Wires manufactured for use in a certain computer system are specified to have resistances between 0.12 ohm and 0.14 ohm, the actual measured resistances of the wires produced by company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm.
(i) What is the probability that a randomly selected wire from company A's production will meet the specifications?
(ii) If 4 such wires are used in the system and are selected from company A, what is the probability that all 4 will meet specifications?
(i) $$P(0.12≤X≤0.14)$$
$$P(\frac{0.12-0.13}{0.005}≤Z≤\frac{0.14-0.13}{0.005})$$
$$P(-2≤Z≤2)$$
$$P(Z≤2) – P(Z≤-2)$$
$$0.9772 – (1-0.9972)$$
$$0.9544$$
(ii) I'm stuck on this part, do you simply put $0.9544^4 = 0.8297?$
Best Answer
Using binomial distribution to solve:
$n=4$, $r=4$, $p=0.954$
$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}] = [4!/4!] [0.954^4] [0.046^0] = 0.954^4 = 0.83$