In proof testing of circuit boards, the probability that any particular diode wi
ID: 3307125 • Letter: I
Question
In proof testing of circuit boards, the probability that any particular diode will fail is 0.01. Suppose a circuit board contains 200 diodes (a) How many diodes would you expect to fail? diodes What is the standard deviation of the number that are expected to fail? (Round your answer to three decimal places) diodes (b) What is the (approximate) probability that at least three diodes will fail on a randomly selected board? (Round your answer to three decimal places.) (c) If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work. Round your answer to four decimal places) You may need to use the appropriate table in the Appendix of Tables to answer this question.Explanation / Answer
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
a.
mean ( np ) = 200 * 0.01 = 2
b.
standard deviation ( npq )= 200*0.01*0.99 = 1.4071
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
c.
P(X < 3) = (3-2)/1.4071
= 198/1.4071= 0.7107
= P ( Z <0.7107) From Standard NOrmal Table
= 0.7614
P(X >= 3) = (1 - P(X < 3))
= 1 - 0.76136 = 0.23864
d.
P(X < 4) = (4-2)/1.4071
= 198/1.4071= 1.4214
= P ( Z <1.4214) From Standard NOrmal Table
= 0.9224
P(X >= 4) = (1 - P(X < 4))
= 1 - 0.92239 = 0.07761