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Tips on Solving Probability Problems


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 
questions I want to ask:

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(b) If one of the sets is temporarily withdrawn for repairs, find the 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) 
is 7.44%. Please help me. Thanks!

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:
I think the answer is

     = 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) 
>is 7.44%. Please help me. Thanks!

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

This is close to your answer, but not exactly the same.

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/   
    
Associated Topics:
High School Probability

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