Question

In the electronic manufacturing assembly line, engineers observe a probability of 0.2 that a cell phone modem chip is defective. Round probabilities to the nearest three decimal places. (a) [6pts] Suppose ten chips are selected for testing, what is the probability that at most one chip is defective? Suppose 600 chips are randomly selected for testing. (b) (6pts] We can approximate the Binomial distribution with a normal distribution. Compute the mean and standard deviation of this normal distribution (c) (pts] What is the probability that at most 105 chips are defective? (d) [9pts) What is the probability that more than 135 chips are defective?

Answer 1

Soln We are given that in electronic manufacturing assembly line, prob. that chip is defective is 0.2 Here we get outputs as wether chip is defective or it is not. And as we conduct it. A times so it becomes binomial distribution. As p= 0.2 19= 1-0.2 = 0.8 wanamumunun (a) we want to find defective if ten prob chips that atmost are selected one chip is at random. Here n=10, p= 0. 2 9 =0.8. npg . Let be no. of defective chips.to we find P(x <1) X UB (102) ... P(X <1) = P(x-AP < 1-?) =plz C -0.625) = '(-0.625 ) = '0.26598 Probability that almost one chip is defectin of ten chips are selected at random is 0.266 (b) Now, 600 chips are randomly selected. - 0 = 600. . we can approximate Binomial distribution with a normal distribution to If XN Bloip) then x N N (np, npg). where np = mean = 1 = 600 x 0.2 = 120. opg - variance = 6 = 600X0 2 X0.8 = 96. mean of this Normal dist is 120 and standard deviation is 9.7979

LDATE 105 chips are (c) defective To find on prob. P(x that almost < 105). ie P(x (105) = P(x-u < 105 - 1201 to g. 7979 - 20 =plzc -1.530g) = (-1.5309) = 0. 06289 V atmost Probability that 105 chips are defective is 0.063 (d) chips are to find prob defective that ie. more than P( X 7135) 135 6 : P(x >135) = 1- P( XL135) L- C P/x-4 o 135 - 120 g. 7979 = 1- P(Z < 11830g) 1- (1.5309) = 1- bag3710 0.06289 Probability defective that more than 135 chips are is 0.063 X 3 UT rematou O