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Chapter 4: Problem 63

Suppose that \(10 \%\) of all steel shafts produced by a process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let \(X\) denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that \(X\) is a. At most 30 ? b. Less than 30 ? c. Between 15 and 25 (inclusive)?

Short Answer

Expert verified
a. 0.9934, b. 0.9875, c. 0.8064

Step by step solution

01

Define the Distribution

We start by determining the type of distribution that applies to the problem. Here, \(X\), the number of nonconforming shafts, follows a binomial distribution. This is because each shaft can be nonconforming with a probability of \(p = 0.1\) and there are a fixed number of independent trials, \(n = 200\). Hence, \(X \sim \text{Binomial}(n=200, p=0.1)\).
02

Apply the Normal Approximation

For large \(n\), the binomial distribution can be approximated by a normal distribution. The mean and variance of \(X\) are \(\mu = np = 200 \times 0.1 = 20\) and \(\sigma^2 = np(1-p) = 200 \times 0.1 \times 0.9 = 18\). Thus, \(X\) can be represented approximately as \(N(20, 18)\).
03

Calculate Probability for Part (a)

For part (a), we need to find \(P(X \leq 30)\). Using the normal approximation, calculate the z-score: \[ z = \frac{30 + 0.5 - \mu}{\sigma} = \frac{30.5 - 20}{\sqrt{18}} \approx 2.48 \] We then find \(P(Z \leq 2.48)\) using the standard normal distribution table, which gives approximately 0.9934.
04

Calculate Probability for Part (b)

For part (b), calculate \(P(X < 30)\). This is \(P(X \leq 29)\) due to continuity correction: \[ z = \frac{29 + 0.5 - 20}{\sqrt{18}} = \frac{29.5 - 20}{\sqrt{18}} \approx 2.24 \] \(P(Z \leq 2.24)\) is approximately 0.9875 from the z-table.
05

Calculate Probability for Part (c)

For part (c), calculate \(P(15 \leq X \leq 25)\). Use the continuity correction to find \(P(14.5 < X < 25.5)\).Calculate z-scores:- For 14.5: \[ z = \frac{14.5 - 20}{\sqrt{18}} \approx -1.30 \]- For 25.5: \[ z = \frac{25.5 - 20}{\sqrt{18}} \approx 1.30 \]Using the z-table,\(P(Z \leq 1.30) \approx 0.9032\) and \(P(Z \leq -1.30) \approx 0.0968\).Hence, the probability is \(0.9032 - 0.0968 = 0.8064\).

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Key Concepts

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

Normal Approximation
When dealing with a binomial distribution, the normal approximation can be a handy tool. This approximation is most useful when the number of trials, denoted by \( n \), is large, and the probability of success, denoted by \( p \), is neither too close to 0 nor too close to 1. In our exercise, we're working with 200 steel shafts, and 10% of these are nonconforming. Given this is a large number, a normal approximation simplifies calculations.

A binomial distribution can be approximated by a normal distribution when \( np > 5 \) and \( n(1-p) > 5 \). In this example, \( np = 20 \) and \( n(1-p) = 180 \), which easily meet these criteria. The mean (\( \mu \)) and variance (\( \sigma^2 \)) of the binomial distribution are \( np \) and \( np(1-p) \) respectively. Thus, the normal distribution's parameters in this scenario are calculated as a mean of 20 and standard deviation calculated as the square root of 18. Now, instead of complex binomial probability calculations, we can use normal distribution tables, which are usually easier to interpret.
Probability Calculation
Probability calculation using normal approximation involves converting binomial data to a standard normal distribution problem. This involves z-scores, which are standardized scores that relate binomial counts to the standard normal distribution.

To find the probability that a variable \( X \) falls within a certain range, you will often compute a z-score using the formula: \[ z = \frac{x - \mu}{\sigma} \] Where:\( x \) is the value of \( X \) you're interested in, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.

In practice, probabilities are determined by checking z-score values against a standard normal distribution table or using technology to find the area under the curve for those z-scores. For instance, to find the probability of \( X \leq 30 \), we determine the z-score and then find the corresponding probability in the z-table. This approach converts an otherwise intricate set of binomial calculations into more manageable normal distribution queries.
Continuity Correction
A crucial factor when using normal approximation for discrete distributions like the binomial is the continuity correction. This adjustment offers a more accurate approximation. When converting from a discrete to a continuous distribution, small adjustments (usually 卤0.5) are added to the value of \( X \) before calculating a z-score.

This means, for the probability \( P(X \leq 30) \), instead, calculate \( P(X \leq 30.5) \). By adding 0.5, you're correcting for the fact that a continuous distribution assumes data between whole numbers. It's like smoothing out the steps between the intervals of a staircase, which better approximates the cumulative nature of a standard normal distribution.

  • For "at most" scenarios like \( P(X \leq 30) \), use \( 30.5 \) instead.

  • For "less than" cases, such as \( P(X < 30) \), use \( 29.5 \) to represent just beneath the threshold.

  • When considering "between" scenarios like \( P(15 \leq X \leq 25) \), apply continuity correction on both bounds: \( 14.5 \) and \( 25.5 \).


This correction helps ensure the approximation closely reflects the true probability, leading to more reliable solutions.

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Most popular questions from this chapter

Let \(X\) have a binomial distribution with parameters \(n=25\) and \(p\). Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases \(p=.5, .6\), and \(.8\) and compare to the exact probabilities calculated from Appendix Table A.1. a. \(P(15 \leq X \leq 20)\) b. \(P(X \leq 15)\) c. \(P(20 \leq X)\)

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