Problem 55 - Applied Statistics and Probability for Engineers

Question

The thickness of photoresist applied to wafers in semiconductor manufacturing at a particular location on the wafer is uniformly distributed between 0.2050 and 0.2150 micrometers. Determine the following: (a) Cumulative distribution function of photoresist thickness (b) Proportion of wafers that exceeds 0.2125 micrometers in photoresist thickness (c) Thickness exceeded by \(10 \%\) of the wafers (d) Mean and variance of photoresist thickness

Step-by-Step Solution

Answer

(a) CDF is given by the piecewise function; (b) 25% exceed 0.2125 micrometers; (c) 0.2140 micrometers; (d) Mean is 0.2100, variance is 0.00000833.

Step 1: Understanding Uniform Distribution

The photoresist thickness is uniformly distributed between 0.2050 and 0.2150 micrometers. This means every value within this range is equally likely. The uniform distribution can be defined by two parameters: the minimum value \(a = 0.2050\) and the maximum value \(b = 0.2150\).

Step 2: Cumulative Distribution Function (CDF)

For a uniform distribution, the cumulative distribution function (CDF) \(F(x)\) for a random variable \(X\) is given by:\[ F(x) = \begin{cases} 0, & \text{if } x < a \ \frac{x-a}{b-a}, & \text{if } a \leq x \leq b \ 1, & \text{if } x > b \end{cases} \]Thus, for the photoresist thickness, the CDF is:\[ F(x) = \begin{cases} 0, & x < 0.2050 \ \frac{x-0.2050}{0.0100}, & 0.2050 \leq x \leq 0.2150 \ 1, & x > 0.2150 \end{cases} \]

Step 3: Proportion of Wafers Exceeding 0.2125 Micrometers

To find the proportion of wafers that exceed 0.2125 micrometers, calculate the probability \(P(X > 0.2125)\) using the compliment of the CDF:\[ P(X > 0.2125) = 1 - F(0.2125) \]\[ F(0.2125) = \frac{0.2125 - 0.2050}{0.0100} = 0.75 \]So, \(P(X > 0.2125) = 1 - 0.75 = 0.25\).

Step 4: Thickness Exceeded by 10% of Wafers

Find the thickness \(x\) such that 10% of wafers exceed this thickness, or \(P(X > x) = 0.10\).This implies \(F(x) = 0.90\).Using the CDF formula:\[ F(x) = \frac{x-0.2050}{0.0100} = 0.90 \ x - 0.2050 = 0.0090 \ x = 0.2140\]Thus, the thickness exceeded by 10% of the wafers is 0.2140 micrometers.

Step 5: Calculating Mean and Variance

For a uniform distribution over \([a, b]\), the mean \(\mu\) is given by:\[ \mu = \frac{a+b}{2} = \frac{0.2050 + 0.2150}{2} = 0.2100 \text{ micrometers} \]The variance \(\sigma^2\) is given by:\[ \sigma^2 = \frac{(b-a)^2}{12} = \frac{(0.2150 - 0.2050)^2}{12} = \frac{0.0001}{12} = 0.00000833 \text{ micrometers}^2 \]

Key Concepts

Cumulative Distribution Function (CDF)

The Cumulative Distribution Function, or CDF, is a fundamental concept in statistics, especially when dealing with continuous probability distributions like the uniform distribution. Think of the CDF as a way to describe how likely it is for a random variable to take on a value less than or equal to a specific number. In simple terms, it accumulates probabilities from the lowest possible value up to any specific value you're interested in.

For a uniform distribution, such as the thickness of photoresist in our example which is uniformly distributed between 0.2050 and 0.2150 micrometers, the CDF is clear-cut. Because all outcomes within this range are equally likely, the formula for the CDF is a nice straight line.

Here's the step-by-step breakdown of the CDF formula for a uniform distribution:

• If our variable of interest is less than the minimum value, the CDF is 0.
• If our variable of interest is between the minimum and maximum value, the CDF increases linearly from 0 to 1.
• If our variable of interest is more than the maximum value, the CDF is 1.

Using these rules helps us quantify probabilities for any threshold within these bounds, creating a simple yet powerful tool for statistical analysis.

Mean and Variance in Statistics

The concepts of mean and variance are pivotal in understanding distributions like the uniform distribution. The mean, often thought of as the average, presents a central point around which the distribution balances. For a uniform distribution between any two points, say 'a' and 'b', the mean is straightforwardly positioned at the middle of these two points.

Mathematically, for a uniform distribution, the mean \[ \mu = \frac{a+b}{2}\]For our photoresist thickness example, the mean is the average of 0.2050 and 0.2150, which calculates to 0.2100 micrometers.

The variance, on the other hand, measures the spread of values around the mean. Essentially, it gives us an idea of how far these values tend to deviate from the mean, showing the extent of dispersion. For uniform distribution, the variance is calculated using the formula:\[\sigma^2 = \frac{(b-a)^2}{12}\]For the photoresist thickness, this evaluates to 0.00000833 micrometers squared. Keep in mind that a lower variance indicates data points are closer to the mean, while a higher variance shows wider dispersal around the mean.

Probability of Exceeding a Value

When working with probabilities, especially in statistics, a common question involves calculating the likelihood that a random variable exceeds a specified value. In our uniform distribution example, we calculate the probability of the photoresist thickness exceeding 0.2125 micrometers.

The quickest way to find this probability is by leveraging the CDF. Once you have the CDF value for the point of interest (in this case, 0.2125 micrometers), simply subtract it from 1 to find the exceeding probability.

• First, compute the CDF at 0.2125 micrometers, where \[F(0.2125) = \frac{0.2125 - 0.2050}{0.0100} = 0.75\]
• Then, calculate probability of exceeding 0.2125 micrometers by 1 minus the CDF value: \[P(X > 0.2125) = 1 - F(0.2125) = 0.25\]

This method highlights how CDFs not only help determine the probability of a variable being below a certain point but also facilitate finding probabilities that surpass specific values—a crucial feature in statistical practice.