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Probability for a Given Distribution of Objects


Date: 7/24/96 at 16:4:8
From: Ariel Di Miro
Subject: Probability for a Given Distribution of Objects

1) A man has a box of light bulbs to sell. 80 percent of these bulbs 
are class 'A', and 20% are class 'B'. The 'A' bulbs have a lifetime of 
2400 hours, while 'B' bulbs last 2600 hours. The man takes a new bulb 
from the box and uses it in his shop for 600 hours. Someone wants to 
buy a bulb, but there are no more, so the man offers him the bulb he 
has used in his shop. Calculate the probability that the bulb will 
burn out in more than 2400 hours. (I think this problem needs more 
data to be solved.)

2) The sizes of some screws are distributed "normally" (a Gaussian 
distribution). They must be between: 10 +- 3 mm (9,7 and 10,3 mm). A 
lot of them are tested with an instrument, and it's found that 5 
percent of them are below the range, and 11 percent are above the 
range (these are refused).

After the test, it is proved that the instrument made a systematic 
error of +0.05 mm. What is the percentage of refused screws if the 
test instrument was right?

3) A company makes wires. A failure in the machine made aleatory 
mistakes in the wires, with an average frequency of 1 mistake in 25 
meters. 30 wires with a length of 100 meters are selected. Which is 
the probability that at least 3 wires have no more than 2 mistakes ?
(I think I have to use Poisson and Bernoulli.)

I would thank you very much if you help me to solve them.


Date: 7/26/96 at 8:50:44
From: Doctor Anthony
Subject: Re: Probability for a Given Distribution of Objects

(Question 1) 
If the life-times have a mean of 2400 hours and 2600 hours, we still 
need to know the standard deviations, if we are to calculate 
probabilities of the bulb from the shop lasting a further 2400 hours.  
Check the question again and see if the standard deviations are given.

(Question 2)
Because of the instrument error, the tail areas are 5 percent below 
9.75 mm, and 11 percent above 10.35 mm. From this information we can 
find the true mean and s.d. of the screws produced.  Let m = mean  
and s = s.d. From tables a 5 percent tail below the mean corresponds 
to z = -1.645   The tail area of 11 percent above the mean corresponds 
to z = 1.2263

We have two equations using z = (x-m)/s for the two values of z and x 
that we know.

          -1.645 = (9.75 - m)/s

          1.2263 = (10.35 - m)/s

-1.645s = 9.75 - m
1.2263s = 10.35 - m

Subtracting these equations we get  2.8713s = 0.6
                                          s = 0.20896

Then from first equation  m = 9.75 + 1.645*0.20896
                            = 10.0937 mm  

The percentage that are actually too small is therefore found from

   z = (9.7 - 10.0937)/0.20896

     = -1.8843 and from Normal tables area = .9702  
                             and tail area = 1 - .9702 = 0.0298
                                           = 2.98 percent
 
The percentage that are actually too large is found from

   z = (10.3 - 10.0937)/0.20896

     = 0.9873 and from Normal tables area = .9382
                            and tail area = 1 - .9382 = 0.0618
                                          = 6.18 percent

Total percentage rejected = 2.98 + 6.18 = 9.16 percent


(Question 3)
We use the Poisson probability model to find probability of 0, 1, 2 
etc. mistakes in 100 metres.  Because of the additive property of the 
means of Poisson probabilities we can say that the mean number of 
mistakes in 100 metres will be 4.

Poisson probabilities are P(0) = e^(-4)     = .018316
                          P(1) = (4/1)*P(0) = .073263
                          P(2) = (4/2)*P(1) = .146525
                          P(3) = (4/3)*P(2) = .195367
                          P(4) = (4/4)*P(3) = .195367  and so on 

We need P(0) + P(1) + P(2) = .238104  and this is the probability that 
one 100 metre wire has no more than 2 mistakes.

Now with 30 wires, at least 3, could mean 3, 4, 5,.... 30, so instead 
we consider 1 - Prob(0 or 1 or 2) with no more than 2 mistakes.

For the binomial probabilities which we now consider

    p = .238104   q = .761896  n = 30

  P(0) = .761896^30 = .000286
  P(1) = 30*.238104*.761896^29 = .002684
  P(2) = 435*.238104^2*.761896^28 = .012165

P(0) + P(1) + P(2) = 0.015135  and subtracting this from 1 we get 
0.984865

So the probability of at least 3 wires with no more than 2 mistakes is 
0.984865

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Probability
College Statistics

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