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Tips on Solving Probability Problems

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Date: 08/27/99 at 03:21:03
From: Karen Tsang
Subject: Probability distribution

Dear Dr. Math,

I have some questions about probability distribution. Below are the

Question 1:
A juice company produces bottles of apple juice. The nominal volume in
a bottle is 250 cm^3. However, due to random fluctuations, 0.5% of the
bottles fall below the nominal volume while 1% of the bottles go above
262.5 cm^3. Furthermore, it is assumed that the distribution is
normal.

(a) Determine the mean and the standard deviation of the volume
distribution.

(b) The production process is modified so that the standard deviation
is changed to 2 cm^3 but the percentage of bottles under nominal
volume remains unchanged. Find the modified mean of the distribution.

(c) The production of the juice is 600,000 bottles and the cost of
producing one bottle is \$(8.5+0.0094x) where x is the actual volume of
apple juice. Find, to the nearest dollar, the average saving in
monthly cost to the manufacturer after the production process is
modified.

My Try:
I know how to solve parts (a) and (b). The mean = 256.5621 cm^3 and
standard deviation is 2.5484 cm^3 and the answer for part (b) is
255.15 cm^3. I think the answer for part (c) is

= 600,000*(8.5+0.0094*256.5621)-600,000*(8.5+0.0094*255.15)
= \$7964.244

However, the suggested answer is \$7970. Is it something wrong in my
equation or in the suggested answer?

Question 2:
An electrical company has a stock of 6 similar television sets, which
they rent out to customers on a monthly basis. It is known that the
monthly demand for the sets has a Poisson Distribution with mean 3.56.

(a) Evaluate the probability that exactly one television set is not
rented in exactly two out of any four months.

(b) If one of the sets is temporarily withdrawn for repairs, find the
expected percentage of unsatisfied demand.

My Try:

Part (a):
Let X be the monthly demand of the television sets

X~Po (3.56)

P(X=5) = (3.565 )( e^-3.56) / 5! = 0.1355

Let Y be the number of months that exactly one set of television is
not rented

Y~b(4, 0.1355)

P(Y=2) = 4 C 2 *(0.1355)^2*[( 3.566 )(e^-3.56) / 6!]^2
= 0.0007

However, the suggested answer is 0.0695. Is it something wrong in my
equation or in the suggested answer?

Sorry. I have no idea on part (b). The suggested answer for part (b)

Question 3:
Two types of error A and B may occur in manufactured cloth. The number
of errors of each type occurring per meter length of the cloth are
independent random variables having Poisson Distribution with mean 0.5
and 1 respectively. Removing a type A error from the cloth costs \$0.1
and removing a type B error costs \$0.2. Find the mean and the standard
deviation of the cost of removing errors per 3-meter length of cloth.

My Try:
Let X be the number of errors of type A and Y be the number of errors
of type B

X~Po (1.5)

Y~Po (3)

Mean = (1.5*\$0.1)+(3*\$0.2)
= \$0.75

But I don't know how to calculate the standard deviation. Also, I want
to know what the meaning of standard deviation is in the general case
and in this specific case. Thanks.

Question 4:
A fighter is intending to drop bombs to hit a target city. It takes 4
hits to destroy the city. From the past records, the fighter has a
probability of 0.6 in hitting the target. Suppose that after a
successful dropping, the gunners will hit the target 65% of the time.
Given that the first shooting hits the target, what is the probability
that three more droppings will destroy the target city?

My Try:

= 0.6 * 9 C 3 * (0.65)^3 *(1-0.65)^6
= 0.0254

But the suggested answer is 0.2746. How can I solve this question?

Thanks a lot.
Karen Tsang
```

```
Date: 08/28/99 at 07:01:39
From: Doctor Mitteldorf
Subject: Re: Probability distribution

>Question 1:
>...the suggested answer is \$7970. Is it something wrong in my
>equation or in the suggested answer?

We need to go back to part (a) and see how you got your answer. The
only way I know how to do this is the tedious way: Find the place on
the normal curve where 1% of the area falls outside. Find the place
where 0.5% falls outside. Then use the results to scale the curve and
see, from your scaling, what the standard deviation is.

From Mathematica, the Error function Erf[1.64498] = 0.98, and
Erf[1.82139] = 0.99. This means that 2% of HALF the area falls outside
1.64498, and that's the same as 1% of the total area. On the other
side, 1% of HALF the area (0.5% of the total area) falls outside
-1.64498. The distance between these two numbers is 3.46637, and the
center of the distribution is 1.82139/3.46637 = 0.525446 from the left
edge. Translating these numbers to our range 250 ml - 262.5 ml, we
find a mean that is 250+(12.5*0.525446) = 256.568074 ml. This differs
a bit from the number you got. How did you get your number?...

>Question 2:
>...the suggested answer is 0.0695. Is it something wrong in my
>equation or in the suggested answer?
>
>Sorry. I have no idea on part (b). The suggested answer for part (b)

First, we'll evaluate the probability that exactly 1 set is unrented
in any one given month. That is the probability that 5 are rented,
which is the Poisson function for n = 5, x = 3.565

(x^n)(e^-x)/n! = 0.1358

The probability that two months in a row have 5 rented is 0.1358^2.
Then the following two months need to have any other number than 5
rented. The probability for that is (1-0.1358)^2. The product of these
two numbers is 0.01377.

But I've just picked one possibility for the choice of months. So, as
you noticed, my number needs to be multiplied by the number of
different ways of choosing 2 months out of 4, which is C(4,2) = 6. I
get 6*0.01377 = 0.08262 for part (a).
------------------------------
(b) To find the "percentage of unsatisfied demand", I think they want
the percentage of months in which the demand exceeds 5. For this, we
need to compute add the Poisson functions for 0, 1, 2, 3, 4 and 5;
then subtract from 1 to see what's left.

>Question 3:
>...I don't know how to calculate the standard deviation. Also, I
>want to know what the meaning of standard deviation is in the general
>case and in this specific case.

In general, think of the standard deviation as a measure of the
breadth of a distribution. How far off the mean (in either direction)
can you expect to be? If you just average "how far off the mean,"
you'll get zero, since you're equally likely to be off to the left or
to the right. So the definition of standard deviation is the average
of the SQUARE of "how far off" you are, which can only be positive.

Specifically, for the Poisson distribution, it happens that the mean
and the standard deviation are equal. (This is not true for other
distributions, of course, but it does work out exactly true for the
Poisson distribution.)

>Question 4:
>...the suggested answer is 0.2746. How can I solve this question?

This is much easier than you're making it. There's no 9C3 involved - I
don't know where you got that. Simply notice that there are 3 more
times he has to hit it. He's already succeeded on try #1. 0.65 of the
time he will hit on the 2d drop. 0.65*0.65 will also succeed on the 2d
and 3d drop, and 0.65*0.65*0.65 will also succeed on the 4th drop.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Probability

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