Calculating Rejected Carbon Rods: Standard Deviation Homework

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Homework Help Overview

The discussion revolves around calculating the percentage of rejected carbon rods based on their diameters, which are normally distributed. The nominal diameter is 1.5 cm, with acceptable limits set between 1.495 cm and 1.505 cm. The mean diameter is given as 1.501 cm, with a standard deviation of 0.003 cm.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the percentage of rods that fall outside the acceptable diameter range. There are attempts to apply standard deviation to determine the limits of acceptance and to calculate rejection percentages. Some participants question the accuracy of the initial calculations and seek clarification on converting diameter statistics to a standard normal distribution.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing guidance on how to approach the conversion to a standard normal distribution. Questions remain about the correct application of statistical methods, and multiple interpretations of the problem are being discussed.

Contextual Notes

One participant notes that their textbook does not provide examples relevant to the conversion process, indicating a potential gap in resources for understanding the statistical methods required for this problem.

matt222
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Homework Statement



Carbon rods with a nominal diamter of 1.5cm, it is only acceptable within the limits of 1.495 to 1.505cm. the actual diamter normaly distributed with a mean of 1.501cm with standard deviation of 0.003cm. what will be the percentage of the rods which are rejected if
1- undersize
2-oversize

Homework Equations





The Attempt at a Solution


from the mean 1.501cm, we have + standard deviation 0.003cm so that going to be 1.504cm
again the mean 1.501cm, we have -standard deviation 0.003cm so that going to be 1.498cm
both are within the limit of acceptable diameter, so that the answer to 1 is 1/1.501-1.495/100=1.67%, and for 2 going to be 0.4%, is it right
 
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matt222 said:

Homework Statement



Carbon rods with a nominal diamter of 1.5cm, it is only acceptable within the limits of 1.495 to 1.505cm. the actual diamter normaly distributed with a mean of 1.501cm with standard deviation of 0.003cm. what will be the percentage of the rods which are rejected if
1- undersize
2-oversize

Homework Equations





The Attempt at a Solution


from the mean 1.501cm, we have + standard deviation 0.003cm so that going to be 1.504cm
again the mean 1.501cm, we have -standard deviation 0.003cm so that going to be 1.498cm
both are within the limit of acceptable diameter, so that the answer to 1 is 1/1.501-1.495/100=1.67%, and for 2 going to be 0.4%, is it right
First off, the value you calculated is wrong. 1/1.501-1.495/100 is about 0.651, nowhere close to the 1.67% that you show.

What you need to do is convert your diameter statistic to a standard normal distribution, and find the interval endpoints that correspond to 1.495 cm and 1.505 cm. Use the normal distribution to find the probability that the diameter is < 1.495 or diameter > 1.505.

Your textbook should have a few examples of how this is done.
 
unfortunately my textbook has no example, how to convert diameter statistic to a standard normal distribution, since i am new for this subject can yoiu refer me to a website, thank you
 
The z statistic is related to your statistic (let's call it x) in this way:
[tex]z = \frac{x - \mu}{\sigma}[/tex]
where [itex]\mu[/itex] is the population mean diameter (1.501 cm) and [itex]\sigma[/itex] is the population standard deviation (.003 cm).

The equation above is equivalent to x = z[itex]\sigma[/itex] + [itex]\mu[/itex]

The first part of your problem is asking you to find P(x < 1.495) and the second part asks you to find P(x > 1.505).

Using the second equation I gave, convert the inequalities in the two probabilities to ones that involve z, and use a table of probabilities for the standard normal distribution to find these probabilities.
 

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