Calculating Rejected Carbon Rods: Standard Deviation Homework

AI Thread Summary
The discussion focuses on calculating the percentage of rejected carbon rods based on their diameters, which are normally distributed with a mean of 1.501 cm and a standard deviation of 0.003 cm. The acceptable diameter range is between 1.495 cm and 1.505 cm. Initial calculations provided an incorrect rejection percentage of 1.67% for undersized rods and 0.4% for oversized rods. The correct approach involves converting the diameter measurements to a standard normal distribution using the z-score formula. Participants emphasized the need to find the probabilities for diameters less than 1.495 cm and greater than 1.505 cm using standard normal distribution tables.
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.
 
unfortunatly 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:
z = \frac{x - \mu}{\sigma}
where \mu is the population mean diameter (1.501 cm) and \sigma is the population standard deviation (.003 cm).

The equation above is equivalent to x = z\sigma + \mu

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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