Question

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 60% 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? What is the probability that 5 or more are extroverts? What is the probability that all are extroverts? (b) In a group of 4 computer programmers, what is the probability that none are introverts? What is the probability that 2 or more are introverts? What is the probability that all are introverts?

Answer 1

(a)

(i)

Binomial Distribution

n = 15

p = 0.75

q = 0.25

P(X10) = P(x = 10) + P(X=11) + P(X=12)+ P(X=13) + P(X=14) + P(X=15)

So,

P(X10) = 0.8516

So,

Answer is:

0.8516

(ii)

P(X5) = P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(x = 10) + P(X=11) + P(X=12)+ P(X=13) + P(X=14) + P(X=15)

So,

P(X5) = 0.9999

So,

Answer is:

0.9999

(iii)

So,

Answer is:

0.0134

(b)

(i)

n = 4

p = 0.6

q = 1 -p = 0.4

So,

Answer is:

0.0256

(ii)

P(X2)= 1 - [P(X=0) + P(X=1)]

So,

P(X2) = 1 - .0256 = 0.8208

So,

Answer is:

0.8208

(iii)

So,

Answer is

0.1296