Anyone ever read an SPC book written by John S. Oakland? Need help to
understand topic discussed in the book.

Yep! But it was a long time ago now (>13-years around the time of the big surge in interest in SPC in the Petrochem Process Industries), and I no longer seem to be able to find a copy of the book on my bookshelf. Nevertheless, you could try with your question.

Kelvin B. Hales Kelvin Hales Associates Limited Consulting Process Control Engineers Web: www.khace.com

I'm referring to page 89 from the 3rd edition book. There is a sentence
in there which points to Appendix A.

In that appendix, there is a table that gives a probability that any item chosen at random from a normal distribution will fall outside a given number of standard deviation from the mean. The table shows that, at a value of µ + 1.96s, only 0.025 or 2.5 per cent of the population will exceed this length. The same proportion will be less than µ - 1.96s. Hence 95 per cent of the population will lie within µ ± 1.96s.

It follows with a case example from steel rods with mean length of 150mm and standard deviation of 5mm where 95% of the rods will have a length between:

150 ± (1.96 X 5) mm

i.e between 140.2 mm and 159.8 mm.

The case example continue by saying that 99.8% of the rod lengths should be inside the range:

µ ± 3.09s

i.e. 150 ± (3.09 X 5) or 134.55 mm to 165.45 mm.

Question is where did the value 3.09 come from? I'd look into the table in Appendix A and could not derive that number at all.

Kelvin Hales wrote:

to

the

and

bookshelf.

In that appendix, there is a table that gives a probability that any item chosen at random from a normal distribution will fall outside a given number of standard deviation from the mean. The table shows that, at a value of µ + 1.96s, only 0.025 or 2.5 per cent of the population will exceed this length. The same proportion will be less than µ - 1.96s. Hence 95 per cent of the population will lie within µ ± 1.96s.

It follows with a case example from steel rods with mean length of 150mm and standard deviation of 5mm where 95% of the rods will have a length between:

150 ± (1.96 X 5) mm

i.e between 140.2 mm and 159.8 mm.

The case example continue by saying that 99.8% of the rod lengths should be inside the range:

µ ± 3.09s

i.e. 150 ± (3.09 X 5) or 134.55 mm to 165.45 mm.

Question is where did the value 3.09 come from? I'd look into the table in Appendix A and could not derive that number at all.

Kelvin Hales wrote:

to

the

and

bookshelf.

: In that appendix, there is a table that gives a probability that any
: item chosen at random from a normal distribution will fall outside a
: given number of standard deviation from the mean. The table shows that,
: at a value of µ + 1.96s, only 0.025 or 2.5 per cent of the population
: will exceed this length. The same proportion will be less than µ -
: 1.96s. Hence 95 per cent of the population will lie within µ ±
: 1.96s.

: The case example continue by saying that 99.8% of the rod lengths : should be inside the range:

: µ ± 3.09s

: i.e. 150 ± (3.09 X 5) or 134.55 mm to 165.45 mm.

: Question is where did the value 3.09 come from? I'd look into the table : in Appendix A and could not derive that number at all.

From the table of standardized normal distribution (You have probably same table)? There you will find Fii(z)-values of .9990 (99.8%, which You are seeking for) in three locations, whereas 3.09 is the middle one.

Any help? Mp

: The case example continue by saying that 99.8% of the rod lengths : should be inside the range:

: µ ± 3.09s

: i.e. 150 ± (3.09 X 5) or 134.55 mm to 165.45 mm.

: Question is where did the value 3.09 come from? I'd look into the table : in Appendix A and could not derive that number at all.

From the table of standardized normal distribution (You have probably same table)? There you will find Fii(z)-values of .9990 (99.8%, which You are seeking for) in three locations, whereas 3.09 is the middle one.

Any help? Mp

Mikko P wrote:

wrote:

any

a

that,

population

table

same

are

The table title that I'm looking at is "Proportions under the tail of the normal distribution". Basically it is to determine Z value which is given by;

Z = (x - µ)/s

However there is no value of .9990 within the table. It shows rows and columns of number with the 1st column begins from 0.0 to 4.0 while top most row with values ranging from .00 to .09. The remaining rows and columns are numbers where 0.00003 is the smallest and 0.5000 is the largest. I think the remaining columns and rows indicates percentage of probability that lies outside at half of the normal distribution curve.

So for the sample case of 99.8%, there are 0.2% that lies outside the normal distribution curve at 0.1% each sides. By converting into fraction and referring the the table I have the Z value that lies between 3.0 and 3.1 corresponding to .0013 adnd .0009 respectively. When I interpolate to find the middle Z value for .001, I couldn't get 3.09.

wrote:

any

a

that,

population

table

same

are

The table title that I'm looking at is "Proportions under the tail of the normal distribution". Basically it is to determine Z value which is given by;

Z = (x - µ)/s

However there is no value of .9990 within the table. It shows rows and columns of number with the 1st column begins from 0.0 to 4.0 while top most row with values ranging from .00 to .09. The remaining rows and columns are numbers where 0.00003 is the smallest and 0.5000 is the largest. I think the remaining columns and rows indicates percentage of probability that lies outside at half of the normal distribution curve.

So for the sample case of 99.8%, there are 0.2% that lies outside the normal distribution curve at 0.1% each sides. By converting into fraction and referring the the table I have the Z value that lies between 3.0 and 3.1 corresponding to .0013 adnd .0009 respectively. When I interpolate to find the middle Z value for .001, I couldn't get 3.09.

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