Question 351943: Approximately 79% of all marketing personnel are extroverts, whereas about 64% of all computer programmers are introverts.
(a) At a meeting of 14 marketing personnel, what is the probability that 9 or more are extroverts? (Use 3 decimal places.)
What is the probability that 4 or more are extroverts? (Use 3 decimal places.)
What is the probability that all extroverts? (Use 3 decimal places.)
(b) In a group of 6 computer programmers, what is the probability that none are introverts? (Use 3 decimal places.)
What is the probability that 4 or more are introverts? (Use 3 decimal places.)
What is the probability that all are introverts? (Use 3 decimal places.)
Found 2 solutions by jrfrunner, edjones:
Answer by jrfrunner(365)   (Show Source):
You can put this solution on YOUR website! Approximately 79% of all marketing personnel are extroverts, whereas about 64% of all computer programmers are introverts.
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(a) At a meeting of 14 marketing personnel, what is the probability that 9 or more are extroverts? (Use 3 decimal places.)
This is a binomial distribution with n=14, x>=9, p=0.79
You want P(X>=9)=1-P(X<=8)=0.9457
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What is the probability that 4 or more are extroverts? (Use 3 decimal places.)
Again a binomial distribution with n=14, x>=4 and p=0.79
you want P(x>=4)=1-P(x<=3)=0.999
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What is the probability that all extroverts? (Use 3 decimal places.)
a binomial with n=14, x=14, p=0.79
P*(all are extroverts)=P(x=14)=0.79^14=0.0369
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(b) In a group of 6 computer programmers, what is the probability that none are introverts? (Use 3 decimal places.)
binomial n=6, x=0, p=.64
P(X=0)=(1-.64)^6=0.00218
What is the probability that 4 or more are introverts? (Use 3 decimal places.)
binomial n=6,x>=4, p=.64
P(x>=4)=1-P(x<=3)=0.6268
What is the probability that all are introverts? (Use 3 decimal places.)
binomial n=6, x=6, p=0.64
P(x=6)=0.64^6=0.0687
Answer by edjones(8007)  (Show Source):
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