Probability for a Given Distribution of ObjectsDate: 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/ |
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