need help thanks Control - ITE-341-W01 ssessmentsv Communication\', Resources, D
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need help thanks
Control - ITE-341-W01 ssessmentsv Communication', Resources, D2L Help ft:1:26:16 Randy Allum: Attempt 1 Question 8 (10 points) After taking many samples of size n-4 of the length of pipe, the sample mean (xbar) and the mean standard deviation (sbar) were determined to be 2 087 and 0 053 respectively The process is in good statistical control and follows a normal distribution. If the product specification is 2 000--0 100 how many of the products produced above the upper specifications 0.189 0 .205 0 221Explanation / Answer
8.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 2.087
standard Deviation ( sd )= 0.053
sample size (n) = 4
P(X > 2.1) = (2.1-2.087)/0.053/ Sqrt ( 4 )
= 0.013/0.027= 0.4906
= P ( Z >0.4906) From Standard Normal Table
= 0.3119
= 0.321
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