaZezima
61
$endgroup$
add a comment |
$begingroup$
Because many passengers who make reservations do not show up, airlines often overbook flights (sell more than there are seats). A certain airplane holds 190 passengers. If the airline believes the rate of passenger no-shows is 9% and sells 208 tickets, is it likely they will not have enough seats and someone will get bumped?
probability statistics normal-distribution
asked Nov 8 '15 at 21:27
ZezimaZezima
61
$endgroup$
Because many passengers who make reservations do not show up, airlines often overbook flights (sell more than there are seats). A certain airplane holds 190 passengers. If the airline believes the rate of passenger no-shows is 9% and sells 208 tickets, is it likely they will not have enough seats and someone will get bumped?
probability statistics normal-distribution
probability statistics normal-distribution
asked Nov 8 '15 at 21:27
ZezimaZezima
61
asked Nov 8 '15 at 21:27
ZezimaZezima
61
asked Nov 8 '15 at 21:27
ZezimaZezima
61
asked Nov 8 '15 at 21:27
ZezimaZezima
61
asked Nov 8 '15 at 21:27
ZezimaZezima
61
61
add a comment |
add a comment |
1 Answer
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$begingroup$
$X$ - numbers of "shows". So, $X in mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) geq 10$ we can make the following approximation, where the approximation was made in the last step
$$ P_{mathrm{Bin}}(X geq 191) = 1- P_{mathrm{Bin}}(X leq 190) = 1- P_{mathrm{Bin}}(X < 191) approx 1- mathrm{N}(frac{190.5-np}{sqrt{np(1-p)}} ). $$
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
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1 Answer
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$begingroup$
$X$ - numbers of "shows". So, $X in mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) geq 10$ we can make the following approximation, where the approximation was made in the last step
$$ P_{mathrm{Bin}}(X geq 191) = 1- P_{mathrm{Bin}}(X leq 190) = 1- P_{mathrm{Bin}}(X < 191) approx 1- mathrm{N}(frac{190.5-np}{sqrt{np(1-p)}} ). $$
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
$endgroup$
add a comment |
$begingroup$
$X$ - numbers of "shows". So, $X in mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) geq 10$ we can make the following approximation, where the approximation was made in the last step
$$ P_{mathrm{Bin}}(X geq 191) = 1- P_{mathrm{Bin}}(X leq 190) = 1- P_{mathrm{Bin}}(X < 191) approx 1- mathrm{N}(frac{190.5-np}{sqrt{np(1-p)}} ). $$
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
$endgroup$
add a comment |
$begingroup$
$X$ - numbers of "shows". So, $X in mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) geq 10$ we can make the following approximation, where the approximation was made in the last step
$$ P_{mathrm{Bin}}(X geq 191) = 1- P_{mathrm{Bin}}(X leq 190) = 1- P_{mathrm{Bin}}(X < 191) approx 1- mathrm{N}(frac{190.5-np}{sqrt{np(1-p)}} ). $$
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
$endgroup$
$X$ - numbers of "shows". So, $X in mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) geq 10$ we can make the following approximation, where the approximation was made in the last step
$$ P_{mathrm{Bin}}(X geq 191) = 1- P_{mathrm{Bin}}(X leq 190) = 1- P_{mathrm{Bin}}(X < 191) approx 1- mathrm{N}(frac{190.5-np}{sqrt{np(1-p)}} ). $$
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
edited Nov 8 '15 at 22:05
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
answered Nov 8 '15 at 21:40
fritzenbauerfritzenbauer
12111
12111
add a comment |
add a comment |
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