Question

An engineering company produces turned round bars. The manufacturing tolerance is 223.92 to 224.08 mm diameter. Given that the standard deviation is 0.03 what percentage of the product would you expect to `color(brown)("not")` be within tolerance?

Answer 1

Think I have got this correct.

As a percentage failure we have `0.768%`

Explanation:

Range of acceptable tolerance is `224.08`
`color(white) ("dddddddddddddddddddddddd")ul(223.92 larr" Subtract")`
`color(white) ("ddddddddddddddfffffffffffffddd.d")0.16`

About the mean value ( `barx` ) this is `barx +-0.16/2 = barx+-0.08`

It is given that the standard deviation `sigma=0.03`

In terms of standard deviations the tolerance is: `barx+-0.08/0.03=+-2.66bar6`

Anything outside this range is 'likely to cause damage'.

The given mean is such that we have `barx=244.00`
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Referencing tables `barx+2.66bar6 = 0.99616`

So the tail is `1-0.99616 = 0.00384`

There are two tails (1 ether end ) giving failure probability of:
`2xx0.00384= 0.00768`

As a percentage failure we have `0.00768/1xx100=0.768%`
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Tables used: Normal distribution

Publisher: Routledge
Author: Henry R Neave