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Lowest bidder

A manufacturer seeks bids on a modification of one of his processing units. Twenty contractors are invited to bid. They bid with probability 0.3, so that the numberof bids N binomial (20,0.3). Assume the bids Y i (in thousands of dollars) form an iid class. The market is such that the bids have a common distributionsymmetric triangular on (150,250). What is the probability of at least one bid no greater than 170, 180, 190, 200, 210? Note that no bid is not a low bid of zero, hence we must use the special case.


P ( V t ) = 1 - g N [ P ( Y > t ) ] = 1 - ( 0 . 7 + 0 . 3 p ) 20 where p = P ( Y > t )

Solving graphically for p = P ( V > t ) , we get

p = [ 23 / 25 41 / 50 17 / 25 1 / 2 8 / 25 ] for t = [ 170 180 190 200 210 ]

Now g N ( s ) = ( 0 . 7 + 0 . 3 s ) 20 . We use MATLAB to obtain

t = [170 180 190 200 210];p = [23/25 41/50 17/25 1/2 8/25];PV = 1 - (0.7 + 0.3*p).^20; disp([t;p;PV]') 170.0000 0.9200 0.3848180.0000 0.8200 0.6705 190.0000 0.6800 0.8671200.0000 0.5000 0.9612 210.0000 0.3200 0.9896
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[link] With a general counting variable

Suppose the number of bids is 1, 2 or 3 with probabilities 0.3, 0.5, 0.2, respectively.

Determine P ( V t ) in each case.


The minimum of the selected Y ' s is no greater than t if and only if there is at least one Y less than or equal to t . We determine in each case probabilities for the number of bids satisfying Y t . For each t , we are interested in the probability of one or more occurrences of the event Y t . This is essentially the problem in Example 7 from "Random Selection", with probability p = P ( Y t ) .

t = [170 180 190 200 210];p = [23/25 41/50 17/25 1/2 8/25]; % Probabilities Y<= t are 1 - p gN = [0 0.3 0.5 0.2]; % Zero for missing value PV = zeros(1,length(t));for i=1:length(t) gY = [p(i),1 - p(i)]; [d,pd]= gendf(gN,gY); PV(i) = (d>0)*pd'; % Selects positions for d>0 and end % adds corresponding probabilitiesdisp([t;PV]')170.0000 0.1451 180.0000 0.3075190.0000 0.5019 200.0000 0.7000210.0000 0.8462

[link] may be worked in this manner by using gN = ibinom(20,0.3,0:20) . The results, of course, are the same as in the previous solution. The fact that the probabilities in this example are lower for each t than in [link] reflects the fact that there are probably fewer bids in each case.

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Batch testing

Electrical units from a production line are first inspected for operability. However, experience indicates that a fraction p of those passing the initial operability test are defective. All operable units are subsequenly tested in a batch under continuousoperation ( a “burn in” test). Statistical data indicate the defective units have times to failure Y i iid, exponential ( λ ) , whereas good units have very long life (infinite from the point of view of the test). A batch of n units is tested. Let V be the time of the first failure and N be the number of defective units in the batch. If the test goes t units of time with no failure (i.e., V > t ), what is the probability of no defective units?


Since no defective units implies no failures in any reasonable test time, we have

{ N = 0 } { V > t } so that P ( N = 0 | V > t ) = P ( N = 0 ) P ( V > t )

Since N = 0 does not yield a minimum value, we have P ( V > t ) = g N [ P ( Y > t ) ] . Now under the condition above, the number of defective units N binomial ( n , p ) , so that g N ( s ) = ( q + p s ) n . If N is large and p is reasonably small, N is approximately Poisson ( n p ) with g N ( s ) = e n p ( s - 1 ) and P ( N = 0 ) = e - n p . Now P ( Y > t ) = e - λ t ; for large n

P ( N = 0 | V > t ) = e - n p e n p [ P ( Y > t ) - 1 ] = e - n p P ( Y > t ) = e - n p e - λ t

For n = 5000 , p = 0 . 001 , λ = 2 , and t = 1 , 2 , 3 , 4 , 5 , MATLAB calculations give

t = 1:5; n = 5000;p = 0.001; lambda = 2;P = exp(-n*p*exp(-lambda*t)); disp([t;P]') 1.0000 0.50832.0000 0.9125 3.0000 0.98774.0000 0.9983 5.0000 0.9998

It appears that a test of three to five hours should give reliable results. In actually designing the test, one should probably make calculations with a number of differentassumptions on the fraction of defective units and the life duration of defective units. These calculations are relatively easy to make with MATLAB.

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Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
Application of nanotechnology in medicine
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RAW Reply
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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The nanotechnology is as new science, to scale nanometric
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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Damian Reply
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how did you get the value of 2000N.What calculations are needed to arrive at it
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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