normal distribution

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Sue0113

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The design tolerance thickness for an automotive part is 55mm to 65mm. If the process mean thickness is 61mm with a standard deviation of 2.3mm, then what percentage of the sheets will be acceptable?(What percentage will fall within the design tolerence?) Assume that the distribution of these parts is normally distributed. Diagram is required

55mm is 61-2.61σ
65mm is 61+1.54σ
Z table amount is
.4955+.4382 = .9337 = 93.37% will pass Is this the correct calculations? Question states a diagram is required, could someone help with this part?
 
You have made an error in converting 65 mm to a z-score, which gives you an error in the percentage that will pass.
 
Sue0113 said:
The design tolerance thickness for an automotive part is 55mm to 65mm. If the process mean thickness is 61mm with a standard deviation of 2.3mm, then what percentage of the sheets will be acceptable?(What percentage will fall within the design tolerence?) Assume that the distribution of these parts is normally distributed. Diagram is required

55mm is 61-2.61σ
65mm is 61+1.54σ
Z table amount is
.4955+.4382 = .9337 = 93.37% will pass Is this the correct calculations? Question states a diagram is required, could someone help with this part?
Click to expand...


I wonder how u came up with calculation 2.61?

Which formula are you applying?
 
missha said:
I wonder how u came up with calculation 2.61?

Which formula are you applying?
Click to expand...

I think i will solve the question like this.

Z is between 55 and 65.

55 < z > 65

z = 55-61 / 2.3 = -2.60

z = 65-61/2.3 = 1.74

Using z-table;

for -2.60 gives 0.0047
for 1.74 gives 0.9592

Total = 0.0047+0.9592 = 0.9639 = 96.39%

Well, anyone who double check my answer if i am correct?
 
If you round to the nearest hundredth, you have:

\(\displaystyle \dfrac{55-61}{2.3}\approx-2.61\)

and:

\(\displaystyle \dfrac{55-61}{2.3}\approx1.74\)

Now, using the table I have, I find the area associated with 2.61 is 0.4955 and with 1.74 is 0.4591, for a total of 0.9546.

Using numeric integration, without rounding the z-scores, I find it is closer to 0.95445.
 
MarkFL said:
If you round to the nearest hundredth, you have:

\(\displaystyle \dfrac{55-61}{2.3}\approx-2.61\)

and:

\(\displaystyle \dfrac{55-61}{2.3}\approx1.74\)

Now, using the table I have, I find the area associated with 2.61 is 0.4955 and with 1.74 is 0.4591, for a total of 0.9546.

Using numeric integration, without rounding the z-scores, I find it is closer to 0.95445.
Click to expand...

My z-table is giving for -2.61 as 0.0045 (towards the negative side), and 1.74 as 0.9591


See for 1.74 the z-table in the image:

http://www.icess.ucsb.edu/gem/material_clima_I/normal01.jpg
 
It appears you need to find the difference rather than the sum. My table is "old school" and gives areas for the standard normal curve from the mean to a z-score.
 
MarkFL said:
It appears you need to find the difference rather than the sum. My table is "old school" and gives areas for the standard normal curve from the mean to a z-score.
Click to expand...

YEs ure right, we got to find out difference
 
this is what I did

P(55<x<=P[(55-61)/2.3<(x-61)/2.3<(65-61)/2.3)]
p=(-2.61<Z<1.74)
p=(0<Z<1.74)+p(0<Z<2.61)
0.45907 + 0.495547
=0.95454
or 95.45% wil;l fall within the design tolerance
Thanks for help people
 
Sue0113 said:
P(55<x<=P[(55-61)/2.3<(x-61)/2.3<(65-61)/2.3)]
p=(-2.61<Z<1.74)
p=(0<Z<1.74)+p(0<Z<2.61)
0.45907 + 0.495547
=0.95454
or 95.45% wil;l fall within the design tolerance
Thanks for help people
Click to expand...

How did u came up with 0.45907 and 0.495547
 
Z tables supplied with textbook

I found the 0.45907 and 0.49547 using the z tables provided with my textbook.
recieved assignment back and recieved full marks.
 
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