91Ó°ÊÓ

Samples of 20 parts from a metal punching process are selected every hour. Typically, \(1 \%\) of the parts require rework. Let \(X\) denote the number of parts in the sample of 20 that require rework. A process problem is suspected if \(X\) exceeds its mean by more than 3 standard deviations. a. If the percentage of parts that require rework remains at \(1 \%,\) what is the probability that \(X\) exceeds its mean by more than 3 standard deviations? b. If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds \(1 ?\) c. If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds 1 in at least one of the next five hours of samples?

Step by step solution

01

Define the Problem

Given that parts have typically a 1% rework rate, the number of reworks in a sample of 20 follows a binomial distribution with parameter \( n = 20 \) and \( p = 0.01 \). We want to find the probability that the number of reworks \( X \) exceeds its mean by more than 3 standard deviations.

02

Calculate Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( np \), and the standard deviation \( \sigma \) is given by \( \sqrt{np(1-p)} \).\[\mu = 20 \times 0.01 = 0.2 \]\[\sigma = \sqrt{20 \times 0.01 \times 0.99} \approx 0.445\]

03

Determine the Critical Value

The mean plus 3 standard deviations is given by \( \mu + 3\sigma \).\[0.2 + 3 \times 0.445 \approx 1.535\]Since the number of reworks \( X \) must be an integer, it must exceed 1.

04

Calculate the Probability for Part A

Since \( X \) needs to exceed 1, calculate \( P(X > 1) \).\[ P(X > 1) = 1 - P(X \leq 1) = 1 - (P(X = 0) + P(X = 1)).\]Using the binomial formula, calculate for \( X = 0 \) and \( X = 1 \).\[ P(X = 0) = \binom{20}{0} (0.01)^0 (0.99)^{20} \approx 0.817\]\[ P(X = 1) = \binom{20}{1} (0.01)^1 (0.99)^{19} \approx 0.165\]\[ P(X \leq 1) \approx 0.817 + 0.165 = 0.982\]\[ P(X > 1) \approx 1 - 0.982 = 0.018\]

05

Interpret Part A Result

The probability that \( X \) exceeds its mean by more than 3 standard deviations when rework rate is 1% is approximately 0.018.

06

Rework Percentage Increases to 4% for Part B

Now the rework percentage is 4%, so \( p = 0.04 \). We want the probability \( P(X > 1) \).Using \( n = 20 \) and \( p = 0.04 \), calculate using binomial distribution.\[ P(X \leq 1) = P(X = 0) + P(X = 1)\]\[ P(X = 0) = \binom{20}{0} (0.04)^0 (0.96)^{20} \approx 0.442\]\[ P(X = 1) = \binom{20}{1} (0.04)^1 (0.96)^{19} \approx 0.369\]\[ P(X > 1) = 1 - (0.442 + 0.369) = 0.189\]

07

Interpret Part B Result

When the rework percentage increases to 4%, the probability that more than 1 part requires rework is approximately 0.189.

08

Probability Exceeds 1 in at Least One of Five Samples for Part C

We are interested in the event that at least one out of five samples has more than one rework when the failure rate is 4%.The probability that a single sample has \( X \leq 1 \) is calculated in Step 6.\[ P(X \leq 1) = 0.811\]The probability that none of the five samples exceed 1 rework is:\[P(X \leq 1 \text{ in all 5 samples}) = (0.811)^5 \approx 0.349\]Thus, the probability of having more than 1 rework in at least one of the five samples is:\[P(\text{At least one sample exceeds 1}) = 1 - 0.349 = 0.651\]

09

Interpret Part C Result

The probability that at least one of the five samples in the next five hours has more than 1 part requiring rework is approximately 0.651.

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Key Concepts

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

Probability Calculation

Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. When it comes to binomial distribution, the probability calculation involves specific parameters. In our example, these parameters include:

• Number of trials ( $n = 20$ parts in each sample)
• Probability of success ( $p = 0.01$ for 1% rework)

To find the probability that the number of parts requiring rework exceeds a given threshold, such as exceeding the mean by more than 3 standard deviations, we use the binomial probability formula. Using this, we calculated the probability of events like less than 1 part needing rework, helping to determine if more than 1 rework in a sample of 20 is unusual.

Standard Deviation

Standard deviation is a measure of how spread out the values in a distribution are. It tells us how much variation exists from the average (mean). In a binomial distribution like ours:

• The formula for standard deviation $$$sigma = \sqrt{np(1-p)}\(
• With \)n = 20\( and \)p = 0.01\(, we found \)sigma \approx 0.445$

This value helps us assess the variability of the rework rate. When we calculate mean plus 3 standard deviations, we find a critical threshold to decide if an unusually high number of parts need rework in a sample, indicating a potential process problem.

Mean of Distribution

The mean of a distribution is the average value expected over a number of trials. In binomial distributions, the mean (\(mu\)) is calculated as:

• \(mu = np\)

In our scenario:

• For a rework rate of 1%, \(mu = 20 \times 0.01 = 0.2\)
• This means that, on average, 0.2 parts in a sample of 20 are expected to require rework.

Understanding the mean helps us make predictions about our distribution. It is crucial in setting benchmarks for assessing whether deviation from the mean might indicate an issue with the process stability.