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A cork intended for use in a wine bottle is considered acceptable if its diameter is between 2.9 cm and 3.1 cm (so the lower specification limit is LSL 5 2.9 and the upper specification limit is USL 5 3.1).

a. If cork diameter is a normally distributed variable with mean value 3.04 cm and standard deviation .02 cm, what is the probability that a randomly selected cork will conform to specification?

b. If instead the mean value is 3.00 and the standard deviation is .05, is the probability of conforming to specification smaller or larger than it was in (a)?

Short Answer

Expert verified

Probability that a randomly selected cork will conform to specification:

\(P(X \in (2.9,3.1)) = 1\)

Probability of conforming to specification smaller or larger than it was in (a):

\(P(X \in (2.9,3.1)) = 0.9544\)

Step by step solution

01

Control Limit

The lower control limit is a line below the centerline that indicates the number below which any individual data point would be considered out of statistical control due to special cause variation.

Upper Control Limit (UCL) means a value greater than the maximum value of a chemical or physical parameter that can be attributed to natural fluctuations and sampling and agree upon by the Administrator and the operator prior to initiation of mining.

02

Answering the above mentioned question:

\(\begin{array}{l}LSL = 2.9\\USL = 3.1\end{array}\)

(a)

\(\begin{array}{l}X:N(3.04,{0.02^2}),\\P(X \in (2.9,3.1))\end{array}\)

\(\begin{array}{l}P(X \in (2.9,3.1)) = P(2.9 \le X \le 3.1) = P\left( {\frac{{2.9 - 3.04}}{{0.02}} \le X \le \frac{{3.1 - 3.04}}{{0.02}}} \right) = P( - 7 \le X* \le 3) \approx 1\\\end{array}\)

(b)

\(\begin{array}{l}X:N(3,{0.05^2}),\\P(X \in (2.9,3.1)) = P(2.9 \le X \le 3.1) = P\left( {\frac{{2.9 - 3}}{{0.05}} \le X* \le \frac{{3.1 - 3}}{{0.05}}} \right) = P( - 2 \le X* \le 2) = 2\phi (2) \approx 0.9544\\X*:N(0,1)\end{array}\)

Hence, the final answer is:

\(\begin{array}{l}P(X \in (2.9,3.1)) = 1\\P(X \in (2.9,3.1)) = 0.9544\end{array}\)

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