Problem 67 - Applied Statistics and Probability for Engineers

Question

In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met. Let \(X\) and \(Y\) denote the thickness of two different layers of ink. It is known that \(X\) is normally distributed with a mean of \(0.1 \mathrm{~mm}\) and a standard deviation of \(0.00031 \mathrm{~mm},\) and \(Y\) is also normally distributed with a mean of \(0.23 \mathrm{~mm}\) and a standard deviation of \(0.00017 \mathrm{~mm}\). Assume that these variables are independent. (a) If a particular lamp is made up of these two inks only, what is the probability that the total ink thickness is less than \(0.2337 \mathrm{~mm} ?\) (b) A lamp with a total ink thickness exceeding \(0.2405 \mathrm{~mm}\) lacks the uniformity of color that the customer demands. Find the probability that a randomly selected lamp fails to meet customer specifications.

Step-by-Step Solution

Answer

Both probabilities are nearly 0, due to large negative Z-scores.

Step 1: Understand the Problem

We need to find the probability of the total ink thickness, which is the sum of the thicknesses of layers \(X\) and \(Y\), having certain characteristics: being less than \(0.2337\) mm (part a) and exceeding \(0.2405\) mm (part b). Both \(X\) and \(Y\) are normally distributed and independent.

Step 2: Define the Sum of Variables

Since \(X\) and \(Y\) are independent, the sum \(T = X + Y\) is also normally distributed. The mean of \(T\) is the sum of the means of \(X\) and \(Y\), \(\mu_T = \mu_X + \mu_Y = 0.1 + 0.23 = 0.33\) mm. The variance of \(T\) is the sum of the variances of \(X\) and \(Y\), \(\sigma_T^2 = \sigma_X^2 + \sigma_Y^2 = 0.00031^2 + 0.00017^2\).

Step 3: Calculate the Standard Deviation for T

Compute the variance of \(T:\)\[\sigma_T^2 = 0.00031^2 + 0.00017^2 = 9.61 \times 10^{-8} + 2.89 \times 10^{-8} = 1.25 \times 10^{-7}\]Then, the standard deviation of \(T\) is \(\sigma_T = \sqrt{1.25 \times 10^{-7}} = 0.000353553\) mm.

Step 4: Calculate Probability for Part (a)

For \(P(T < 0.2337)\), first calculate the Z-score:\[Z = \frac{0.2337 - 0.33}{0.000353553} = \frac{-0.0963}{0.000353553} \approx -272.43\]Using standard normal distribution tables, the probability associated with this Z-score is essentially 0, indicating it's nearly impossible for the total thickness to be under \(0.2337\) mm.

Step 5: Calculate Probability for Part (b)

For \(P(T > 0.2405)\), calculate the Z-score:\[Z = \frac{0.2405 - 0.33}{0.000353553} = \frac{-0.0895}{0.000353553} \approx -253.17\]Again, this Z-score is so large in the negative direction that the probability is practically 0, meaning it's almost impossible for the total thickness to exceed \(0.2405\) mm.

Key Concepts

Normal Distribution

The normal distribution is one of the most important concepts in statistics. It is characterized by its bell-shaped curve and is defined by two parameters: the mean and the standard deviation. The mean determines the center of the distribution, while the standard deviation measures the spread. A normally distributed dataset is symmetric around its mean, meaning that the probabilities of deviations on either side of the mean are equal.

In our exercise, the thickness of the ink layers, denoted by the variables \(X\) and \(Y\), are both normally distributed. This means that most of the data will cluster around a central point, with fewer occurrences as we move away from the mean. This attribute helps in making predictions about a range of possible thickness values.

• The mean of \(X\) is \(0.1\, \text{mm}\) with a standard deviation of \(0.00031\, \text{mm}\).
• The mean of \(Y\) is \(0.23\, \text{mm}\) with a standard deviation of \(0.00017\, \text{mm}\).

In estimating probabilities like those in our example, the normal distribution helps determine the likelihood that a certain ink thickness falls within a specified range.

Independent Random Variables

Independent random variables are a set of variables where the outcome of one variable does not influence or affect the outcome of another. In other words, knowing the value of one of these variables gives no information about the other. This property simplifies combinatorial calculations involving these variables because we can treat each variable separately when calculating probabilities.

In our symmetrical analysis of the ink layer thicknesses in the lamps, \(X\) and \(Y\) are independent. Therefore, the probability distribution of the combined layer thickness \(T = X + Y\) can be calculated using the properties of independent random variables.

• The sum \(T\) is also normally distributed with a mean \(\mu_T = \mu_X + \mu_Y\).
• The variance of \(T\) is the sum of the variances of \(X\) and \(Y\), calculated as \(\sigma^2_T = \sigma^2_X + \sigma^2_Y\).

This mathematical independence makes it possible to derive the distribution parameters for \(T\) directly as a step in calculating probabilities for thicknesses meeting or not meeting specifications.

Z-score

The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a Z-score of 0 indicates the value is identical to the mean, a positive Z-score indicates a value above the mean, while a negative Z-score signals a value below it.

Z-scores are crucial when calculating probabilities from a normal distribution. They allow us to use standard normal distribution tables, which provide probabilities for a range of Z-scores.

In our problem, Z-scores help calculate the probability of ink thickness calculations:

• For a total thickness less than \(0.2337 \text{mm}\), a Z-score calculation shows the extreme improbability of this event.
• The calculation for a thickness exceeding \(0.2405 \text{mm}\) similarly reveals it's almost impossible, indicating how the colors will consistently meet customer specifications.

This context translates the problem into simpler components for determining probabilities against the central trend imposed by the standard normal distribution.

Standard Deviation

The standard deviation is a measure of how spread out numbers in a data set are. It quantifies the amount of variation or dispersion from the average. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation indicates that the data points are spread out over a wider range of values.

For normally distributed data, about 68% of the samples will fall within one standard deviation of the mean, about 95% within two, and about 99.7% within three standard deviations, forming what is known as the empirical rule or the 68-95-99.7 rule.

In the context of this problem, understanding standard deviation helps assess:

• The layer thicknesses' degree of consistency, where
  \(\sigma_X = 0.00031 \text{mm}\) and \(\sigma_Y = 0.00017\text{mm}\).
• The variance of the total thickness \(\text{T}\) helps in determining its standard deviation \(\sigma_T = 0.000353553\, \text{mm}\).

The corresponding probabilities related to the ink thickness can then be accurately derived, using the standard deviation in the calculation of Z-scores.