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Chapter 3: Problem 48

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) 0.8918, b) 0.0121, c) 0.7105, d) 0.2775, e) Expected value: 1.25, Standard deviation: 1.091.

Step by step solution

01

Understanding the Binomial Distribution

The problem gives a binomial distribution with parameters \( n = 25 \) and \( p = 0.05 \). Here, \( X \) represents the number of defective boards in a sample of 25. We need to calculate probabilities for \( P(X \leq 2) \), \( P(X \geq 5) \), \( P(1 \leq X \leq 4) \), and the probability of no defects.
02

Calculating P(X 鈮� 2)

To find \( P(X \leq 2) \), we sum the probabilities \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For \( X \sim \text{Bin}(25, 0.05) \):- \( P(X = 0) = \binom{25}{0} (0.05)^0 (0.95)^{25} = 0.2775 \)- \( P(X = 1) = \binom{25}{1} (0.05)^1 (0.95)^{24} = 0.3656 \)- \( P(X = 2) = \binom{25}{2} (0.05)^2 (0.95)^{23} = 0.2487 \)Thus, \( P(X \leq 2) = 0.2775 + 0.3656 + 0.2487 = 0.8918 \).
03

Calculating P(X 鈮� 5)

To find \( P(X \geq 5) \), we use the complement rule: \[ P(X \geq 5) = 1 - P(X < 5) \]We calculate \( P(X < 5) = P(X \leq 4) \) as the sum of probabilities \( P(X = 0) \), \( P(X = 1) \), \( P(X = 2) \), \( P(X = 3) \), and \( P(X = 4) \). We have already calculated \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \):- \( P(X = 3) = \binom{25}{3} (0.05)^3 (0.95)^{22} = 0.0785 \)- \( P(X = 4) = \binom{25}{4} (0.05)^4 (0.95)^{21} = 0.0177 \)Summing these, \( P(X < 5) = 0.8918 + 0.0785 + 0.0177 = 0.9879 \).Thus, \( P(X \geq 5) = 1 - 0.9879 = 0.0121 \).
04

Calculating P(1 鈮� X 鈮� 4)

We have already calculated \( P(X = 1) \), \( P(X = 2) \), \( P(X = 3) \), and \( P(X = 4) \). Therefore, \[ P(1 \leq X \leq 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \]\[ P(1 \leq X \leq 4) = 0.3656 + 0.2487 + 0.0785 + 0.0177 = 0.7105 \].
05

Calculating P(X=0)

The probability that none of the 25 boards is defective is \( P(X = 0) \). This was calculated as part of \( P(X \leq 2) \):\[ P(X = 0) = 0.2775 \].
06

Calculating Expected Value and Standard Deviation

The expected value \( E(X) \) for a binomial distribution is given by \( n imes p = 25 imes 0.05 = 1.25 \).The standard deviation \( \sigma \) is calculated as:\[ \sigma = \sqrt{n imes p imes (1-p)} = \sqrt{25 imes 0.05 imes 0.95} = 1.091 \].

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Key Concepts

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

Expected Value
The expected value, often denoted as \( E(X) \), is a fundamental concept in statistics that represents the average number of successes you would expect in a given number of trials. For a binomial distribution, it's calculated using the formula: \( E(X) = n \times p \). Here, \( n \) is the number of trials (boards tested), and \( p \) is the probability of a single success (a board being defective). In our scenario, with 25 boards and a defective rate of 5% (or 0.05), the expected value is:
  • \( E(X) = 25 \times 0.05 = 1.25 \)
This means that, on average, we expect 1.25 boards out of 25 to be defective.
Standard Deviation
Standard deviation provides a measure of the amount of variation or dispersion in a set of values. It tells us how much the number of defective boards is likely to deviate from the expected value. For a binomial distribution, the standard deviation \( \sigma \) is calculated using:
  • \( \sigma = \sqrt{n \times p \times (1-p)} \)
With \( n=25 \) and \( p=0.05 \), the calculation is:
  • \( \sigma = \sqrt{25 \times 0.05 \times 0.95} \)
  • \( \sigma = \sqrt{1.1875} \approx 1.091 \)
This indicates that the number of defective boards will typically lie within about 1.091 of the expected value, 1.25.
Probability Calculation
Calculating probabilities for certain outcomes in a binomial distribution involves using the binomial probability formula:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
where \( n \) is the number of trials, \( k \) is the number of successful outcomes you are interested in, \( p \) is the probability of success on an individual trial, and \( \binom{n}{k} \) is the binomial coefficient. This formula can help you compute various probabilities such as:
  • The probability of having exactly \( k \) defectives
  • The cumulative probability of having up to \( k \) defectives, or more than \( k \) defectives using complementary probability
This approach is used in the exercise to find probabilities like \( P(X \leq 2) \), \( P(X \geq 5) \), and \( P(1 \leq X \leq 4) \). By understanding how to plug numbers into this formula, you can tackle a wide array of probability questions in binomial distributions.
Defective Rate
The defective rate is the probability that a single item will be defective in any given trial. In our scenario, it is set at 5% or 0.05. This rate is a crucial part of calculating both expected values and probabilities in a binomial distribution. It provides insight into:
  • How often defects occur in production, helping inform quality control measures.
  • The reliability of the production process, as a lower defective rate suggests better quality.
Additionally, the defective rate is used along with the size of the sample (\( n \)) to compute both the expected value \( E(X) = n \times p \) and standard deviation \( \sigma = \sqrt{n \times p \times (1-p)} \), which inform us about likely outcomes and their variability.

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Most popular questions from this chapter

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