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Statistics / Statistics for Engineers and Scientists 4 / Chapter 4.5 / Problem 14E

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

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Shafts manufactured for use in optical storage devices

Chapter 4, Problem 14E

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QUESTION:

Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean \(\mu\) = 0.652 cm and standard deviation \(0\) = 0.003 cm. The specification for the shaft diameter is 0.650 \(\pm\) 0.005 cm.

a. What proportion of the shafts manufactured by this process meet the specifications?

b. The process mean can be adjusted through calibration. If the mean is set to 0.650 cm, what proportion of the shafts will meet specifications?

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

Equation transcription:

$\mu{}$

$\pm{}$

$\sigma{}$

Text transcription:

\mu

\pm

0

Questions & Answers

QUESTION:

Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean \(\mu\) = 0.652 cm and standard deviation \(0\) = 0.003 cm. The specification for the shaft diameter is 0.650 \(\pm\) 0.005 cm.

a. What proportion of the shafts manufactured by this process meet the specifications?

b. The process mean can be adjusted through calibration. If the mean is set to 0.650 cm, what proportion of the shafts will meet specifications?

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

Equation transcription:

$\mu{}$

$\pm{}$

$\sigma{}$

Text transcription:

\mu

\pm

0

ANSWER:

Answer:

Step 1 of 3:

(a)

In this question, we are asked to find the proportion of the shafts manufactured by this process meet the specifications.

Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean $\mu{}\ =\ 0.652\ cm$ and standard deviation $\sigma{}\ \ =\ 0.003\ cm$ .

The specification for the shaft diameter is $0.650\pm{}0.005\ cm$ .

Let $X$ represent the diameter of a randomly chosen shaft.

We need to find $P(0.645<X<0.655)$ .

Here we can write,

$X\ \sim{}N(0.652\ ,\ {0.003}^{2})$

Now we will calculate the $Z$ score because our distribution is approximately normal.

$z\ =\ \left({\frac{X\ -\ {\mu{}}_{\ }}{{\sigma{}}_{\ }}}\right)$

The z-score of 0.645 is z = $\left({\frac{0.645-\ 0.652}{0.003}}\right)\ =\ -2.33$

The z-score of 0.655 is z = $\left({\frac{0.655-\ 0.652}{0.003}}\right)\ =\ 1$

The area to the left of $z$ = $-2.33$ is 0.0099, and the area to the left of $z$ = $1$ is 0.8413.

The area between $\ z$ = $-2.33$ and $z$ = $1$ is 0.8413 − 0.0099 = 0.8314.

Therefore $P(0.645<X<0.655)$ = 0.8314.

Hence 83.14% of the diameters will meet the specification.

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