Question

In proof testing of circuit boards, the probability that any particular diode will fail is . Suppose a circuit board contains diodes. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? What is the (approximate) probability that at least four diodes will fail on a randomly selected board? 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.)

Answer 1

Complete 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. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? What is the (approximate) probability that at least four diodes will fail on a randomly selected board?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)

Answer:

  The number of diode expected to fail is

             $\mu =2$

 The standard deviation is

          $\sigma = 1.407$

  The probability that at least four diodes will fail on a randomly selected board is  

$P(X \ge 4) = 0.1419$

Step-by-step explanation:

Generally the testing follows a binomial distribution because

1 The number of trials is finite

2 There is only two outcomes( which (a) A diode fail (b) a diode does not fail)

3 The each individual trial is independent of each other.

4 The chances of success is constant for every trial

From the question we are told that

The probability that a diode will fail is $p = 0.01$

The sample size is n = 200

Generally the number of diode expected to fail is mathematically represented as

$\mu = n * p$

=> $\mu = 200 * 0.01$

=> $\mu =2$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{ np * (1 - p)}$

=> $\sigma = \sqrt{ 200* 0.01 * (1 - 0.01 )}$

=> $\sigma = 1.407$

Generally the probability that at least 4 diode will fail is mathematically represented as

$P(X \ge 4) = 1 - P(X < 4 )$

Generally

$P(X < 4 ) = P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3)$

=> $P(X < 4 ) = ^nC_0 p^0 * (1-p)^{n-0} +^nC_1 p^1 * (1-p)^{n-1}+^nC_2 p^2 * (1-p)^{n-2}+ ^nC_3 p^3 * (1-p)^{n-3}$

=> $P(X < 4 ) = ^{200}C_0 p^0 * (1-0.01)^{200-0} +^{200}C_1 (0.01)^1 * (1-0.01)^{200-1}+^{200}C_2 (0.01)^2 * (1-0.01)^{200-2}+ ^{200}C_3 (0.01)^3 * (1-0.01)^{200-3}$

=> $P(X < 4 ) = 0.134 + 0.2707 + 0.2727 + 0.1814$

=> $P(X$

So

      $P(X \ge 4) = 1 -0.8581)$

=> $P(X \ge 4) = 0.1419$