Question 457880: Can you please help me solve this problem. Thank you!
Approximately 73% of all marketing personnel are extroverts, whereas about 69% of all computer programmers are introverts.
(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts? (Use 3 decimal places.)
What is the probability that 5 or more are extroverts? (Use 3 decimal places.)
What is the probability that all extroverts? (Use 3 decimal places.)
(b) In a group of 5 computer programmers, what is the probability that none are introverts? (Use 3 decimal places.)
What is the probability that 3 or more are introverts? (Use 3 decimal places.)
What is the probability that all are introverts? (Use 3 decimal places.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Approximately 73% of all marketing personnel are extroverts,
about 69% of all computer programmers are introverts.
------------------------
(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts? (Use 3 decimal places.)
Ans: 1- binomcdf(15,0.73,9) = 0.804
========================================
What is the probability that 5 or more are extroverts? (Use 3 decimal places.)
Ans: 1- binomcdf(15,0.73,4) = 0.999
-----------------------------------------
What is the probability that all extroverts? (Use 3 decimal places.)
Ans: (0.73)^15 = 0.009
---------------------------
(b) In a group of 5 computer programmers, what is the probability that none are introverts? (Use 3 decimal places.)
Ans: 0.31^3 = 0.030
---------------------------
What is the probability that 3 or more are introverts? (Use 3 decimal places.)
Ans: 1- binomcdf(5,0.69,2) = 0.823
--------------------------------
What is the probability that all are introverts? (Use 3 decimal places.)
Ans: 0.69^5 = 0.156
=============================
Cheers,
Stan H.
|
|
|
