You can
put this solution on YOUR website!In a recent study 90 percent of the homes in the United States were found to have color TVs. In a sample of nine homes, what is the probability that:
p(have color TV)=0.9
--------------------------
a. All nine have color TVs?
0.9^9=0.3874
-----------------------
b. Less than five have color TVs?
binomcdf(9,0.9,4)=0.0008909
----------------------
c. More than five have color TVs?
1-binomcdf(9,0.9,5)=0.99166
--------------------
d. At least seven homes have color TVs?
1-binomcdf(9.0,9,6)=0.94703
------------
Cheers,
Stan H.