Question 1013208: A manufacturer of precision tools specifies that the diameter of a circular disc must be 10cm with a tolerance of 1cm. The machine produces these discs with diameters that are normally distributed with a standard deviation of 5mm. If the mean diameter is set to 10cm, what proportion of discs will be rejected as outside the tolerance limits? Discs with diameters above the upper limit can be reworked to within the specifications, but those that are below the lower limit have to be scrapped. If it is desired that at most 1% output can be scrapped, what should be the setting for the mean? When the mean is so set, what proportion of discs will have to be reworked? If the cost of production of a disc is $10, and the cost of reworking a disc is $2, how much would it cost to have 1000 discs ready for shipping?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the discs must have a diameter of 10 cm plus or minus 1 cm.
that means the tolerance limits are between 9 cm and 11cm.
the machine is set to produce discs with a mean of 10 cm and a standard deviation of 5 mm.
since 5mm is equivalent to .5 cm, then the machine is set to produce discs with a mean of 10 cm and a standard deviation of .5 cm.
the tolernace limits are a diameter of 10 cm plus or minus 1 cm.
the low tolerance limit is therefore 9 cm.
the high tolerance limit is therefore 11 cm.
any disc produced below 9 cm diameter will be rejected and thrown away.
any disc produced above 11 cm diameter will be re-worked to get it back under 11 cm.
z-scores can be calculated to determine percents above and below the threshold limits.
the z-score formula is z = (x-m)/s
z is the z-score
x is the threshold limit.
m is the mean
s is the standard deviation.
when x = 9, the formula becomes z = (9-10)/.5 = -2.
when x = 11, the formula becomes z = (11-10)/.5 = 2.
with z = -2, 2.275% will be rejected because they're too small.
with z = 2, 2.275% will be rejected because they're too large.
you only want 1% to be rejected.
to get 1% or less to be rejected requires a z-score of -2.33.
we can use the z-score formula to find the new mean.
z = (x-m)/s
z = -2.33
x = 9
s = .5
m is what we are looking for.
z = (x-m)/s becomes -2.33 = (9-m)/.5.
solve for m to get m = 9 + 2.33*.5 = 10.165.
the new mean that we want to set the machine to produce should be 10.165 cm.
when the machine is set to a mean of 10.165 with a standard deviation of .5, the low threshold of 9 will result in a z-score of (9-10.165)/.5 = -2.33.
when the machine is set to a mean of 10.165 with a standard deviation of .5, the high threshold of 11 will result in z-score of (11-10.165)/.5 = 1.67.
a z-score of -2.33 results in approximately 1% being rejected for being too small.
a z-score of 1.67 results in approximately 4.75% having to be re-worked because their diameter is too large.
when the machine is set to a mean of 10.165, therefore:
1% will be rejected for having a diameter that is too small.
4.75% will need to be re-worked for having a diameter that is too large.
you want to have 1000 ready to be shipped.
with a 1% rejection rate, you will need to produce 1000 / .99 = 1011 discs.
when you produce 1011 discs:
1% of 1011 = 10.11 which means that 11 will be rejected.
4.75% of 1011 = 48.0225 which means that 49 will need to be re-worked.
the cost to produce 1000 discs ready for shipping is therefore:
1011 * 10 + 49 * 2 = 10,110 + 98 = 10,208.
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