A.  According to a survey of American households, the probability that
the residents own two cars if annual household income is over $25,000
is 80%. Of the households surveyed, 60% had incomes over $25,000 and
70% had two cars. The probability that the residents of a household
own two cars and have an income less than or equal to $25,000 a year
is

B.  A company has two machines that produce widgets. An older machine
produces 23% defective widgets, while the new machine produces only 8%
defective widgets. In addition, the new machine produces three times
as many widgets as the older machine does. What is the probability
that a randomly chosen widget was produced by the older machine?

C.  A company has two machines that produce widgets. An older machine
produces 23% defective widgets, while the new machine produces only 8%
defective widgets. In addition, the new machine produces three times
as many widgets as the older machine does. What is the probability
that a randomly chosen widget produced by the company is defective?

D.  Blossom’s Flowers purchases roses for sale for Valentine’s Day.
The roses are purchased for $10 a dozen and are sold for $20 a dozen.
Any roses not sold on Valentine’s Day can be sold for $5 per dozen.
The owner will purchase one of three amounts of roses for Valentine’s
Day: 100, 200, or 400 dozen roses. If the probability of selling 100
dozen roses is 0.2 and 200 dozen roses is 0.5, then the probability of
selling 400 dozen roses is?

E.  Blossom’s Flowers purchases roses for sale for Valentine’s Day.
The roses are purchased for $10 a dozen and are sold for $20 a dozen.
Any roses not sold on Valentine’s Day can be sold for $5 per dozen.
The owner will purchase one of three amounts of roses for Valentine’s
Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the
probabilities for the sale of 100, 200, or 400 dozen roses,
respectfully, then the optimal arithmetic mean for buying roses is?

F.  In a local cellular phone area, company A accounts for 60% of the
cellular phone market, while company B accounts for the remaining 40%
of the market. Of the cellular calls made with company A, 1% of the
calls will have some sort of interference, while 2% of the cellular
calls with company B will have interference. If a cellular call is
selected at random, the probability that it will have interference is?

G.  A catalog company that receives the majority of its orders by
telephone conducted a study to determine how long customers were
willing to wait on hold before ordering a product. The length of time
was found to be a random variable best approximated by an exponential
distribution with a mean equal to 3 minutes. What proportion of
customers having to hold more than 4.5 minutes will hang up before
placing an order?

H. Let X represent the amount of time it takes a student to park in
the library parking lot at the university. If we know that the
distribution of parking times can be modeled using an exponential
distribution with a mean of 4 minutes, find the probability that it
will take a randomly selected student more than 10 minutes to park in
the library lot.

I.  The owner of a fish market determined that the average weight for
a catfish is 3.2 pounds with a standard deviation of 0.8 pound.
Assuming the weights of catfish are normally distributed, above what
weight (in pounds) do 89.80% of the weights occur? What is the
probability that a randomly selected catfish will weigh less than 2.2
pounds?

J.  At a computer manufacturing company, the actual size of computer
chips is normally distributed with a mean of 1 centimeter and a
standard deviation of 0.1 centimeter. A random sample of 12 computer
chips is taken What is the probability that the sample mean will be
below 0.95 centimeters?

K. Records of an automobile insurance company show that 10% of its
policyholders were involved in an accident during the past year. A
random sample of 400 policyholders is to be selected. So, 95% of all
the sample percentages will fall between ________ and ________,
symmetric about the population proportion.

L.  A manufacturer of power tools claims that the average amount of
time required to assemble their top-of-the-line table saw is 80
minutes with a standard deviation of 40 minutes. Suppose a random
sample of 64 purchasers of this table saw is taken. The mean of the
sampling distribution of the sample mean is __________ minutes.

M.  Suppose a random sample of 100 restaurant owners are asked their
opinion of the California smoking ban. The number of restaurant owners
against the ban has a binomial distribution with p = 0.8. The
probability that the proportion of restaurant owners against the ban
is at least 0.88 is __________.

O.  The Wall Street Journal recently ran an article indicating
differences in perception of sexual harassment on the job between men
and women. The article claimed that women perceived the problem to be
much more prevalent than did men. One question asked to both men and
women was: “Do you think sexual harassment is a major problem in the
American workplace?” Some 24% of the men compared to 62% of the women
responded “Yes.” Assuming W designates women’s responses and M
designates men’s, what hypothesis should the Wall Street Journal test
in order to show that its claim is true?

