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a. If a normal distribution has \(\mu=30\) and \(\sigma=5\), what is the 91 st percentile of the distribution? b. What is the 6th percentile of the distribution? c. The width of a line etched on an integrated circuit chip is normally distributed with mean \(3.000 \mu \mathrm{m}\) and standard deviation \(.140\). What width value separates the widest \(10 \%\) of all such lines from the other \(90 \%\) ?

Short Answer

Expert verified
a) 36.7, b) 22.25, c) 3.1792 \(\mu \mathrm{m}\).

Step by step solution

01

Understand the problem

You need to calculate percentiles from a normal distribution with given mean (\(\mu\)) and standard deviation (\(\sigma\)). The formula used for finding the percentile involves the inverse of the cumulative distribution function (CDF), commonly looked up using a Z-table or a calculator.
02

Identify given parameters for part (a)

For part (a), we know \(\mu = 30\) and \(\sigma = 5\). We need to find the 91st percentile. Percentiles in a normal distribution can be found using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \]Where \(X\) is the value corresponding to the desired percentile.
03

Find Z-score for 91st percentile

The Z-score corresponding to the 91st percentile is approximately 1.34 (from a standard normal distribution table or calculator).
04

Calculate the value for the 91st percentile

Use the Z-score to find \(X\) with the equation: \[ X = \mu + Z \cdot \sigma \]Substituting in the values gives: \[ X = 30 + 1.34 \cdot 5 = 36.7 \]
05

Perform similar steps for the 6th percentile

For part (b), find the 6th percentile. Identify that the Z-score corresponding to 6% is approximately -1.55, using a Z-table.
06

Calculate the value for the 6th percentile

Substitute the values into the equation: \[ X = \mu + Z \cdot \sigma \]Thus, \[ X = 30 - 1.55 \cdot 5 = 22.25 \]
07

Identify parameters for part (c)

For part (c), the mean \(\mu = 3.000\ \mu m \) and \(\sigma = 0.140\). We need to find the value where the top 10% starts.
08

Find Z-score for the top 10%

Top 10% implies the 90th percentile. The Z-score corresponding to the 90th percentile is approximately 1.28.
09

Calculate the width value

Use the formula with the given values: \[ X = \mu + Z \cdot \sigma \]So, \[ X = 3.000 + 1.28 \cdot 0.140 = 3.1792 \ \mu m \]

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Key Concepts

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

Normal Distribution
The concept of normal distribution is fundamental in statistics and relates to how data spreads across a set of values. This type of distribution is characterized by its symmetric bell-shaped curve.
The center of this curve is defined by the mean, while the spread is defined by the standard deviation.
A normal distribution is useful as a model because many natural phenomena follow this pattern. This includes anything from student test scores to biological measurements like heights or blood pressure.
  • The mean is located at the peak of the bell curve.
  • The standard deviation determines the width of the distribution.
  • About 68% of values fall within one standard deviation from the mean.
  • Approximately 95% of values are within two standard deviations.
Knowing this distribution helps in estimating probabilities and calculating percentiles, as every point on the curve corresponds to a probability of occurrence. It results in powerful insights, helping with making predictions and identifying trends.
Z-score
A Z-score is a statistical measurement 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.
Knowing the Z-score of a value helps to understand how unusual or typical this value is in the context of the dataset.
  • Z-scores can be positive or negative.
  • A Z-score of 0 indicates the value is exactly at the mean.
  • Positive Z-scores denote values above the mean.
  • Negative Z-scores denote values below the mean.
To calculate a Z-score, the formula is \[Z = \frac{X - \mu}{\sigma}\] Here, \(X\) refers to the data value, \( \mu \) is the mean, and \( \sigma \) represents the standard deviation.
This standardization allows for comparisons between normally distributed datasets that might have different means or standard deviations.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a way to describe the probability that a random variable takes on a value less than or equal to a specific value. In simpler terms, it gives the probability of finding observations below a particular threshold.
The CDF is crucial when assessing percentiles in any distribution.
  • The CDF begins at 0 at the farthest left tail of the distribution.
  • It approaches 1 (or 100% probability) as you move to the right.
  • This function is non-decreasing, meaning it never decreases as \(X\) increases.
For a standard normal distribution, the CDF operates with Z-scores. This translates data values to a position on the standard normal curve, where tables or calculators can be employed to determine exact probabilities or percentiles.
Mean and Standard Deviation
The mean and standard deviation are key concepts in understanding data distribution, especially in statistics and probability. They are essential for calculating probabilities, understanding variability, and determining relative standing in datasets.
The mean is the average of a set of numbers and serves as a central point of an expected value for the data. It is calculated by summing all data points and dividing by the total number of points.
  • The formula is \( \mu = \frac{1}{n} \sum_{i=1}^{n} X_i \)
Equally important, the standard deviation measures the dispersion of data points. A small standard deviation indicates that data points tend to be close to the mean, while a large standard deviation suggests data is spread out over a wide range.
  • The standard deviation formula is \( \sigma = \sqrt{ \frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2 } \)
These two statistics are integral to determining how normal a distribution is, and they form the foundation for calculating Z-scores and interpreting the cumulative distribution function.

