hi,

consider the following problem: find a standard deviation signa such
that 95% of all values of X (a normally distributed variable) are
within .1 of the mean mu. here's how i thought of doing it. we want to
find a sigma that'll give P(mu - .1 <= X <= mu + .1) = .95. we
standardize X to get: P((mu - .1 - mu)/sigma <= Z <= (mu + .1 - mu)/
sigma) = .95. simplifying, we get: P(-.1/sigma <= Z <= .1/sigma) = .
95. At this point, i am stuck. i can see by trial and error that if .1/
sigma = 1.96 then we get the right answer, since P(Z <= 1.96) - P(Z <=
-1.96) = .95 -- but what is the principled way to come up with 1.96?
again, i can see this by trial and error, but would like to find a
method for deriving it formally.

a second, more general question: how do we know that the answer will
necessarily be of the form of finding a value c such that P(-c <= Z <=
c) = .95? in other words why couldn't it be that some percent M of the
distribution is such that only two values c, d where c != d give P(c
= Z <= d) = M/100 ?

On 16 Mar 2007 20:00:03 -0700, perfr...@yahoo.com wrote:

hi,

consider the following problem: find a standard deviation signa such
that 95% of all values of X (a normally distributed variable) are
within .1 of the mean mu. here's how i thought of doing it. we want to
find a sigma that'll give P(mu - .1 <= X <= mu + .1) = .95. we
standardize X to get: P((mu - .1 - mu)/sigma <= Z <= (mu + .1 - mu)/
sigma) = .95. simplifying, we get: P(-.1/sigma <= Z <= .1/sigma) = .
95. At this point, i am stuck. i can see by trial and error that if .1/
sigma = 1.96 then we get the right answer, since P(Z <= 1.96) - P(Z <=
-1.96) = .95 -- but what is the principled way to come up with 1.96?
again, i can see this by trial and error, but would like to find a
method for deriving it formally.

Can you be happy with the complement? And symmetry?
P(Z >= .1/sigma) = 0.025 falls out immediately.

a second, more general question: how do we know that the answer will
necessarily be of the form of finding a value c such that P(-c <= Z <=
c) = .95? in other words why couldn't it be that some percent M of the
distribution is such that only two values c, d where c != d give P(c
= Z <= d) = M/100 ?

The *problem* does arise for distributions that are not
symmetrical, and it is worse when you add in being
discrete, such as Fisher's Exact Test.

The most popular convention may be to assume that half
the test size belongs to each end. That gives exact
limits that are not always symmetrical.