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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is \(5 \%\). Let \(X=\) the number of defective boards in a random sample of size \(n=25\), so \(X \sim \operatorname{Bin}(25, .05)\). a. Determine \(P(X \leq 2)\). b. Determine \(P(X \geq 5)\). c. Determine \(P(1 \leq X \leq 4)\). d. What is the probability that none of the 25 boards is defective? e. Calculate the expected value and standard deviation of \(X\).

Short Answer

Expert verified
a. \(P(X \leq 2)\) is the sum of probabilities for \(X=0,1,2\). b. \(P(X \geq 5) = 1 - P(X \leq 4)\). c. \(P(1 \leq X \leq 4)\) is the sum for \(X = 1,2,3,4\). d. \(P(X=0)\) is calculated directly. e. Expected value is \(1.25\); standard deviation is \(1.092\).

Step by step solution

01

Understanding the Problem

We are given a binomial distribution problem where the number of defective boards follows a binomial distribution. This means we can apply the binomial probability formula to solve the various parts of the problem. Recall that the binomial distribution is defined as \(X \sim \text{Binomial}(n, p)\), where \(n\) is the number of trials, and \(p\) is the probability of success in each trial.
02

Determine \(P(X \leq 2)\)

Use the binomial probability formula: \(P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\). We need to calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\) and sum these probabilities:1. \(P(X = 0) = \binom{25}{0} (0.05)^0 (0.95)^{25}\).2. \(P(X = 1) = \binom{25}{1} (0.05)^1 (0.95)^{24}\).3. \(P(X = 2) = \binom{25}{2} (0.05)^2 (0.95)^{23}\).Adding these gives \(P(X \leq 2)\).
03

Determine \(P(X \geq 5)\)

Instead of calculating \(P(X \geq 5)\) directly, calculate its complement \(P(X \leq 4)\). This is because \(P(X \geq 5) = 1 - P(X \leq 4)\). Use the binomial sum for \(k = 0\) to \(k = 4\) and subtract from 1.
04

Determine \(P(1 \leq X \leq 4)\)

Calculate the probabilities for \(P(X = 1)\), \(P(X = 2)\), \(P(X = 3)\), and \(P(X = 4)\) using the binomial probability formula and sum them up. \(P(1 \leq X \leq 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)\).
05

Probability that None is Defective

This is \(P(X = 0)\). Use the binomial probability formula for \(X = 0\):\(P(X = 0) = \binom{25}{0} (0.05)^0 (0.95)^{25}\).
06

Calculate Expected Value and Standard Deviation

For a binomial distribution \(X \sim \text{Binomial}(n, p)\), the expected value \(E(X) = np\) and the standard deviation \(\sigma = \sqrt{np(1-p)}\). So:1. Expected value \(E(X) = 25 \times 0.05\).2. Standard deviation \(\sigma = \sqrt{25 \times 0.05 \times 0.95}\).

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Key Concepts

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

Probability Calculation
In this exercise, calculating the probability of different outcomes for defective boards involves understanding and computing various probabilities. The primary tool used here is the binomial probability formula, which helps us calculate the likelihood of obtaining exactly a certain number of defects, as each trial (testing a board) results in either a defective or non-defective board. We want to find the probability that at most 2 boards, at least 5 boards, and between 1 to 4 boards are defective. Also, calculate the probability of having no defective boards.
  • For probabilities like \(P(X \leq 2)\), we find the sum of probabilities for 0, 1, and 2 using the formula: \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\).
  • When calculating \(P(X \geq 5)\), we find the complement \(P(X \leq 4)\) and subtract it from 1.
  • To obtain probabilities like \(P(1 \leq X \leq 4)\), calculate each probability from 1 to 4 and add them up, which illustrates how likely it is to see these outcomes.
This method allows for precise determination of probabilities in a binomial distribution setup, highlighting the calculation of possibilities using the defined success (defective board) rate.
Expected Value
The expected value in a binomial distribution gives us a snapshot of the average or mean number of successes across many trials. It is a vital measure that helps understand what to anticipate when repeating an experiment many times.
To compute the expected value for our scenario, use the formula:
\[E(X) = np\]
Here, \(n\) is the total number of trials (25 boards), and \(p\) is the probability of one trial being a success (5% defect rate).
  • This results in an expected number of defective boards as \( E(X) = 25 \times 0.05 \), yielding 1.25 defective boards.
The expected value provides a clear indication of the number of defects you might "expect" on average every time you test 25 boards, though actual outcomes can vary around this average due to random variation.
Standard Deviation
Standard deviation is a critical concept in statistics that quantifies the amount of variation or dispersion in a set of values in a distribution.
For the binomial distribution in this context, the formula for standard deviation is:
\[\sigma = \sqrt{np(1-p)}\]
This measure helps you understand how much actual results can deviate from the expected value due to variability in the data.
  • Plugging in our values (\(n = 25\), \(p = 0.05\)), we calculate \( \sigma = \sqrt{25 \times 0.05 \times 0.95} \).
  • This yields a standard deviation of approximately 1.089.
A larger standard deviation indicates more variation from the expected number of defects. In our case, this helps in assessing risk and variability, assisting in making informed decisions based on the data's spread.
Binomial Probability Formula
The binomial probability formula is the backbone of calculating probabilities in scenarios where you have a fixed number of independent experiments, each with the same probability of success. It is given by:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Here's how it breaks down:
  • \(n\) is the number of trials.
  • \(k\) is the number of successful trials (defective boards, in our case).
  • \(p\) is the probability of success in a single trial.
  • \((1-p)\) is the probability of failure.
  • \(\binom{n}{k}\) is a combination function calculating the different ways \(k\) successes can occur in \(n\) trials.
This formula helps in evaluating probabilities for any specific number of successes in the context of binomial distribution, such as testing circuit boards in our example. Mastering this formula allows for versatile application across various real-world scenarios where similar conditions of fixed trials and constant probabilities apply.

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Most popular questions from this chapter

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