Calculating Mu for Unacceptable Dye Discharge in Paint

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To determine the appropriate setting for mu in dye discharge for paint, the goal is to ensure that only 1% of cans exceed 6 mL. This involves using the normal distribution and identifying the z-score for the 99th percentile, which is approximately 2.33. By applying the formula mu = x - (z * sigma), where x is 6 mL and sigma is 0.4 mL, the calculation yields mu = 5.068 mL. Setting mu at 5.068 mL will ensure that only 1% of the cans will have unacceptable dye discharge. This approach effectively balances the dye output within acceptable limits.
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OK, here is the question... its probably simple but i can't figure it out.

A machine used to regualte the amount of dye dispensed for mixing shades of paint can be set so that it discharges an average of mu milliliteres(mL) of dye per can of paint. The amount of dye discharged is known to have a normal distribution with standard deviation of 0.4 mL. If more than 6 mL of dye are discharged when making a certain shade of blue paint, thd shade is unacceptable. Determine the setting for mu so that only 1% of the cans of paint will be unacceptable.

Thanks a lot!
 
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Ok, so you first want to find how many standard deviations you can be away from your given mu so that only 1% come above the limit of 6 mL, given that standard deviation is two sided, you must find the number of standard deviations you can be away from the mean such that only 2% of the data does not fall within the given area. Then, you can take this number, which is the number of standard deviations the data is allowed to deviate from the median and multiply it by the given standard deviation, .4 mL This should give you a mu for which only 1% of your results will come above 6 mL.

~Lyuokdea
 


To calculate mu for unacceptable dye discharge in paint, we need to use the normal distribution formula and set the probability of unacceptable cans to 1%. This means that we are looking for the value of mu that corresponds to the 99th percentile of the normal distribution curve.

Using a z-score table, we can find that the z-score for the 99th percentile is approximately 2.33. We can then plug this value into the formula for the normal distribution:

z = (x - mu)/sigma

where z is the z-score, x is the value we are looking for (in this case, 6 mL), mu is the mean (which we are trying to find), and sigma is the standard deviation (which is given as 0.4 mL).

Rearranging the formula, we get:

mu = x - (z * sigma)

Plugging in the values, we get:

mu = 6 mL - (2.33 * 0.4 mL)

Simplifying, we get:

mu = 5.068 mL

Therefore, the setting for mu should be 5.068 mL to ensure that only 1% of cans of paint will have unacceptable dye discharge.
 
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