---

Request for Question Clarification by answerguru-ga on 10 Jul 2003 19:09 PDT

Hi cosmotini-ga,

Considering that wonko-ga has completed the first six questions, I can
do the remaining questions for $30 if you split this into two
questions as suggested by wonko :)

answerguru-ga

---

Clarification of Question by cosmotini-ga on 11 Jul 2003 06:57 PDT

Hello,

Please answer the questions as posted or unlock for another
researcher.  The tip will be $5 as long A-0 are completed correctly.

If no other researcher completes the problems by saturday night.  I
will reconsider your offer.  Thanks for your quick attention to the
questions :-)

Hello cosmotini!

Here are the answers to your questions.

A - First of all let me introduce some notation. P(A) means
"probability of event A", and P(A|B) means "probability of event A
conditional on event B". We also know that, by Bayes, P(A|B)=P(A and
B)/P(B).

Stats: Conditional Probability
http://www.richland.cc.il.us/james/lecture/m170/ch05-cnd.html

This question can be solved as follows. We have that the probability
of having 2 cars given that your income is over $25,000 is 80%. That
is, P(2 cars|income>25000)=0.8

Using Bayes, and the fact that 60% of the households have income over
25,000 we find that

P(2 cars AND income>25000)=0.8*0.6= 0.54

Finally, 70% of the population has 2 cars and any income. 54% of the
population has 2 cars and income over 25,000. Therefore 70-54 = 16% of
the population has 2 cars and income under 25,000.


B - In this question, the percentage of defective widgets in each
machine is irrelevant. The important fact is that the new one produces
3 times as many as the other one. That is, for every 1 widget produced
by the old machine, the new one produces 3 widgets. Thus, for every 4
widgets, 1 is produced by the old machine and 3 are produced by the
new one. So 25% of the widgets are produced by the old one and 75% of
the widgets are produced by the new one. The answer to this question
is then 0.25.


C - To simplify notation, let's call D to "deffective", O to "produced
by old machine" and N to "produced by new machine". We know that

P(deffective) = P(deffective AND produced by old machine)
                 + P(deffective AND produced by new machine)

P(D) = P(D and O) + P(D and N)
     = P(D|O)*P(O) + P(D|N)*P(N)   (by Bayes)
     =  0.23 *0.25 + 0.08  *0.75   (data given by the problem)
     =  0.1031

The probability of a randomly chosen widget produced by the company
being deffective is 0.1031


D - The answer to this question depends on the assumptions we make
about the demand for flowers. If we assume that Blossom's could sell
any number of flowers (100, 120, 247, 351, etc.) then there's no way
to know the probability of selling 400 flowers. However, if we assume
that the demand for flowers is either 100, 200 or 400 flowers (there's
no chance of anything else than that happening), then we can compute
it simply by:

P(100)+P(200)+P(400) = 1 (because these are all the possible events) 

P(400) = 1-P(100)-P(200)
       = 1 - 0.2 - 0.5
       = 0.3

The probability of selling 400 flowers is in this case 0.3.


E - We have to analyze each case:

Strategy 1: Buy 100 flowers
Given that in any case the demand for flowers will be equal or greater
than 100 (recall that Blossom's sells either 100, 200 or 400 flowers),
in this strategy Blossom's sells 100 flowers on Valentine's day with
probability 1. Therefore, its profits are:

$20*100 - $10*100 = $1000

Strategy 2: Buy 200 flowers
The probability of selling 100 flowers is .2 the probability of
selling 200 flowers or more is 0.8 (because the probability of selling
200 is 0.4 and the prob. of selling 400 is 0.4). Therefore, if it
sells 100 flowers on Valentine's Day (and the other 100 later) the
firm's profits are:

A) $20*100 + $5*100 - $10*200 = $500

If the demand is 200 or more, then its profits are:

B) $20*200 - $10*200 = $2000

Finally, we know that A happens with probability 0.2 and B happens
with probability 0.8, so the expected profits are:

0.2*$500 + 0.8*$2000 = $1700

Strategy 3: Buy 400 flowers
Using a similar argument, we find that, if the demand is 100, then the
profts are:

A) $20*100 + $5*300 - $10*400 = -$500

If the demand is 200:

B) $20*200 + $5*200 - $10*400 = $1000

If the demand is 400:

C) $20*400 - $10*400 = $4000

Therefore the expected profits are:

0.2*($-500) + 0.4*($1000) + 0.4*($4000) = $1900

So the expected profits are maximized when buying 400 roses, as it
yields an expected profit of $1900.


