The length of an injection-modeled olastic case that holdsmagnetic tape is norma

Question

The length of an injection-modeled olastic case that holdsmagnetic tape is normally distributed with a length of 90.2millimeters and a standard deviation of 0.1 millimeter. a) What is the probability that a part is longer than 90.3millimeters or shorter than 89.7 millimeters? b) What should the process mean be set at to obtain thegreatest number of parts between 89.7 and 90.3 millimeters? c) If parts that are not between 89.7 and 90.3 millimeters arescrapped, what is the yield for the process mean that you selectedin part (b)? Assume that the process is centered so that the mean is 90millimeters and the standard deviation is 0.1 millimeters. Supposethat 10 cases are measured, and they are assumed to beindependent. d) What is the probability that all 10 cases are between 89.7and 90.3 millimeters? e) What is the expected number of the 10 cases that arebetween 89.7 and 90.3 millimeters? Thanks in advance! The length of an injection-modeled olastic case that holdsmagnetic tape is normally distributed with a length of 90.2millimeters and a standard deviation of 0.1 millimeter. a) What is the probability that a part is longer than 90.3millimeters or shorter than 89.7 millimeters? b) What should the process mean be set at to obtain thegreatest number of parts between 89.7 and 90.3 millimeters? c) If parts that are not between 89.7 and 90.3 millimeters arescrapped, what is the yield for the process mean that you selectedin part (b)? Assume that the process is centered so that the mean is 90millimeters and the standard deviation is 0.1 millimeters. Supposethat 10 cases are measured, and they are assumed to beindependent. d) What is the probability that all 10 cases are between 89.7and 90.3 millimeters? e) What is the expected number of the 10 cases that arebetween 89.7 and 90.3 millimeters? Thanks in advance!

Explanation / Answer

X~N(=90.2,=.1) Define z=(X-)/ the Z~N(0,1) a)1-P(89.7<X<90.3)=1-P(X<90.3)-P(X<89.7)=1-P(Z<(90.3-90.2)/.1)-P(Z<(89.7-90.2)/.1)=1-.8413=.1587 b)Center the process at the midpoint 90. c)Now process mean is =90. Following same steps asin (a) with =90 the process yield is P(89.7<X<90.3)=.9973   or 99.73% yield d)binomial win n=10, p=.9973 P(10 0ut of10)=C(10,10)p10(1-p)10-10=(.9973)10=.973326 e)Expected number =np=10(.9973)=9.973

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