Twenty-percent of the residents of Boston were born outside of the United States. A random sample of 400 residents is to be selected. Without using the Z table, compute the approximate probability that at least 22% of those in the sample were born outside of the United States.
+118677 Solveit asked me to look at this question.
I saw it when it was first posted and I wasn't really sure. I am still not sure.
I'd guess the answer is 50% That seems logical to me .......
Would anyone like to comment?
+33661 My reasoning is as follows:
However, since it's a question of probabilities, I don't guarantee that my reasoning is correct!!
+118677 Well Alan, at least you have some reasoning. I think that beats my guess LOL.
+2498 But what if i wanna to find probability that there is only 88 residents outside of US:
nCr(400, 88)*0.8^312*0.2^88 = 0.0294786177861719029113601772508066971031878720367673319179859403
nCr(400, 312)*0.8^312*0.2^88 = 0.0294786177861719029113601772508066971031878720367673319179859403
0.8 and 0.2 how it is posible i can t understand i know tha one of them is Pin and the other Pout
sorry that i am asking proisaic questions i just wanna to understand
