Question

Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean μ = 0.652 cm and standard deviation σ = 0.003 cm. The specification for the shaft diameter is 0.650±0.005 cm.
A) What proportion of the shafts manufactured by thisprocess meet the specifications?B) The process mean can be adjusted through calibration. If the mean is set to 0.650cm, what proportion of the shaftswill meet specification?C) If the mean is set to 0.650cm, what must the standarddeviation be so that 99% of the shafts will meetspecifications?

Answer 1

Answer:

a) 83.14% of the shafts manufactured by thisprocess meet the specifications

b) 90.50% of the shafts will meet specification.

c) 0.002

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 0.652, \sigma = 0.003$

The specification for the shaft diameter is 0.650±0.005 cm.

A) What proportion of the shafts manufactured by this process meet the specifications?

This is the pvalue of Z when X = 0.650 + 0.005 = 0.655 subtracted by the pvalue of Z when X = 0.650 - 0.005 = 0.645.

X = 0.655

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.655 - 0.652}{0.03}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 0.645

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.645 - 0.652}{0.03}$

$Z = -2.33$

$Z = -2.33$ has a pvalue of 0.0099

0.8413 - 0.0099 = 0.8314

83.14% of the shafts manufactured by thisprocess meet the specifications

B) The process mean can be adjusted through calibration. If the mean is set to 0.650cm, what proportion of the shaftswill meet specification?

Now we have that $\mu = 0.65$

X = 0.655

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.655 - 0.65}{0.03}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525

X = 0.645

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.645 - 0.65}{0.03}$

$Z = -1.67$

$Z = -1.67$ has a pvalue of 0.0475

0.9525 - 0.0475 = 0.9050

90.50% of the shafts will meet specification.

C) If the mean is set to 0.650cm, what must the standard deviation be so that 99% of the shafts will meet specifications?

This means that 0.655 will have a pvalue of 0.995 and 0.645 will have a pvalue of 0.005.

I will use 0.655. So when X = 0.655, Z = 2.575.

$Z = \frac{X - \mu}{\sigma}$

$2.575 = \frac{0.655 - 0.65}{\sigma}$

$2.575\sigma = 0.005$

$\sigma = \frac{0.005}{2.575}$

$\sigma = 0.002$