/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 102 In proof testing of circuit boar... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. 2 diodes expected to fail, SD is 1.41. b. Probability is ~0.144. c. Probability is ~0.0094.

Step by step solution

01

Calculate the expected number of failed diodes

The expected number of failed diodes, also known as the mean of the distribution, can be found using the formula \( \mu = np \), where \( n \) is the total number of diodes and \( p \) is the probability of failure for a single diode.Substitute \( n = 200 \) and \( p = 0.01 \) into the formula:\[ \mu = 200 \times 0.01 = 2 \] So, we expect 2 diodes to fail.
02

Calculate the standard deviation of failed diodes

The standard deviation of a binomial distribution can be calculated using \( \sigma = \sqrt{np(1-p)} \).Using \( n = 200 \) and \( p = 0.01 \), we have:\[\sigma = \sqrt{200 \times 0.01 \times (1-0.01)} = \sqrt{200 \times 0.01 \times 0.99} = \sqrt{1.98} \approx 1.41\]The standard deviation is approximately 1.41.
03

Approximate probability at least four diodes fail

We can use the normal approximation to the binomial distribution since \( n \) is large and \( p \) is small. First, calculate the z-score for 3.5 diodes failing (continuity correction):\[ z = \frac{3.5 - \mu}{\sigma} = \frac{3.5 - 2}{1.41} \approx 1.06 \]Check standard normal distribution tables or use software to find the probability \( P(X > 3.5) = P(Z > 1.06) \). This gives approximately 0.144.So, the probability that at least four diodes fail is about 0.144.
04

Calculate probability a board works

For a board with 200 diodes to work properly, all diodes must work, i.e., 0 diodes fail. The probability that a single diode works is \(1 - 0.01 = 0.99\).Thus, the probability that all 200 diodes work is:\[ 0.99^{200} \approx 0.135 \] So, a single board has approximately 0.135 probability of working properly.
05

Calculate probability at least four boards work properly

Define the random variable \( Y \) as the number of boards that work properly out of 5. \( Y \) follows a binomial distribution with \( n = 5 \) and \( p = 0.135 \).We want \( P(Y \geq 4) \), which is the sum of probabilities that 4 or 5 boards work properly:\[ P(Y \geq 4) = P(Y = 4) + P(Y = 5) \]Calculate each using \( P(Y = k) = \binom{5}{k} p^k (1-p)^{5-k} \):\[P(Y = 4) = \binom{5}{4} (0.135)^4 (0.865)^1 \approx 0.009\]\[P(Y = 5) = \binom{5}{5} (0.135)^5 (0.865)^0 \approx 0.0004\]Thus, \( P(Y \geq 4) \approx 0.009 + 0.0004 = 0.0094 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability
Probability is a fundamental concept in statistics that measures the likelihood of a certain event occurring. In the context of our circuit board problem, we deal with a specific type of probability called binomial probability. It is applicable when there are only two possible outcomes, such as success or failure, for each trial.

For example, the probability that a single diode will fail is given as 0.01, or 1%. This is a simple scenario where the diode can either fail (which happens with 1% probability) or work (which happens with 99% probability).
  • **Independent Trials**: Each diode’s operation is independent. The success of one does not affect another.
  • **Fixed Number of Trials**: With 200 diodes, the probability calculations consider all diodes together.
These aspects make the situation a perfect match for binomial probability, allowing for calculations using formulas specific to the binomial distribution.
Standard Deviation
Standard deviation is a measure of how spread out the outcomes are in a probability distribution. In the binomial context, it tells us how much variation to expect from the mean number of events (like failing diodes in circuit boards).

To find the standard deviation of diodes failing, you can use the formula:\[\sigma = \sqrt{np(1-p)}\]where \( n \) is the total number of diodes (200), and \( p \) is the probability of failure (0.01). This results in a standard deviation of approximately 1.41.
  • **Interpretation**: A small standard deviation means most diode failures will be close to the mean of 2.
By understanding this concept, you gain insight into how predictable the number of failing diodes is on any given board.
Normal Approximation
Normal approximation is a technique used to make binomial probability calculations easier. When you have a large number of trials, like 200 diodes, and the probability of success is small, the binomial distribution resembles a normal distribution.

We apply this method by calculating the z-score which represents how far a certain raw score (like the number of failing diodes) lies from the mean, adjusted for standard deviation:\[ z = \frac{x - \mu}{\sigma}\]where \( x \) is your value of interest (such as 3.5), \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • **Why 3.5?**: When calculating probabilities like "at least 4 failures," a continuity correction (next topic) slightly shifts this to accommodate a discrete-to-continuous distribution transition.
This approximation simplifies complex binomial probability calculations, making them more manageable using normal distribution tables or software.
Continuity Correction
Continuity correction is an adjustment technique when applying normal approximation to a discrete distribution like the binomial distribution. Since the binomial outcomes (number of diode failures) are discrete, while the normal distribution is continuous, a small correction helps align the two.

For example, if you want the probability of "at least four diodes failing," you would technically look for \( P(X \geq 4) \). Using a continuity correction, you adjust this to \( P(X > 3.5) \), because it smooths the binomial step-function to better fit the normal curve.
  • **Adjustment Explained**: This correction involves decimals to bridge the gap between whole numbers (diodes) and the continuous nature of the normal distribution, improving accuracy.
By implementing continuity correction, the approximation aligns more closely with the actual probabilities, especially when binomial and normal distributions overlap.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that the number of plants of a particular type found in a rectangular region (called a quadrat by ecologists) in a certain geographic area is an rv \(X\) with pmf $$ p(x)=\left\\{\begin{array}{cc} c / x^{3} & x=1,2,3, \ldots \\ 0 & \text { otherwise } \end{array}\right. $$ Is \(E(X)\) finite? Justify your answer (this is another distribution that statisticians would call heavy-tailed).

Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices, with \(P(A)=.5, P(B)=.2\), and \(P(C)=.3 .\) a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

Starting at a fixed time, each car entering an intersection is observed to see whether it turns left \((L)\), right \((R)\), or goes straight ahead \((A)\). The experiment terminates as soon as a car is observed to turn left. Let \(X=\) the number of cars observed. What are possible \(X\) values? List five outcomes and their associated \(X\) values.

Grasshoppers are distributed at random in a large field according to a Poisson distribution with parameter \(\alpha=2\) per square yard. How large should the radius \(R\) of a circular sampling region be taken so that the probability of finding at least one in the region equals \(.99\) ?

If \(X\) is a hypergeometric rv, show directly from the definition that \(E(X)=n M / N\) (consider only the case \(n

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.