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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. (a) Determine the cumulative distribution function of photoresist thickness. (b) Determine the proportion of wafers that exceeds 0.2125 micrometers in photoresist thickness. (c) What thickness is exceeded by \(10 \%\) of the wafers? (d) Determine the mean and variance of photoresist thickness.

Short Answer

Expert verified
a) CDF: as given b) Proportion: 0.25 c) Thickness: 0.2140 d) Mean 0.2100, Variance 0.00000833

Step by step solution

01

Identify the distribution parameters

The thickness of the photoresist is uniformly distributed between \(0.2050\) and \(0.2150\) micrometers. Thus, the minimum value \(a = 0.2050\) and the maximum value \(b = 0.2150\).
02

Determine the cumulative distribution function (CDF)

For a uniform distribution, the cumulative distribution function \(F(x)\) is given by:\[ F(x) = \begin{cases} 0, & \text{if } x < a \ \frac{x-a}{b-a}, & \text{if } a \leq x \leq b \ 1, & \text{if } x > b \end{cases} \]Plugging in \(a = 0.2050\) and \(b = 0.2150\), we have:\[ F(x) = \begin{cases} 0, & x < 0.2050 \ \frac{x - 0.2050}{0.0100}, & 0.2050 \leq x \leq 0.2150 \ 1, & x > 0.2150 \end{cases} \]
03

Calculate the proportion exceeding 0.2125 micrometers

The proportion of wafers exceeding a certain thickness \(x_0\) is given by \(1 - F(x_0)\). Using the CDF from Step 2, calculate:\[ F(0.2125) = \frac{0.2125 - 0.2050}{0.0100} = 0.75 \]Thus, the proportion exceeding 0.2125 micrometers is:\[ 1 - F(0.2125) = 1 - 0.75 = 0.25 \]
04

Determine the thickness exceeded by 10% of the wafers

To find the thickness \(x\) that is exceeded by 10% of the wafers, solve for \(x\) in the equation:\[ F(x) = 0.9 \]This means \(\frac{x - 0.2050}{0.0100} = 0.9\), leading to:\[ x - 0.2050 = 0.0090 \]\[ x = 0.2050 + 0.0090 = 0.2140 \]
05

Calculate the mean and variance of the distribution

The mean of a uniform distribution is given by \(\mu = \frac{a + b}{2}\), and the variance is given by \(\sigma^2 = \frac{(b-a)^2}{12}\). Calculate the mean:\[ \mu = \frac{0.2050 + 0.2150}{2} = 0.2100 \]Calculate the variance:\[ \sigma^2 = \frac{(0.2150 - 0.2050)^2}{12} = \frac{0.0001}{12} \approx 0.00000833 \]

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Key Concepts

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

Probability
Probability is the cornerstone of statistical distribution, and the uniform distribution is no exception. In a uniform distribution, the probability that any interval within the distribution's range will occur is equal, assuming equal length intervals. This makes it a straightforward model where each outcome has the same likelihood of happening. For the case at hand, where the photoresist thickness varies uniformly between 0.2050 and 0.2150 micrometers, each thickness within this range is equally likely to occur. This simplicity is one of the features that makes the uniform distribution both elegant and useful in modelling scenarios where all outcomes within a given range are equally probable. Understanding this idea helps demystify why calculating probabilities with uniform distribution is more direct compared to other distributions.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) provides significant insight into the probability distribution of a random variable by representing the probability that a variable takes a value less than or equal to a specific value. For a uniform distribution, this function has distinct segments:

  • For any value less than the minimum (0.2050 micrometers in our example), the CDF is zero because the variable cannot take on values less than this minimum.
  • If the value lies between the minimum and maximum, the CDF increases linearly from 0 to 1. This linearity reflects the constant slope of probability increase within the range due to uniform distribution. The CDF is calculated using the formula \(F(x) = \frac{x - a}{b - a}\), which gives the cumulative probability up to a certain point \(x\).
  • For values greater than the maximum (0.2150 micrometers), the CDF is one, indicating the probability of encountering a value less than or equal to the maximum boundary is guaranteed.
Understanding the CDF helps to determine probabilities over intervals and is essential for calculating probabilities of certain outcomes being exceeded, such as determining what thicknesses exceed 0.2125 micrometers.
Mean and Variance
The mean, also known as the expected value, provides the central location of the distribution where you can expect the average observation to lie. In a uniform distribution, the mean is the midpoint between the minimum and maximum values. For the photoresist thickness, the mean is \(\mu = \frac{0.2050 + 0.2150}{2} = 0.2100\) micrometers, offering a sense of the central tendency.

Variance, on the other hand, measures the spread of the distribution, indicating how much values differ from the mean. In uniform distributions, the variance is computed using \(\sigma^2 = \frac{(b-a)^2}{12}\). For this example, the variance is \(\sigma^2 \approx 0.00000833\) micrometers squared. This low variance reflects that measurements of the thickness do not deviate widely from the mean, indicating consistency in the wafer's photoresist thickness. Mean and variance together help describe both the central tendency and the reliability of the distribution, providing essential insights for quality control in manufacturing processes.

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