F - Using the same reasoning as in question C, we find that:

P(interference)=P(interf|A)*P(A) + P(interf|B)*P(B)
               = 0.01*0.6       +     0.02*0.4
               = 0.014

Therefore, the probability that a cellular call selected at random
will have interference is 0.014


G - The formula for the cummulative distribution function (cdf) of an
exponential distribution is:

F(x) = P(X < x) = 1 - exp(-x/beta)

where "beta" is the distribution mean. Check the following pages for
more information.

Exponential Distribution
http://www.itl.nist.gov/div898/handbook/eda/section3/eda3667.htm

http://csep1.phy.ornl.gov/mc/node18.html

Here, beta=3. We want to calculate the probability of having to wait
more than 4.5 minutes. Using the cdf, we can find that probability of
having to wait less than that time is:

1 - exp(-4.5/3) = 0.776

Therefore, the probability of having to wait MORE than 4.5 minutes is
simply

1 - 0.776 = 0.224


H - Again we're facing an exponential distribution, this time with
mean equal to 4 minutes. Therefore, beta=4. Using the cdf provided
above, the probability of taking less than 10 minutes to find a
parking lot is:

1 - exP(-10/4) = 0.9179

Thus, the probability of taking MORE than 10 minutes is:

1 - 0.9179 = 0.0821


I - In order to answer this question, we must use a standard normal
table. This table gives the probability that a random variable
following the standard normal distribution will be equal or less than
a specified number. The following page explains this very well.

Normal Curves
http://www1.hollins.edu/faculty/clarkjm/Stat140/normalcurves.htm

A standard normal table can be found at

Standard Normal Table
http://www.stat.psu.edu/~herbison/stat200/stat200_model_demo/supplements/NormalTable.html

So, above what weight do 89.8% of the weights occur? Well, the same
weight under which 10.2% of the weights occur. So, given that X is
distributed normally with mean 3.2 and SD 0.8, then

P(X < a) = 0.102

We have to solve this for "a". In order to use the standard normal
table, we must first convert X to a standard normal by substracting
the mean and dividing by the SD. Thus,

 P(X < a)
=P(X - 3.2 < a -3.2)
=P( (X-3.2)/0.8 < (a-3.2)/.8 )
=P(   Z < (a-3.2)/.8 ) = 0.102

(Z stands for a random variable that is standard normally distributed)

Now, checking the table for 0.102, we find that the corresponding
z-value is -2.32. So

(a-3.2)/.8 = -2.32
    a = 1.344

Therefore, 89.8% of the weights occur above 1.344 pounds.

What is the probability that a randomly selected catfish will weight
less than 2.2 pounds?

 P( X < 2.2)
=P(X-3.2 < 2.2-3.2)
=P( (X-3.2)/.8 < (-1)/.8 )
=P(   Z        <   -1.25)

Looking up -1.25 in the table, one finds out that this probability is
0.1056. That's the probability that a randomly selected catfish will
weigh less than 2.2 pounds. Note that in both cases I manipulated the
equations in order to get the prob. of Z LESS than something. This is
because the table I provided above gives the probability of Z being
less than some number. If you have a table in your book that gives the
probability of Z being greater than a number, then you should
manipulate them in order to get P(Z>a).


J - We have 12 random variables here: X1, X2,... X12 (the 12
observations). These 12 observations are independent random variables:
observing a number does not affect the probability distribution of the
subsequent observations (because the sample is a random sample). The
sample mean is then another random variable which is obtained by
summing the values of X1 through X12 and dividing them by 12:

Mean = (X1+X2+...+X12)/12

In the following page you can see that the sum of normal distributions
is itself a normal distribution with mean and variance equal to the
sum of the means and the sum of the variances.

