please explain question B PLEASE SHOW WORK TO GET FULL CREDT Problem 2) The life

Question

please explain question B

PLEASE SHOW WORK TO GET FULL CREDT Problem 2) The life length of semiconductor lasers is normaly distributed with a me deviation of 500 hours. a) (@ points) What is tho probabiythat the lie engith of arandomly chosen laser ls less han 600 hours Round your answer to four decimal places). (6000-100 0) p( -2) 0. 0223 (8 points) If4 lasers are randomly selected, independent of each other, what is the probability that at least one of these 4 lasers has We length exceeding 7000 hours? (Round your answer to four decimal places) b) P (X 7000 1- P( 2) to): (1)(0.022 7 s)(o, a 7125), 0.0 7725o 0, 912 05S MATH 333 Midterm Exam 2 November 16, 2016

Explanation / Answer

(b) If 4 lasers are randomly selected.

All are random and independent lasers

Here mean life length of an laser is = 7000 hours

Here Pr(any random laser will exceed 7000 hours) = 0.5

as the probability on either side of mean is 0.5

so Now this question will behave like a binomial distribution where n = 4 and p= 0.5

now we have to find that at least one of these 4 lasers have life length above 7000 hours

so Pr(X > 0) = BIN (X >0 ; 4 ; 0.5) = 1 - BIN (X = 0; 4 ; 0.5)

where X is the number of lasers whose life length exceeds 7000 hours.

Pr(X > 0) = BIN (X >0 ; 4 ; 0.5) = 1 - BIN (X = 0; 4 ; 0.5) = 1- 0.54 = 0.9375

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