Question

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. (Round your answers to three decimal places.)

(a) What is the probability that at most 4 of the calls involve a fax message?
(b) What is the probability that exactly 4 of the calls involve a fax message?
(c) What is the probability that at least 4 of the calls involve a fax message?
(d) What is the probability that more than 4 of the calls involve a fax message?

Answer 1

@mathmale

Answer 2

I was up until midnight trying to figure out c and d. I got answers, but I am not understanding something I will show you when you get here

Answer 3

okay so for part a...

Answer 4

Hello! Am I correct in assuming that this is binomial probability? If so, what is n? p?

Answer 5

give me a sec. I know I am suppose to use the binomial distribution but I don't want to worry about that now. Let me upload a pic

Answer 6

So does P(X=4) = P(X<=4) - P(<=3) is this a true statement?

Answer 7

I agree with your approach to Problem 11a.
Before I answer your question: do you have a TI-83 or -84 calculator on hand?

Answer 8

I have a TI-89 on hand

Answer 9

I don't have an 83 or 84 just an 89.

Answer 10

I got that answer from previous work. You want to see?

Answer 11

Fancy, fancy. In my old (retired) age I have only a TI-83 to my name. But your TI-89 will surely calculate binomial probabilities. May i assume you've used it for that purpose before? No, your Problem 11a answer is clear, correct and appropriate.

Answer 12

calculator will do both the binomial cumulative probability and the probability of ONE particular x value.

Answer 13

Yes, I was shown from another tutor how to use nCy(y,n) command.

Answer 14

when you say cumulative you mean continuous?

Answer 15

my calc has these 2 commands that are relevant here: binompdf and binomcdf. You familiar with both?

Answer 16

Here is my work for 11a

Answer 17

cumulative =?= continuous? No.

Answer 18

cumulative and density? I have the formulas. I can do by hand. I don't know yet. I have not looked at solving these in my calculator. For parts C and D you can use a table in the back of the text. Lets move on to part b, c, and d

Answer 19

First, I'd like to test y ou. If n=25, p=0.25 and x=3, what is p(x)?

Answer 20

got PX=3) = 0.06410

Answer 21

P(X=3)

Answer 22

Looks OK, but I want to check; hold just a few sec, please.

Answer 23

Good. You've got it.

Answer 24

Just a sec, please

Answer 25

I am skipping (a) and starting with (b) at your request, OK?
(b) What is the probability that exactly 4 of the calls involve a fax message?
Find binompdf(25,.25,4), please.

Answer 26

This applies to "exactly 4."

Answer 27

P(X=4) = 0.11756

Answer 28

great. Part (b) done.

Answer 29

Explain in your own words what "(c) What is the probability that at least 4 of the calls involve a fax message?" means. Could you give an example?

Answer 30

P(X>=4)

Answer 31

um... thinking

Answer 32

are we looking above 4 so looking at 5,6,7,8,9,10,..... 25

Answer 33

In this case, no. "at least 4" includes 4, as well as all the other integers up to and including 25.

Answer 34

"at least 4" has a diff meaning than "above 4."

Answer 35

most and least are the two I get confused with the most.

Answer 36

least is implying we are looking at 4 and below?

Answer 37

No, "4 or greater, up to and possibily including 25."

Answer 38

I want you to buy me at least 1 ice cream cone of thanks, but preferably 2.

Answer 39

0 ice cream cones would be a deal breaker.
1 cone would be OK (minimally acceptable)
2 cones would be great.

Answer 40

this is implying X<=4 we are looking at random variable X above starting with 4.... so 4,5,6,7,8.... 25

Answer 41

Exactly. Rather than calculate the probabilities for 4 through 25,
we could calculate the probs P(x=0), P(x=1), etc., up thru P(x=3).

That's 4 calculations.

Or, y ou could find the cumulative probability for 0 thru 3 calls.

Know how to do that on your TI-89?

Answer 42

I get 0.09621. Most imp't is that you know how to use binompdf( ) and binomcdf( ).

Answer 43

I said above... Lol you just implied below 0,1,2,3

Answer 44

Nope, I've implied the prob. that you'd get 0, 1, 2 or 3 calls. I got 0.09621, which y ou should have gotten also. Now subtract this result from 1, please tell me your result, and please explain why we're subtracting here.

Answer 45

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

Answer 46

^^ I don't get this statement. We are subtracting the 1 because it's the complement. it represnts the 100%

Answer 47

Actually, friend, that'd be 1-(P(X=0) + P(X=1) + P(X=2) + P(X=3))

Answer 48

yes...

Answer 49

Yes. you want the prob. of "at least 4." That covers most of the cases we're looking at, right? meaning that this prob is close to 1.

We cheat a bit, start with 1, and then subtr. P(3 or fewer). Result is the prob. you wanted, that of 4 or more calls.

Answer 50

can you draw this?

Answer 51

1 => all cases
.09 something => P(fewer than 4)
(1-0.09 something) => P (4 or more calls are fax calls)

Answer 52

I've cheated again by drawing this verbally.