Question

A process has a mean of 9 grams and a standard deviation of 0.25 grams.  The lower spec limit is 8.00 grams and the upper spec limit is 10.00 grams.

- Compute Cp.

- What is the probability the process will operate within the specification limits?

- P{out of spec} for Wonderdisk

- μ = 10, σ = 2.3

qP{X ≤ 6 or X ≥ 12}

qP{8 ≤ X ≤ 13}

Answer 1

Answer:

PAGE NO. DATE 6 (an) The process has a mean of a grams and a standard deviation of 0.25 gramu. 7 Given:- 439_qramy z XNN (9,0.25 0.25 LSL=8.00 a 10.00 % gramu. =) e co(Process = USL - LSL capability) USL = 66 10-8 6x0.25 2 6X0.25 Cp 4 3 • Prob [ process within =P[SL <x<USL] specification limit =P[8<x<10] =P 8-9 <0-4 <10-9 -0.25 0.25 0.25 1 4 22 -4 <ZI 0.25 • PC-4<Zc4) 0.9999 38319

PAGE NO // The values can be found wring Stanishal aber Plout of specification) = |- P ( within specification limit =1 -0.999938314 Plout of specificalton) = 6.1686X10 If u = 10, 6=2.3 P ( X56.or X7112 } = P(X<6) + P(X>,12) P(X=4 5640) +P/x-07, 12-10 2.3 2.3 G 2.3 = P(Z <-1.739))+ P( z7,0.86956) -0.041009 +1 - PC Z <0.86956) = 0.041009 + - 0.80773 0.233279 : P ( X = 6 or x7,124 = 0.233279

PAGEHO PC:85x5133 =P(8-10 SX510 13-10 2.3. 2.3 2.3 = P(-0.869565€ Z 41.30434) = P(1.30434) - P(-0.869565) = 0.90394 0.19227 18(85X513) 0.71167 Every where we have to me standard Normal distribution by transforming Normal is to standard Normal distibution se XC Normal ) = Z=) CX-mean Standard deviation.

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