SOLUTION: In 1995, 70% of all children in the U.S. were living with both parents. If 25 children were selected at random in the U.S., what is the probability that at least 15 of them will be

Algebra ->  Probability-and-statistics -> SOLUTION: In 1995, 70% of all children in the U.S. were living with both parents. If 25 children were selected at random in the U.S., what is the probability that at least 15 of them will be      Log On


   

Question 669134: In 1995, 70% of all children in the U.S. were living with both parents. If 25 children were selected at random in the U.S., what is the probability that at least 15 of them will be living with both of their parents? Round your answer to 4 decimal places.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In 1995, 70% of all children in the U.S. were living with both parents. If 25 children were selected at random in the U.S., what is the probability that at least 15 of them will be living with both of their parents? Round your answer to 4 decimal places.
------
Binomial Problem with n = 25 ; p(living with) = 0.7 ; p(not liv with) = 0.3
----------------
P(15 <= x <= 25) = 1 - binomcdf(25,0.7,14) = 0.9022
--------------------
Cheers,
Stan H.
=================