You appear to be dealing with one of those half-tables of the normal
distribution printed by textbook publishers who want to save one or two
pages (out of 500 or 600) at the expense of clarity. Consult it more
carefully.
One comment is embedded in the query, below:
On Fri, 11 Jul 2003, Albert wrote:
> Q. There is a Chinese steel company ... the rods produced are
> normally distributed with a mean length of 300cm. The standard
> deviation is 8cm. If 22% of the sample means are more than specific
> length K, what is the value of K?
>
> Solution:
> Given: u = 300cm, sigma = 8cm, n = 14, p(>K)= 0.22
> Let K = be the length of the steel rod
>
> Since u (population) = u (sample) = 300,
> sigma (sample) = sigma /squareroot(14) = 2.138099
2.138090, actually; compare more precise value below.
> P(bar X > K) = [K - u(sample)]/2.138099 = Zvalue(0.22)
This sentence is ungrammatical. One way of stating the case
correctly:
P{bar X > K} = P{Z > [K - u(sample)]/sigma(sample)} = 0.22,
whence Z = <value read from table>, and [continue with your equation,
below].
Now consult your table again. Is it reporting P{Z > Z*}, P{Z < Z*},
P{0 < Z < Z*}, or something else; where Z* is the tabled value?
> (K - 300) / 2.138089935 = Zvalue(0.22)
> (K - 300) / 2.138089935 = 0.58
> K = 2.138089935(0.58) + 300
> K = 301.2401cm
I'm glad to see you restore the units of measurement, even if only at
the end of the computation. I sometimes have great difficulty
persuading students that "301.24", e.g., is a meaningless number if it
is not accompanied by the units in which the value referred to is
measured.
< snip, the rest >
-----------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816
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