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satellite73
 one year ago
Best ResponseYou've already chosen the best response.03 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]

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

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0what about the second one?

kropot72
 one year 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
 one year 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
 one year 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.

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0because there two values 17% and 24%
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