Question 853158: I need help solving this problem. Can you please help me?
A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20 adult women are randomly selected, find the probability that fewer than 15 of them have had a physical examination in the past two years.
Round to three decimal places.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20 adult women are randomly selected, find the probability that fewer than 15 of them have had a physical examination in the past two years.
-------------------------------
Binomial Problem with n = 20 ; p(visit) = 0.7 ; p(not visit) = 0.3
-----------------------
P(0<= x <=14) = binomcdf(20,0.7,14) = 0.5836
===============================================
Cheers,
Stan H.
=================
|
|
|
