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anonymous
 3 years ago
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anonymous
 3 years ago
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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.03 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think you can use the formula (n x) . p^x . q ^(nx)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what about the second one?

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.0For the third one you use the fact that the binomial distribution can be approximated by the normal distribution when n is large and p is neither small nor near 1.\[\mu=np=175\times 0.21\] \[\sigma=\sqrt{np(1p)}=\sqrt{175\times 0.21\times 0.79}\] Then you find the required probability by means of a standard normal distribution table.

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.0Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.0You need to find one value of zscore for the third one and two values of zscore for the fourth.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because there two values 17% and 24%
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