) Assume that each light bulb bought at a shop has probability 0.1 of being defe
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Question
) Assume that each light bulb bought at a shop has probability 0.1 of being defective. It a customer buys ten light bulbs, (a (i) what is the probability that less than two are defective? (ii) what is the expected number of defective light bulbs? ) Lace curtain fabric manufactured as a continuous roll at a factory has two possible kinds of defects: Colour defects which appear on average 2 times per 10 meters of fabric, an weaving defects which appear on average 1 time per 10 meters of fabric. The two types defects appear independently of each other. of () What is the probability that in one 10 meter length of fabric there is one colour de and one weaving defect? (i) What is the expected total number of defects per 10 metres of fabric? (ii) If a customer buys a two-meter length of the curtain fabric, what is the probability that there are no defects at all in the fabric?Explanation / Answer
(a) Number of sampled bulbs n = 10
Probability a bulb being defective p = 0.1
(i) if x is the number of defective bulbs out of 10.
Pr(x < 2) = Pr( x =0) + Pr( x = 1) = 10C0 (0.1)0(0.9)10 + 10C1 (0.1)1(0.9)9 = 0.3487 + 0.3874 = 0.7361
(ii) Expected number of defective light bulbs= 10 * 0.1 = 1
(b) Here these two events are independent, so, if we assume that if x is the number of color defects and y are number of weaving defects per 10 meters of fabric. BOth are poisson distribution variable
Now, it is asked that
Pr(x = 1 ; y = 1) = POISSON (x = 1 ; =2 ) * POISSON (y = 1 ; = 1) = e-2 *2 * e-1 = 0.0498 *2 = 0.0996
(ii) Expected total number of defects = 1 + 2 = 2 +1 = 3
(iii) He buys two meter of curtain fabric, so expected number of Color defects = 2 * 2/10 = 0.4 and expected number of weaving defects = 2 * 1/10 = 0.2
Pr(x = 0 , y = 0 ; 1 = 2; 2 = 1) = e-0.4 * e-0.2 = e-0.6 = 0.5488
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