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Most popular questions from this chapter

Let \(X=\) the time (in \(10^{-1}\) weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is \(\gamma=3.5\) and that the excess \(X-3.5\) over the minimum has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=1.5\) (see the Industrial Quality Control article referenced in Example 4.26). a. What is the cdf of \(X\) ? b. What are the expected return time and variance of return time? [Hint: First obtain \(E(X-3.5)\) and \(V(X-3.5)\).] c. Compute \(P(X>5)\). d. Compute \(P(5 \leq X \leq 8)\).

Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The paper "Effects of \(2,4-\mathrm{D}\) Formulation and Quinclorac on Spray Droplet Size and Deposition" (Weed Technology, 2005: \(1030-1036\) ) investigated the effects of herbicide formulation on spray atomization. A figure in the paper suggested the normal distribution with mean \(1050 \mu \mathrm{m}\) and standard deviation \(150 \mu \mathrm{m}\) was a reasonable model for droplet size for water (the "control treatment") sprayed through a \(760 \mathrm{ml} / \mathrm{min}\) nozzle. a. What is the probability that the size of a single droplet is less than \(1500 \mu \mathrm{m}\) ? At least \(1000 \mu \mathrm{m}\) ? b. What is the probability that the size of a single droplet is between 1000 and \(1500 \mu \mathrm{m}\) ? c. How would you characterize the smallest \(2 \%\) of all droplets? d. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds \(1500 \mu \mathrm{m}\) ?

In commuting to work, I must first get on a bus near my house and then transfer to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with \(A=0\) and \(B=5\), then it can be shown that my total waiting time \(Y\) has the pdf $$ f(y)=\left\\{\begin{array}{cl} \frac{1}{25} y & 0 \leq y<5 \\ \frac{2}{5}-\frac{1}{25} y & 5 \leq y \leq 10 \\ 0 & y<0 \text { or } y>10 \end{array}\right. $$ a. Sketch a graph of the pdf of \(Y\). b. Verify that \(\int_{-x}^{x} f(y) d y=1\). c. What is the probability that total waiting time is at most 3 min? d. What is the probability that total waiting time is at most 8 min? e. What is the probability that total waiting time is between 3 and \(8 \mathrm{~min}\) ? f. What is the probability that total waiting time is either less than \(2 \mathrm{~min}\) or more than \(6 \mathrm{~min}\) ?

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean \(3 \mathrm{~cm}\) and standard deviation \(.1 \mathrm{~cm}\). The second machine produces corks with diameters that have a normal distribution with mean \(3.04 \mathrm{~cm}\) and standard deviation \(.02 \mathrm{~cm}\). Acceptable corks have diameters between \(2.9 \mathrm{~cm}\) and \(3.1 \mathrm{~cm}\). Which machine is more likely to produce an acceptable cork?

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, 1984: 1169-1174) suggests the uniform distribution on the interval \((7.5,20)\) as a model for depth \((\mathrm{cm})\) of the bioturbation layer in sediment in a certain region. a. What are the mean and variance of depth? b. What is the cdf of depth? c. What is the probability that observed depth is at most 10 ? Between 10 and 15 ? d. What is the probability that the observed depth is within 1 standard deviation of the mean value? Within 2 standard deviations?

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