Question

An inspection process is used to reject manufactured parts when a thickness dimension is not with a blueprint tolerance. It is known that the inspection of the thickness dimension is normally distributed with a mean of 1.50 and standard deviation of 0.2. The specification limit is the expected thickness plus or minus some tolerance, d. Determine the value of d so that 95% of the parts pass the inspection. Illustrate the probability distribution and the specification limits, and shade the probability of a failed inspection.

Answer 1

@radar @whpalmer4

Answer 2

@ybarrap

Answer 3

Since you have a normal distribution, the expected thickness is the same as the mean of the distribution, \(1.50\). So you want to find a confidence interval \(1.50\pm d\) with 95% confidence.

Let \(T\) denote the parts' thickness. You want to find the limits \(a\) and \(b\) such that
\[P(a\le T\le b)=0.95\]
Alternatively, this is the same as finding \(a\) and \(b\) for
\[P(T<a)=0.05~~~\text{and}~~~P(T>b)=0.05\]

Answer 4

Find the corresponding \(z\)-values:
\[P(T<a)=P\left(\frac{T-1.50}{0.2}<\frac{a-1.50}{0.2}\right)=0.05~~\Rightarrow~~\frac{a-1.50}{0.2}=-1.65\]
(\(z\) values were taken from this table, http://facstaff.unca.edu/dohse/Stat225/Images/Table-z.JPG)

Solve for \(a\). Do the same for \(b\), taking care to account for the difference between \(P(T<a)\) and \(P(T>b)\).

Answer 5

@SithsAndGiggles , so a=b?

Answer 6

No, not quite. It doesn't look like you accounted for the difference I mentioned at the bottom of my last comment.

Answer 7

\[P(T>b)=1-P(T<b)=0.05~~\Rightarrow~~P(T<b)=0.95\]
Checking the \(z\) table, you find that
\[\frac{b-1.50}{0.2}=1.65\not=-1.65\]
So no, \(a\not=b\).

Answer 8

thank you

Answer 9

yw