Question

A particular manufacturing design requires a shaft with a diameter of 23.000 mm, but shafts with diameters between 22.992 mm and 23.008 mm are acceptable. The manufacturing process yields shafts with diameters normally distributed, with a mean of 23.005 mm and a standard deviation of 0.004 mm. Complete parts (a) through (d) below. (Round to four decimal places as needed.) c. For this process, what is the diameter that will be exceeded by only 0.5% of the shafts? The diameter that will be exceeded by only 0.5% of the shafts is 23.0153 mm. (Round to four decimal places as needed.) d. What would be your answers to parts (a) through (c) if the standard deviation of the shaft diameters were 0.003 mm? to parts (a) through (c) if the standard de If the standard deviation is 0.003 mm, the proportion of shafts with diameter between 22.992 mm and 23.000 mm is 0.0478. (Round to four decimal places as needed.) If the standard deviation is 0.003 mm, the probability that a shaft is acceptable is (Round to four decimal places as needed.)

Answer 1

Solution

given : mean (u) = 23.005, standard deviation (0) = 0.003

X -M1 formula : Z=-

d) P(between 22.992 and 23.000) = P(22.992 < X < 23.000) =?

X - > P 22.992 - 23.005 0.003 23.000 - 23.005 0.003

$\Rightarrow P\left ( -4.33<Z<-1.67 \right )$

$\mathbf{Note}: P(a<Z<b)=P\left ( Z<b \right )-P(Z<a)$

$\Rightarrow P\left ( Z<-1.67 \right )-P(Z<-4.33)$

Refer Z-table to find the probability or use excel formula "=NORM.S.DIST(-1.67, TRUE)" & "=NORM.S.DIST(-4.33, TRUE)" to find the probability.

$\Rightarrow 0.0475-0.0000$

$\Rightarrow \mathbf{{\color{Red} 0.0475}}$

If the standard deviation is 0.003mm, the population of shafts with a diameter between 22.992 mm and 23.000 mm is $\mathbf{{\color{Red} 0.0475}}$

$\mathbf{\\P(shaft\;is\;acceptable)\Rightarrow P(between\;22.992\;and\;23.008)\Rightarrow P(22.992<X<23.008)=?}$

$\Rightarrow P\left ( \frac{22.992-23.005}{0.003}<\frac{X-\mu}{\sigma}<\frac{23.008-23.005}{0.003} \right )$

$\Rightarrow P\left ( -4.33<Z<1\right )$

$\mathbf{Note}: P(a<Z<b)=P\left ( Z<b \right )-P(Z<a)$

$\Rightarrow P\left ( Z<1\right )-P(Z<-4.33)$

Refer Z-table to find the probability or use excel formula "=NORM.S.DIST(1, TRUE)" & "=NORM.S.DIST(-4.33, TRUE)" to find the probability.

$\Rightarrow 0.8413-0.0000$

合0.8413

If the standard deviation is 0.003mm, the probability that a shaft is acceptable is $\mathbf{{\color{Red} 0.8413}}$