I am having difficulty with the follwoing question,
A sample of measurements shows a mean of 2.000 in. and 0.004 in., respectively. These results are for the observed Values. A separate error-of-measurment study indicates a precision of 0.002 in. (1s). There is no bias error. The dimension has a specification of 2.000 +/- 0.006 in. What percentage of the population has true dimensions outside the specification?
Here is what I have learnd so far,
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We have,
?O2 = ?P2 + ?E2,
where,
?O is the standard deviation of the observed data
?P is the standard deviation of the product and
?E is the standard deviation of the measuring method
Thus,
(0.004)2 = ?P2 + (0.002)2
That is, P2 =(0.004)2 - (0.002)2 = 0.000012
That is P = 0.0035
Now, the percentage of the population has true dimensions outside the specification is given by,
P (X > 2.006 or X < 1.994) = P(X > 2.006) + P(X < 1.994)
= P[(X-2.000)/0.0035 > (2.006-2)/0.0035] + = P[(X-2.000)/0.0035 < (1.994-2)/0.0035]
= P(Z > 1.714) + P(Z < -1.714)
= 2*P(Z < -1.714)
= 2 (0.0433)
= 0.0865 or 8.65%
Note that the probability P(Z < -1.714) is calculated using the Excel formula =NORMSDIST(-1.714)
=======================================================
And this:
Since the dimension has a specification of 2.000 +/- 0.006 in., the true dimensions outside the specification are
X 2.000 + 0.006
That is, X 2.006
Now, the percentage of the population has true dimensions outside the specification is given by,
P (X > 2.006 or X 2.006) + P(X 2.006) = P[(X-2)/0.0035 > (2.006-2)/0.0035] = P(Z > 1.714)
P(X < 1.994) = P[(X-2)/0.0035 < (1.994-2)/0.0035] = P(Z 1.714) = P(Z 2.006 or X < 1.994) = 2*P(Z < -1.714)
= 2 (0.0433)
= 0.0865 or 8.65%
Note that the probability P(Z < -1.714) is calculated using the Excel formula =NORMSDIST(-1.714)
====================================================================================
So here is my question,
I still trying to understand the second part of the above answer. I get lost in the percentage calculation and the Z number.
Could someone explain this to me? I only have a shallow understanding of the concepts.
Thank you,
A sample of measurements shows a mean of 2.000 in. and 0.004 in., respectively. These results are for the observed Values. A separate error-of-measurment study indicates a precision of 0.002 in. (1s). There is no bias error. The dimension has a specification of 2.000 +/- 0.006 in. What percentage of the population has true dimensions outside the specification?
Here is what I have learnd so far,
===================================================================
We have,
?O2 = ?P2 + ?E2,
where,
?O is the standard deviation of the observed data
?P is the standard deviation of the product and
?E is the standard deviation of the measuring method
Thus,
(0.004)2 = ?P2 + (0.002)2
That is, P2 =(0.004)2 - (0.002)2 = 0.000012
That is P = 0.0035
Now, the percentage of the population has true dimensions outside the specification is given by,
P (X > 2.006 or X < 1.994) = P(X > 2.006) + P(X < 1.994)
= P[(X-2.000)/0.0035 > (2.006-2)/0.0035] + = P[(X-2.000)/0.0035 < (1.994-2)/0.0035]
= P(Z > 1.714) + P(Z < -1.714)
= 2*P(Z < -1.714)
= 2 (0.0433)
= 0.0865 or 8.65%
Note that the probability P(Z < -1.714) is calculated using the Excel formula =NORMSDIST(-1.714)
=======================================================
And this:
Since the dimension has a specification of 2.000 +/- 0.006 in., the true dimensions outside the specification are
X 2.000 + 0.006
That is, X 2.006
Now, the percentage of the population has true dimensions outside the specification is given by,
P (X > 2.006 or X 2.006) + P(X 2.006) = P[(X-2)/0.0035 > (2.006-2)/0.0035] = P(Z > 1.714)
P(X < 1.994) = P[(X-2)/0.0035 < (1.994-2)/0.0035] = P(Z 1.714) = P(Z 2.006 or X < 1.994) = 2*P(Z < -1.714)
= 2 (0.0433)
= 0.0865 or 8.65%
Note that the probability P(Z < -1.714) is calculated using the Excel formula =NORMSDIST(-1.714)
====================================================================================
So here is my question,
I still trying to understand the second part of the above answer. I get lost in the percentage calculation and the Z number.
Could someone explain this to me? I only have a shallow understanding of the concepts.
Thank you,