Question

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 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 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?

Answer 1

Solution:

a) p = 0.75( probability that mktg. Pers. are extt.)

q= 1-p = 1- 0.75 = 0.25 (prob. Not are extrovert)

n=15

Using bionomial formula

b) probability that 5 or more are extrovert =

1 - P(x<5)

P(x=0)=approx. 0

P(x=1)=app. 0

P(x=2)= 0

P(x=3)=0

P(x=4)=0.0001

So

probability that 5 or more are extrovert =

1 - P(x<5) = 1-0.0001 = 0.999

c) p(X= 15) = 0.0134....(calculated in part a)

B). p= 0.55, q= 0.45 ,n= 5

a) p(none introvert) =

b)

c) p(x=5) = 0.05