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Chapter 3: Problem 88

In proof testing of circuit boards, the probability that any particular diode will fail is .01. Suppose a circuit board contains 200 diodes. a. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? b. What is the (approximate) probability that at least four diodes will fail on a randomly selected board? 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.)

Short Answer

Expert verified
a) Expect 2 diodes to fail, with SD 鈮� 1.41. b) Probability 鈮�4 fail 鈮� 0.146. c) Likelihood 鈮�4 boards work 鈮� 1.

Step by step solution

01

Determine Expected Number of Fails

For part (a), we first calculate the expected number of diodes that would fail using the formula for expected value in a binomial distribution: \(E(X) = np\), where \(n\) is the number of trials (200 diodes) and \(p\) is the probability of failure for each diode (0.01). Thus, \(E(X) = 200 \times 0.01 = 2\). So, we expect 2 diodes to fail.
02

Calculating Standard Deviation

The standard deviation \(\sigma\) for a binomial distribution can be found using the formula \(\sigma = \sqrt{np(1-p)}\). Substituting \(n = 200\) and \(p = 0.01\), we have \(\sigma = \sqrt{200 \times 0.01 \times 0.99} = \sqrt{1.98} \approx 1.41\). So, the standard deviation is approximately 1.41.
03

Probability That At Least Four Diodes Fail

For part (b), we use the normal approximation to the binomial distribution since \(n\) is large. The z-score for 3.5 (continuity correction) is calculated as \(z = \frac{3.5 - 2}{1.41} \approx 1.06\). Using the standard normal table, \(P(Z < 1.06) \approx 0.854\). Therefore, \(P(X \geq 4) = 1 - P(X < 4) \approx 1 - 0.854 = 0.146\).
04

Probability At Least 4 Boards Work Properly

In part (c), keep in mind a proper board means all 200 diodes work. The probability of all diodes working is \(0.99^{200}\). For at least 4 boards (out of 5) working, we calculate \(1 - P(\text{at most 1 board fails})\). Using binomial probabilities, calculate \(P(0) = (0.99^{200})^5\) and \(P(1) = \binom{5}{1} (0.99^{200})^4 (0.01)\), and subtract from 1 for at least 4 proper \(P(\geq 4) \approx 1 - (P(0) + P(1))\), which is very close to 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The concept of expected value is essential in probability and statistics. When dealing with a binomial distribution, such as the failure rate of diodes on a circuit board, the expected value helps us predict the average outcome. For instance, if a circuit board has 200 diodes and each diode has a 1% probability of failing, the expected number of failed diodes can be calculated using the formula:
  • \(E(X) = np\)
Where:
  • \(n\) is the number of trials (200 diodes)
  • \(p\) is the probability of failure for each diode (0.01)
Replacing these values, we find:
  • \(E(X) = 200 \times 0.01 = 2\)
Thus, we expect 2 diodes to fail on average. Calculating the expected value gives us a clear idea of what to anticipate in repeated trials.
Standard Deviation
Standard deviation in a binomial distribution measures the spread or dispersion of the data around the expected value. It tells us how much variation or "uncertainty" exists. Using the standard deviation formula for a binomial distribution, we can determine this spread:
  • \(\sigma = \sqrt{np(1-p)}\)
Substituting the values from our diode exercise:
  • \(n = 200\)
  • \(p = 0.01\)
  • \(\sigma = \sqrt{200 \times 0.01 \times 0.99} = \sqrt{1.98} \approx 1.41\)
A standard deviation of approximately 1.41 indicates how much variation we can expect from the expected number of failed diodes. This knowledge is crucial in many applications where reliability and consistency are important.
Normal Approximation
Normal approximation comes in handy when we deal with large sample sizes. It allows us to approximate the binomial distribution with a normal distribution. In our exercise, where we want to calculate the probability of at least 4 diode failures, this approximation simplifies calculations greatly. Performing a continuity correction, we use:
  • Convert \(X \geq 4\) to \(X > 3.5\) for better precision
We calculate the z-score as follows:
  • \(z = \frac{3.5 - 2}{1.41}\)
  • \(z \approx 1.06\)
Using a standard normal distribution table, this z-score tells us how likely it is to have more than the expected number of failures. In our case, this probability is approximately 0.146, indicating a fairly unusual situation compared to what's expected.
Probability Calculation
Probability calculation is central to understanding the likelihood of different outcomes. In part (c) of the exercise, we need to determine the probability that at least 4 out of 5 shipped boards work properly. This requires us to consider not just individual diode probabilities but also combinations of board outcomes.For a board to work properly:
  • All 200 diodes must function, which has a probability of \(0.99^{200}\)
We calculate the probability that "at most" 1 board fails:
  • \(P(0) = (0.99^{200})^5\)
  • \(P(1) = \binom{5}{1} \times (0.99^{200})^4 \times 0.01\)
Subtracting these from 1 gives us the probability that at least 4 boards are successful, which is very close to 1. Thus, we are very confident that nearly all boards will work properly, demonstrating the high reliability of these components.

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Most popular questions from this chapter

A consumer organization that evaluates new automobiles customarily reports the number of major defects in each car examined. Let \(X\) denote the number of major defects in a randomly selected car of a certain type. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<0 \\ .06 & 0 \leq x<1 \\ .19 & 1 \leq x<2 \\ .39 & 2 \leq x<3 \\ .67 & 3 \leq x<4 \\ 92 & 4 \leq x<5 \\ .97 & 5 \leq x<6 \\ 1 & 6 \leq x\end{cases} $$ Calculate the following probabilities directly from the cdf: a. \(p(2)\), that is, \(P(X=2)\) b. \(P(X>3)\) c. \(P(2 \leq X \leq 5)\) d. \(P(2

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Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\mu=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

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