Hey Guys, how are you? I'm working on this homework for my statistics class and it is truly driving my crazy. I excelled in other math courses, but something about these problems is really just throwing me off. Anyway, just wondering if you guys could help me out with a few problems. Anything at all is greatly appreciated. Anyway here are the problems:

1. Each salesperson in a large department store chain is rated on their sales ability and their potential for advancement. The data for the 500 sampled salespeople are summarized in the table below:

Potential For Advancement If: Sales Ability is:

Below Average: 16 Fair Chance, 12 Good Chance, 22 Excellent Chance
Average: 45 Fair Chance, 60 Good Chance, 45 Excellent Chance
Above Average: 93 Fair Chance, 72 Good Chance, 135 Excellent Chance


a. What is the probability that a salesperson selected at random has above average sales ability and is an excellent potential for advancement?

b. What is the probability that a salesperson selected at random will have average sales ability an and good potential for advancement?

c. What is the probability that a salesperson selected at random will have below average sales ability and fair potential for advancement?

d. What is the probability that a salesperson selected at random will have an excellent potential for advancement given they also have above average sales ability?

e. What is the probability that a salesperson selected at random will have an excellent potential for advancement given they also have average sales ability?




2.An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of four e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution.

a. What is the probability Linda Lahey, company president, received exactly one e-mail between 4pm and 5pm yesterday?

b. What is the probability she did not receive any e-mail during this period?

c. What is the probability she received ten or more e-mails during the same period?




3. According to a recent survey, 50% of all customers will return to the same grocery store. Suppose nine customers are selected at random, what is the probability that:

a. Exactly five of the customers will return?

b. All nine will return?

c. At least eight will return?

d. At least one will return?

e. How many customers would be expected to return to the same store, i.e., what is the mean of the distribution?




4. A study of long distance phone calls made from the corporate offices of Pepsi Bottling Group, Inc., in Somers, New York, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.2 minutes and the standard deviation was 0.50 minutes.

a. What fraction of the calls last between 3.2 and 4 minutes?

b. What fraction of the calls last more than 4 minutes?

c. What fraction of the calls last between 4 and 4.5 minutes?

d. What fraction of the calls last between 3 and 4.5 minutes?

I think I understand the first one as it seems to be based off the table: Here is what I got:

A) 135/500=.27
B)60/500=.12
C)16/500=.032
D)135/500 =.27
E)45+60+45+=150 150/500=.30

The other ones though are really driving me crazy... lol. Thanks again guys!

For 2 you have to use a Poisson distribution

p(k) = (lambda^k(e^lambda))/k!

Where lambda is the expected value

Lets say you expect 20 letters on a given day.

The chance you receive 14 letters is

p(k) = ((20^14)(e^-20))/14!

Don't have a calculator at the moment and can't find e on the Microsoft one.


3 looks like a binomial distribution

And 4 of course is a normal distribution

These should be pretty easy to find in your textbook.

Sorry man, I failed history and Spanish.

Ok, here are my answers on the other three. How do they look and the first one as well? Thanks so much again!


A) P(4,1)= [e^(-4)*4^1]/1!= .07326
B P(4,0)=[e^(-0)*0^4]/4!= .01832
C) P(4,10)=[e^(-40)*40^4]/4!=.0053

A)5C9 .50^9(1-.50)^5-9=.7461
B)9C9 .50^9(1-.50)^9-9=1
C)8C9 .50^9 (1-.50)^8-9= .9980
D)1C9 .50^9(1-.50)^1-9= .0195
E)Mean= 9*.50= 4.5


A)3.2-3.2/.50=0 zscore for 0=0 4-3.2/.50= 1.6 zscore=0.000-.4432= .4432
B)4-3.2/.50=1.6 zscore= .4432-.5000=.0568
C)4-3.2/.50=1.6 zscore=.4432 4.5-3.2/.50=2.6 zscore=.4953 .4432-.4953=.0521
D) 3-3.2/.50=-.4 zscore= .1554 4.5-3.2/.50=2.6 zscore= .4953 .4953+.1554=.6507

Oh god, this reminds me of last year. Still got a 92% in my final, and the exam is EXACTLY the same each year, so it's not THAT bad.

