Question

In a grinding operation, there is an upper specification of 3.150 in. on a dimension of a certain part after grinding. Suppose that the standard deviation of this normally distributed dimension for parts of this type ground to any particular mean dimension LaTeX: \mu\:is\:\sigma=.002 μ i s σ = .002 in. Suppose further that you desire to have no more than 3% of the parts fail to meet specifications. What is the maximum (minimum machining cost) LaTeX: \mu μ that can be used if this 3% requirement is to be met?

Answer 1

Answer:

Step-by-step explanation:

Let X denote the dimension of the part after grinding

X has normal distribution with standard deviation $\sigma=0.002 in$

Let the mean of X be denoted by $\mu$

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 $\mu$ such that can be used if this 3% requirement is to be meet

$\Rightarrow P(\frac{X- \mu}{\sigma}$

We know from the Standard normal tables that

$P(Z\leq -1.87)=0.0307\\\\P(Z\leq -1.88)=0.0300\\\\P(Z\leq -1.89)=0.0293$

So, the value of Z consistent with the required condition is approximately -1.88

Thus we have

$\frac{3.15- \mu}{0.002} =-1.88\\\\\Rrightarrow \mu =1.88\times0.002+3.15\\\\=3.15$