Question 1168391: Suppose 40% of TV sets use in Canada on particolare night into 7 game of stably cup playoffs
A) if we were take a sample of six in use tv , what is probability exactly three are tuned to stably cup playoffs ?
B) if instead the sample consisted of 10 in use tv
What is probability 4 or more are turned to stably cup playoffs
What is the probability that almost 7 of them are tuned to stably cup playoffs
C) if instead sample consider of 100 I uses tv what is probability that atleast 55 of them are tuned to stably cup playoffs ?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! this is 6C3*0.4^3*0.6^3=0.2765
this is 4 or more would be 1-(3 or fewer): binomcdf(10,.4,3)=0.3822
0.2150 for 3
0.1209 for 2
0.0403 for 1
0.006 for 0
that is 0.3822
at most (not almost) 7 would be everything fewer than or equal to 6. That is 0.9452.
Normal approximation
np=100*0.4=40 mean
np(1-p)=40*0.6=24=variance
SD = sqrt (V)=4.90
at least 55: z>(55-40)/4.9 or >3.06. that has probability of 0.0011
exact is 0.0017 (1-binomcdf(100,0.4,54))ENTER
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