This is clearly wrong. If the probability of a success is less than 1, the probability of n successes isn't going to be 1



Given that X ~ B(9, 0.5), we want to find P(X=5)

P(X=5) = ⁹C₅ 0.5⁵ (1-0.5)^9-5 = 63/256 ≈ 0.246

This. Looks like you're using the cumulative binomial distribution function. You need to calculate it for just one number. Try calculating the cumulative of the number before it and subtract that from the cumulative of the one you want.

I feel pretty good on 1 and 4... I think at least. But 2 and 3 are really bothering me. I understand that on the 3B, the probability cannot be 1. Still, I'm lost to the max lol. 3A seems to make more sense, but I seem to have trouble applying the formula. How are my other answers?


A) P(4,1)= [e^(-4)*4^1]/1!= .07326
B P(4,0)=[e^(-0)*0^4]/4!= .01832
C) P(4,10)=[e^(-40)*40^4]/4!=.0053

A)5C9 .50^9(1-.50)^5-9=.7461 (wrong)
B)9C9 .50^9(1-.50)^9-9=1 (wrong)
C)8C9 .50^9 (1-.50)^8-9= .9980
D)1C9 .50^9(1-.50)^1-9= .0195
E)Mean= 9*.50= 4.5

I can't thank you guys enough for all your help so far. I'm just really struggling with this stuff, but slowly getting there lol. Thanks again!

Here, I'll be a little more specific on these ones. You're using the cumulative formula so to figure out
a: you need to figure out the answer for 5 and then subtract the answer for 4.
b: Figure out the answer for 9(Obviously 1 as it includes all possibilities) and subtract the answer for 8
c: Just do the answer for 7 and invert it (e.g. .65 would become .35, .24 would become .76) You want the outcome of 8 or 9 so taking 1 and subtracting the outcome for 0-7 will do it
d: Do the answer for 0 and invert it

To expand a bit on what Cecilff2 said.


In all of these, let X ~ B(0.5, 9) (X is distributed as a Binomial distribution with a probability of 0.5 and 9 tests)

a:
P(X = 5)

(this can be calculated using the PMF function, which I think you got slightly wrong when you were trying to use it)

b:
P(X = 9)

(Again, PMF function)

c:
P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10) + ...

However, we know that P(X = 10), P(X = 11) etc are all 0 (They cannot be anything else, as if we have 9 tests, the chance that we get 10 successes is 0)

So:
P(X >= 8) = P(X = 8) + P(X = 9) + 0 + 0 + ...
P(X >= 8) = P(X = 8) + P(X = 9)

So using the PMF functions you should be able to work it out.

d:
e need to find P(X >= 1), which is 1 - P(X < 1) = 1 - [P(X = 0) + P(X = -1) + P(X = -2) + ...]

We can flip it around like that as we know that the probabilites must add up to 1, so
P(X < 1) + P(X = 1) + P(X > 1) = 1
= P(X < 1) + P(X >= 1)

So rearranging we get P(X >= 1) = 1 - P(X < 1) = 1 - P(X <= 0).

Bet as we know that we can't get a number of successes less than 0, P(X = -1), P(X = -2) etc are all 0. Hence
P(X >= 1) = 1 - P(X = 0).

e:
Your answer was fine.

I may have got some of that slightly wrong, but you should get the basic idea.

Very interesting thus far. Really helped to clarify some more. In general, is anyone able to post what they got for the problems so I can compare to mine. Regardless though, you guys really do rock. The help and dedication you guys show for a fellow forum member is incredible. I really cannot thank you all enough! THANK YOU!