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Chapter 3: Problem 84

Suppose that only , \(10 \%\) of all computers of a certain type experience CPU failure during the warranty period. Consider a sample of 10,000 computers. a. What are the expected value and standard deviation of the number of computers in the sample that have the defect? b. What is the (approximate) probability that more than 10 sampled computers have the defect? c. What is the (approximate) probability that no sampled computers have the defect?

Short Answer

Expert verified
Expected value is 1,000; standard deviation is 30; probability of more than 10 defects is 0.3632; probability of zero defects is approximately 0.

Step by step solution

01

Identify the Distribution

The problem involves the number of computers experiencing CPU failure, which can be modeled using a binomial distribution since there are a fixed number of trials (n = 10,000) and a fixed probability of success (p = 0.10) for each trial.
02

Calculate Expected Value

For a binomial distribution, the expected value \( E(X) \) is given by the formula \( np \). Here, \( n = 10,000 \) and \( p = 0.10 \), so \( E(X) = 10,000 \times 0.10 = 1,000 \).
03

Calculate Standard Deviation

The standard deviation \( \sigma \) for a binomial distribution is calculated using \( \sqrt{np(1-p)} \). Substituting the values, \( \sigma = \sqrt{10,000 \times 0.10 \times 0.90} = \sqrt{900} = 30 \).
04

Approximate Probability Using Normal Distribution

Since \( n \) is large, the binomial distribution can be approximated by a normal distribution. We use \( N(1,000, 30) \) to approximate. We are looking for \( P(X > 1,010) \), using continuity correction, it's \( P(X > 1,010.5) \).
05

Calculate Probability Using Z-Score

Calculate the Z-score for \( X = 1,010.5 \) using the formula \( Z = \frac{X - \mu}{\sigma} \). Here \( \mu = 1,000 \) and \( \sigma = 30 \), so \( Z = \frac{1010.5 - 1000}{30} \approx 0.35 \). Use a standard normal distribution table to find \( P(Z > 0.35) \approx 0.3632 \).
06

Calculate Probability for Zero Defects

For part (c), calculate \( P(X = 0) \) directly using the binomial probability formula, \( \binom{n}{k} p^k (1-p)^{n-k} \). \( P(X = 0) = \binom{10,000}{0} (0.10)^0 (0.90)^{10,000} = (0.90)^{10,000} \approx 0 \). As \( n \times q \to \infty \) (i.e., \( q \) close to 1 for large \( n \)), the probability is practically zero.

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Key Concepts

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

Expected Value
In statistics, when we talk about the expected value, we are referring to the average or mean outcome you would anticipate from a probability distribution if you could repeat the experiment a large number of times. For a binomial distribution, where you have a fixed number of trials and outcomes of either success or failure, the expected value helps in predicting how many successes you can expect in those trials.
In the given problem, we have 10,000 computers examined, with the probability of failure being 10%. The formula to find the expected value for a binomial distribution is \( E(X) = np \). This formula multiplies the number of trials \( n \) with the probability of success \( p \).
  • For our example, \( n = 10,000 \) and \( p = 0.10 \), so \( E(X) = 10,000 \times 0.10 = 1,000 \).
The expected value here is 1,000. This indicates that out of the 10,000 computers, we expect 1,000 will have experienced CPU failure during the warranty period if repeated over many samples.
Standard Deviation
The standard deviation is a measure of the amount of variation or spread in a set of values. For probability distributions, it tells us how much the values tend to deviate from the expected value. When discussing a binomial distribution, the standard deviation can help understand the variability of our expected results.
For a binomial distribution, we calculate the standard deviation \( \sigma \) using the formula \( \sqrt{np(1-p)} \). This formula incorporates:
  • \( n \), the number of trials,
  • \( p \), the probability of success,
  • and \( 1-p \) which represents the probability of failure.

Let's look at our case:
  • Using \( n = 10,000 \), \( p = 0.10 \), and \( 1-p = 0.90 \), the standard deviation becomes \( \sigma = \sqrt{10,000 \times 0.10 \times 0.90} = \sqrt{900} = 30 \).
This result of 30 tells us how much the number of computers with CPU failure is likely to differ from the expected number, which is 1,000.
Normal Approximation
When dealing with large sample sizes in a binomial distribution, we often use a normal approximation. This is because calculating probabilities directly from the binomial distribution can become complex. If \( n \) is large and neither \( p \) nor \( 1-p \) is extremely close to zero, the distribution of sample proportions can be approximated by a normal distribution.
In our case, the sample size is 10,000, which is sufficiently large to use this method. We approximate the binomial distribution with the normal distribution \( N(\mu, \sigma) \), where \( \mu = np \) and \( \sigma = \sqrt{np(1-p)} \).
With \( \mu = 1,000 \) and \( \sigma = 30 \), our normal distribution becomes \( N(1,000, 30) \). This is used to find probabilities, such as the chance of having more than a certain number of failures, using tools like Z-scores for calculations.
This approximation simplifies calculations and helps predict the likelihood of different outcomes without manually crunching through complex binomial probability formulas.

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Most popular questions from this chapter

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt., 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

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