The Normal Distribution
://www.google.com.ar/search?q=cache:WXNRXZw0D74J:library.thinkquest.org/10030/7ndndpnd.htm+properties+of+normal+distribution+sum+independent&hl=es&ie=UTF-8

Therefore, X1+X2+...+X12 is normally distributed with mean equal to
1+1+...+1=12, and variance equal to 0.01+0.01+...+0.01=0.12. Finally,
if X follows a normal distribution with mean u and variance s^2, then
X/n (where n is any number) also follows a normal distribution, with
mean u/n, and variance s^2/n^2. Thus, (X1+X2+...+X12)/12 follows a
normal distribution with mean 12/12=1 and variance 0.12/12^2= 0.0008.

So, the SD is the square root of 0.0008, that is 0.0288.

Now we have all the data we need. Let's call S=(X1+X2+...)/12 the
sample mean. Then

 P( S < 0.95)
=P( (S-1)/0.0288 < (0.95-1)/0.0288 )
=P( Z < -1.73 )
= 0.0418 (looking in the table)

So the probability that the sample mean is below 0.95 cm is 0.0418


K - Here we can use the normal approximation to a binomial
distribution. Check the following pages for more information on this
subject:

Normal Approximation to Binomial
http://stat.tamu.edu/stat30x/notes/node89.html

http://www.ruf.rice.edu/~lane/stat_sim/binom_demo.html

Hence, since the sample size is big (400 observations), we can
approximate the distribution of sample percentages with a normal
distribution with mean 0.1 (because we assume that the 10% mentioned
in the problem is the population proportion) and variance equal to
0.1*(1-0.1)/400=0.000225. Therefore, the SD is 0.0150.

As you can see in the page that follows, 95% of the observations from
a normally distributed variable fall "inside" two standard deviations
around the mean. Two standard deviations is 0.03. Therefore, 95% of
the sample percentages fall between 0.07 and 0.13.

A normal distribution...
http://www.stoller-eser.com/raddeviation.htm


L - Here we use proporties of the mean. As in question J, we have here
64 independent, identically distributed random variables. The sample
mean is (X1+X2+...+X64)/64. We want to calculate here the mean (or
expected value) of this variable:

E( (X1+X2+...+X64)/64 )

Using standard properties of the expected value, we find that

 E( (X1+X2+...+X64)/64 )
=1/64*E(X1+X2+...+X64)
=1/64*[E(X1)+E(X2)+...+E(X64)]
=1/64*[80+80+...+80]
=80

Thus the mean of the sampling distribution of the sample mean is 80
minutes (the same as the population mean).


M - This problem is very similar to K. The distribution of proportions
can be approximated with a normal distribution, with mean equal to 0.8
(the population mean according to the problem) and variance equal to
0.8*(1-0.8)/100=0.0016. Therefore, the SD is 0.04. Now we have to
calculate. Let X be the observed proportion, whose distributed is
approximated with a normal distribution with mean=0.8 and SD=0.04.

Then,

 P(X > 0.88)
=1-P(X<0.88)
=1-P(X-0.8 < 0.88-0.8)
=1-P( (X-0.8)/0.04 < 0.08/0.04)
=1-P(Z < 2)
=1-0.9772
=0.0228

That's the probability that at least 0.88 of the owners are against
the
ban.


O - The hypothesis the Wall Street Journal should test is the null 
hypothesis that the population proportion of men who think sexual
harassment is a serious problem is equal to the population of women
who do; versus the alternative hypothesis that the population means of
both groups are statistically different. Let me illustrate with an
example. If the results of the study were that 25% of the women think
sexual harassment is a serious problem, while 24% of the men think the
same, should one conclude that they have differnent perceptions on the
subject? Probably not. The small difference could be due to the fact
that we are not asking the entire population about their views, but
only a sample of it. But then, where's the limit for a "small"
difference? Why should 1 or 3% "small" while 10% should be "big"?
That's what the test for different means (the one the WSJ should do)
is about. The WSJ has to see wether 62% is *statistically* different
from 24%, and this will depend on the variance of the responses. The
test would be to ask "if the proportions were equal (that is, if men
and women had the same perceptions), what is the probability that a
sample would show 62% for women and 24% for men?". If this probability
turns out to be very small, then one could conclude that it is highly
unlikely that the assumption that the populations proportions are
equal is correct. Thus the conclusion would be that men and women do
have different perceptions on the subject.


That's all the answers. If you have any doubts regarding any of them,
don't hesitate to request a clarification; I'll answer it as soon as
possible. Otherwise, I await your rating and final comments.


