The required probabilities are as follows:
a) P(X=5) = 0.123
b) P(X6) = P(X=6+X=7+X=8+X=9+X=10)
c) P(X4)=P(X=0), P(X=1), P(X=2), and P(X=3)
Using a binomial random variable, the following is modeled:
There are always ten (10) persons that are being questioned.
From trial to trial, the likelihood of success (i.e., very little faith in newspapers) remains consistent (p=0.68).
The trials are impartial (a reasonable assumption, since the response of 1 of the people should not affect the response of any other)
There are just two possible results for each trial: low faith in newspapers or not.
X counts how many of the 10 trials were successful.
P(X=k) = is the probability mass function for the binomial random variable.
= [tex]^nC_k\times(p^k)\times(1-p)^{n-k}[/tex]
a) P(X=5) =[tex]^{10}C_5\times(0.68)^5\times(0.32)^5[/tex]≈ 0.123
b) P(X6) = P(X=6+X=7+X=8+X=9+X=10)
These may all be calculated individually, then summed together.
c) P(X4)=P(X=0), P(X=1), P(X=2), and P(X=3)
the same concept as in (b) above.
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