Suppose approximately 75% of all marketing personnel are extroverts, whereas about 55% of all computer programmers are introverts. (For each answer, enter a number. Round your answers to the three decimal places.) A. At the meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?
What is the probability that 5 or more are extroverts?
What is the probability that all are extroverts?
B. In a group of 5 computer programmers, what is the probability that none are introverts?
What is the probability that 3 or more are introverts?
What is the probability that all are introverts?

A.)

0.852 ; 0.999 ; 0.013

B.)

0.018 ; 0.593 ; 0.050

Step-by-step explanation:

Given :

Proportion of marketing personnel that are extrovert, p = 0.75

1 - p = 1 - 0.75 = 0.25

Proportion of computer programmers that are introverts, p = 0.55

1 - p = 1 - 0.55 = 0.45

According to binomial distribution formula :

P(x =x) = nCx * p^x * (1 - p)^(n - r)

A.) n = 15

Probability that 10 or more are extrovert

P(x ≥ 10) = p(x = 10) + p(x = 11) +... p(x = 15)

To save computation time, we can use a binomial probability calculator ;

P(x ≥ 10) = 0.8516 = 0.852

Probability that 5 or more are extrovert :

P(x ≥ 5) = p(x = 5) + p(x = 6) +... p(x = 15)

To save computation time, we can use a binomial probability calculator ;

P(x ≥ 5) = 0.9999 = 0.999

Probability that all are extrovert :

P(x = 15) = 15C15 * 0.75^15 * 0.25^0

P(x = 15) = 1 * 0.75^15 * 0.25^0

P(x = 15) = 0.01336 = 0.013

B.

What is the probability that all are introverts?

n = 5

None are introvert :

P(x = 0) = 5C0 * 0.55^0 * 0.45^5

P(x = 0) = 1 * 1 * 0.0184528125

P(x = 0) = 0.01845 = 0.018

P(x ≥ 3) = p(3) + p(4) + p(5)

To save computation time, we can use a binomial probability calculator ;

P(x ≥ 3) = 0.5931 = 0.593

Probability that all are introvert

P(x = 5) = 5C5 * 0.55^5 * 0.45^0

P(x = 5) = 1 * 0.55^5 * 0.45^0

P(x = 5) = 0.0503 = 0.050

the answer is c. 23/3

c

c

c

c

c

c

c

cc

c

step-by-step explanation: