Approximately 65% of all marketing personnel are extroverts, whereas about 70% o

Question

Approximately 65% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introverts. (Round your answers to three decimal places.)

(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

b) What is the probability that 5 or more are extroverts?

c) What is the probability that all are extroverts?

(d) In a group of 5 computer programmers, what is the probability that none are introverts?

e) What is the probability that 3 or more are introverts?

f) What is the probability that all are introverts?

Explanation / Answer

Let X be the random variable that the number of marketing personnel are extroverts.

X ~ Binomial (n=15, p = 65% = 0.65)

And let Y be the random variable that  the number of computer programmers are introverts.

Y ~ Binomial (n=5, p = 70% = 0.70)

a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

That is here we have to find P(X >=10).

P(X>=10) = P(X=10) + P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15)

These probabilities we can find by using EXCEL.

syntax is,

=BINOMDIST(number_s, trials, probability_s, cumulative)

where number_s are x-values.

trials are number marketing personnel = 15.

probability_s is probability of success = 0.65.

P(X>=10) = 0.2123 + 0.1792 + 0.1110 + 0.0476 + 0.0126 + 0.0016 = 0.5643

b) What is the probability that 5 or more are extroverts?

That is here we have to find P(X>=5) = 1- P(X < 5)

= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) ]

= 1 - [ 0.0000 + 0.0000 + 0.0001 + 0.0004 + 0.0024 + 0.0028 ]

= 1 - 0.0028

P(X>=5) = 0.9972

c) What is the probability that all are extroverts?

That is here we have to find P(X=15).

P(X=15) = 0.0016

d) In a group of 5 computer programmers, what is the probability that none are introverts?

n = 5

That is here we have to find P(Y = 0) = 0.0024

e) What is the probability that 3 or more are introverts?

Here we have to find P(X >=3).

P(X>=3) = P(X=3) + P(X=4) + P(X=5)

= 0.3087 + 0.3602 + 0.1681

P(X>=3) = 0.8369

f) What is the probability that all are introverts?

Here we have to find P(X = 5).

P(X=5) = 0.1681

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