Problem 28 - Applied Statistics and Probability for Engineers

Question

A manufacturer of electroluminescent lamps knows that the amount of luminescent ink deposited on one of its products is normally distributed with a mean of 1.2 grams and a standard deviation of 0.03 gram. Any lamp with less than 1.14 grams of luminescent ink fails to meet customers' specifications. A random sample of 25 lamps is collected and the mass of luminescent ink on each is measured. (a) What is the probability that at least one lamp fails to meet specifications? (b) What is the probability that five or fewer lamps fail to meet specifications? (c) What is the probability that all lamps conform to specifications? (d) Why is the joint probability distribution of the 25 lamps not needed to answer the previous questions?

Step-by-Step Solution

Answer

(a) ≈ 0.458, (b) ≈ 0.9998, (c) ≈ 0.542, (d) Independence assumption allows use of binomial distribution.

Step 1: Determine Individual Lamp Probability

First, find the probability that one lamp has less than 1.14 grams of luminescent ink, which is normally distributed with \( \mu = 1.2 \) and \( \sigma = 0.03 \). Use the cumulative distribution function (CDF) of the normal distribution to find \( P(X < 1.14) \). Compute the Z-score: \( Z = \frac{1.14 - 1.2}{0.03} = -2 \). Look up this Z-score in the standard normal distribution table to find the probability: \( P(Z < -2) \approx 0.0228 \).

Step 2: Calculate Probability At Least One Fails

Given the probability that one lamp fails is 0.0228, the probability that one lamp meets specifications is 0.9772. For 25 lamps, the probability that all lamps meet specifications is \(0.9772^{25}\). Therefore, the probability that at least one fails is \( 1 - 0.9772^{25} \approx 0.458 \).

Step 3: Probability Five or Fewer Fail to Meet Specs

We need to find the probability of \( X \leq 5 \) where \( X \) is binomially distributed with \( n = 25 \) and \( p = 0.0228 \). Use the binomial CDF to find \( P(X \leq 5) \). Use either statistical software or tables to compute this probability. The calculation yields approximately 0.9998.

Step 4: Probability All Lamps Conform

The probability that all lamps conform to specifications is the same as previously calculated in Step 2, which is \( P(X = 0) = 0.9772^{25} \approx 0.542 \).

Step 5: Justification of Independent Probabilities

The joint probability distribution of the 25 lamps is not needed because we are assuming each lamp's mass of luminescent ink is independent of the others. This means we can use the binomial distribution for our probability calculations, since the probability of failure is the same for each lamp and independent of others.

Key Concepts

Normal Distribution

The Normal Distribution is a key concept in statistics, and it's often visualized as a bell-shaped curve. This is because the distribution describes data that clusters around a mean or average value. For our specific problem, the manufacturer knows the amount of luminescent ink follows a normal distribution with mean (\( \mu \) = 1.2 grams) and standard deviation (\( \sigma \) = 0.03 grams). This tells us that most lamps will have ink amounts close to 1.2 grams, with fewer lamps showing much smaller or greater amounts.
Using the normal distribution, we calculated probabilities related to individual lamps and their ink content. For example, to find the probability of a lamp having less than 1.14 grams of ink, we standardized the problem to find a Z-score. The Z-score is a measure of how many standard deviations an element is from the mean. By using the standard normal distribution table, we found a probability of about 0.0228.
This method allows engineers to make informed predictions about product quality and ensure consistency in manufacturing through statistical analysis.

Probability Theory

Probability Theory is fundamental in this exercise and helps determine how likely it is that an event will occur. In this context, we determined the probabilities of certain numbers of lamps not meeting specifications.
To find out if at least one lamp fails, we leveraged probability rules. Specifically, if the probability of success (a lamp meeting specifications) is known, the probability of at least one failure is simply one minus the probability of all succeeding: \[ P(\text{at least one fails}) = 1 - P(\text{all succeed}) \].
This simplifies calculations and provides engineers with vital information without needing to know the specific probability of every possible outcome. Probability theory, through simple rules and calculators, provides a powerful tool to handle such otherwise complex situations.

Binomial Distribution

The Binomial Distribution is used to model the number of successes in a series of independent experiments. In this exercise, it helps us understand how many lamps might fail to meet specifications within a sample of 25.
In this case, each lamp either fails or meets specifications, forming a binary outcome. With a failure probability of 0.0228 for each lamp, we use the binomial distribution to compute the probability of five or fewer failures across 25 lamps. This calculation indicates a high likelihood (~0.9998) of five or fewer failures.
The binomial distribution here is particularly useful as it handles discrete data and can provide probabilities for ranges of outcomes, such as "five or fewer failures," rather than single fixed points. This distribution allows engineers to assess risk and make adjustments according to the probability of diverse outcomes occurring in the production process.