/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 If the mesh tension of individua... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the mesh tension of individual monit ors follows a Normal distribution, we can describe capability by giving the percentage of monitors that meet specifications. The old specifications for mesh tension are 100 to \(400 \mathrm{mV}\). The new specifications are 150 to \(350 \mathrm{mV}\). Because the process is in control, we can estimate that tension has mean \(275 \mathrm{mV}\) and standard deviation \(38.4 \mathrm{mV}\). a. What percentage of monitors meet the old specifications? b. What percentage meet the new specifications?

Short Answer

Expert verified
a. About 99.99% meet the old specifications. b. About 91.02% meet the new specifications.

Step by step solution

01

Define the Problem

The task is to find the percentage of monitors that meet specified mesh tension ranges, given that mesh tension follows a Normal distribution with a mean (\(\mu\) = 275 mV) and a standard deviation (\(\sigma = 38.4 \, \text{mV}\)). We need to calculate this for both the old specifications (100 to 400 mV) and the new specifications (150 to 350 mV).
02

Convert Old Specifications to Z-Scores

For the old specification range (100 to 400 mV), first calculate the Z-score for each boundary value using the formula \(Z = \frac{X - \mu}{\sigma}\).
03

Calculate Z-Scores for Old Specifications

- Low boundary (100 mV):\[ Z_\text{old,low} = \frac{100 - 275}{38.4} \approx -4.5573 \]- High boundary (400 mV):\[ Z_\text{old,high} = \frac{400 - 275}{38.4} \approx 3.2552 \]
04

Compute Percentage for Old Specifications

Using the Normal distribution table or a calculator, find the percentage of data between the Z-scores \(-4.5573\) and \(3.2552\). This area represents the percentage of monitors meeting the old specifications, which is approximately 99.99%.
05

Convert New Specifications to Z-Scores

For the new specifications (150 to 350 mV), calculate the Z-score for each boundary value.
06

Calculate Z-Scores for New Specifications

- Low boundary (150 mV):\[ Z_\text{new,low} = \frac{150 - 275}{38.4} \approx -3.2552 \]- High boundary (350 mV):\[ Z_\text{new,high} = \frac{350 - 275}{38.4} \approx 1.9531 \]
07

Compute Percentage for New Specifications

Find the percentage of data between the Z-scores \(-3.2552\) and \(1.9531\) using the Normal distribution table or a calculator. This area represents the percentage of monitors meeting the new specifications, which is approximately 91.02%.

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Key Concepts

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

Z-Scores
Z-scores provide a way to understand how far a particular value is from the mean in terms of standard deviations. In other words, a Z-score will tell you how many standard deviations away a datapoint is from the average or mean of a data set. To calculate a Z-score, you use the formula: \[ Z = \frac{X - \mu}{\sigma} \] Where:
  • \(X\) is the value in the dataset,
  • \(\mu\) (mu) is the mean of the dataset,
  • \(\sigma\) (sigma) is the standard deviation.
In the original exercise, Z-scores help determine what percentage of monitors fall within specified voltage limits. By transforming these limits into Z-scores, you can use the normal distribution to find probabilities or percentages about how likely monitors are to meet those specifications. Calculating Z-scores for the boundaries of 100 mV and 400 mV gives values of approximately -4.5573 and 3.2552 respectively, encompassing nearly all monitors, about 99.99% under the old specification. For the new specification of 150 mV to 350 mV, the calculated Z-scores are around -3.2552 and 1.9531, corresponding to about 91.02% of monitors meeting these specifications.
Standard Deviation
The standard deviation is a measure of how spread out the values in a data set are around the mean. When data points are tightly clustered around the mean, the standard deviation is small. Conversely, if data points are spread widely, the standard deviation is large. Mathematically, standard deviation is represented by the symbol \(\sigma\) (sigma) and is calculated using the formula: \[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2} \] Where:
  • \(N\) is the number of data points,
  • \(X_i\) are the data points,
  • \(\mu\) is the mean of the data points.
In the context of the problem, the standard deviation of 38.4 mV tells us about the variability of the monitor mesh tension around the mean value of 275 mV. Understanding standard deviation allows us to utilize the normal distribution by converting boundaries to Z-scores, to find the percentage of monitors meeting certain tension specifications.
Mean
The mean, often called the average, is a central measure of a data set. It is calculated by summing up all data points and then dividing by the number of points. In mathematical terms, it is represented as \(\mu\). The formula is: \[ \mu = \frac{1}{N} \sum_{i=1}^{N} X_i \] Where:
  • \(N\) is the total number of observations,
  • \(X_i\) are the data values.
In this exercise, the mean mesh tension is 275 mV. The mean gives a central value around which other data points are distributed. It acts as a reference point when calculating Z-scores, determining how data points relate to the average tension, which in turn helps us figure out the variability and how common different measurement ranges are within our normal distribution.

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Most popular questions from this chapter

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A manager who knows no statistics asks you, "What does it mean to say that a process is in control? Is being in control a guarantee that the quality of the product is good?" Answer these questions in plain language that the manager can understand.

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