Best wishes!
elmarto


Google search strategy
normal table
normal approximation to binomial
normal distribution properties

---

Request for Answer Clarification by cosmotini-ga on 11 Jul 2003 09:50 PDT

Hi Elmarto,

Thanks for the quick solutions.  I took these problems out of my
prentice hall study guide and a few of your answers do not match mine
or the solutions manual.

I will give you the multiple choice options and based on that could
you provide simple steps to your answer?  Thanks so much for all your
help :-)


A.  0.12    0.18  0.22   0.48

C.  I got 0.310

E.  700  900 1500 1600

I. above what weight (in pounds) do 89.80% of the weights occur? 4.21 
3.91  2.48  2.18


O.  Could you simplify e.g  H0: pW-pM>=0 versus H1: pW-pM<0

---

Clarification of Answer by elmarto-ga on 11 Jul 2003 10:58 PDT

Hi cosmotini!
Let's see the troublesome questions.

A - I did make a mistake here, although it was a calculation error,
not an error in the reasoning. As you can see, I posted

"P(2 cars AND income>25000)=0.8*0.6= 0.54"

Which is obviously wrong: 0.8*0.6=0.48. The last paragraph of this
question, then, should read:

"Finally, 70% of the population has 2 cars and any income. 48% of the
population has 2 cars and income over 25,000. Therefore 70-48 = 22% of
the population has 2 cars and income under 25,000."

So 0.22 is the correct answer to this one.


C - 0.310 is not right. However, the answer I provided is also wrong,
again, due to a calculation error. I assume that in order to come to
this result, you summed 0.23 and 0.08, which is not the right way to
calculate this. The intuition is that 8% out of the 75% of the total
number of widgets are deffective. Also, 23% of the other 25% of the
total number of widgets are deffective. Therefore, the correct answer
comes from the formula I gave:

0.08*0.75 + 0.23*0.25 = 0.1175

The right answer is 0.1175.


E - This is one question I checked and re-checked, and I'm almost
certain the right answer was the one I gave you. It's possible that I
misunderstood the assumptions of this problem, so I'll tell you what I
understood, you tell me if I'm wrong about anything.

- The firm must choose either 100, 200 or 400 dozens of flowers. No
other number is possible.
- The problem states: "0.2, 0.4, and 0.4 are the probabilities for the
sale of 100, 200, or 400 dozen roses". I assume this means that the
DEMAND for roses will be one of those numbers, regardless of the
number of flowers the firm has in stock. This implies, for example,
that if the firm has 100 dozen roses it will always sell 100 dozen
roses, because when demand is 200 or 400, it will sell 100 and leave
the other demandants "unsatisfied". Same when it has 200 and demand is
400: there will be demand for 200 more, but the firm won't have
anything else in stock.
- The problem asks you to answer what is the expected profit when the
firm is already choosing the optimal amount of flowers to buy.

Those are the crucial assumptions. Based on those, my answer is
correct (if you don't understand my answer, please ask). Have you
checked that you copied the statement of this question exactly as it
is? Also, if you have the solutions manual, maybe if you tell me the
correct answer I can find where the problem is.


I - Another error of mine (I truly apologize...). The reasoning is
correct, however, I looked up the wrong number in the standard normal
table. The z-value that corresponds to probability 0.102 is -1.27 and
not -2.32. So,

(a-3.2)/0.8 = -1.27
  a = 2.184

So 2.18 is the correct answer here


O - To make it short, the test should be:
H0: pW-pM=0
H1: pW-pM>0

Again, I'm truly sorry for the calculation mistakes. If there's
anything you don't understand, just tell me.


Best wishes!
elmarto

---

Request for Answer Clarification by cosmotini-ga on 11 Jul 2003 12:31 PDT

Thanks elmarto,

no prob about the errors becuause you are helping me undertand the
procedures.  About "E"  my study group partner said that the book must
be wrong,,,so I am going stick with your answer :-)  Thanks again for
your quick reply.

I'll post the rating & tip by this eve.

Ciao

---

Clarification of Answer by elmarto-ga on 11 Jul 2003 16:51 PDT

Thank you for the rating and tip! I'm looking forward to answering
your future questions.
 
 
Best luck! 
elmarto

Thank you.

I have A-F solved.  Would you like to repost this as 2 questions and
have me answer A-F?

Sincerely,

Wonko

Thank you for the rating and tip! I'm looking forward to answering
your future questions.


Best luck!
elmarto