Answer:
Step-by-step explanation:
Let X denote the dimension of the part after grinding
X has normal distribution with standard deviation [tex]\sigma=0.002 in[/tex]
Let the mean of X be denoted by [tex]\mu[/tex]
there is an upper specification of 3.150 in. on a dimension of a certain part after grinding.
We desire to have no more than 3% of the parts fail to meet specifications.
We have to find the maximum [tex]\mu[/tex] such that can be used if this 3% requirement is to be meet
[tex]\Rightarrow P(\frac{X- \mu}{\sigma} <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{0.002} )\leq 0.03[/tex]
We know from the Standard normal tables that
[tex]P(Z\leq -1.87)=0.0307\\\\P(Z\leq -1.88)=0.0300\\\\P(Z\leq -1.89)=0.0293[/tex]
So, the value of Z consistent with the required condition is approximately -1.88
Thus we have
[tex]\frac{3.15- \mu}{0.002} =-1.88\\\\\Rrightarrow \mu =1.88\times0.002+3.15\\\\=3.15[/tex]
