Z-score from Pvalue What is the Z-score for a 10% confidence level (i.e. 0.1 pvalue)?
I want the standard answer used for including in my thesis write up. I googled and used excel to calculate as well but they are all slightly different. 
Thank You. 
 A: The term 'P-value' is usually used in connection with hypothesis
testing, and the term 'confidence level' is usually used in connection with confidence intervals.
The title of your question notwithstanding, let's get to the
question itself.
I believe you want a confidence interval (CI) with 90% confidence
level. Here are two examples.
(1) Confidence for normal mean. You have a sample $X_1, X_2, \dots X_n$ from a normal
population with unknown population mean $\mu$ and known
population standard deviation $\sigma_0$. Then a 95% CI
for (estimating) $\mu$ is 
$$\bar X \pm 1.645 \frac{\sigma_0}{\sqrt{n}},$$
where $\bar X$ is the mean of your sample. 
This is obtained (with a little algebra) from the following probability equation:
$$ P\left(-1.645 < \frac{\bar X - \mu}{\sigma_0/\sqrt{n}} = Z < 1.645\right) = 0.90 = 90\%,$$
Where $Z$ is a standard normal random variable. The number 1.645
cuts the probability (area) 5% from the upper tail of the standard normal distribution, and the number -1.645 cuts probability 5% from the lower tail. This leaves 90% probability in the central part
of the distribution. 
Using normal tables. Looking at a printed table of standard normal probabilities,
find 1.64 and 1.65 in the margins of the table, and notice
that the corresponding probabilities in the body of the table
are  0.9495 and 0.9505, which average to 0.9500. That means
95% of the probability is the the left of 1.645 and 5% is above.
By symmetry, -1.645 cuts 5% from the lower tail of the distribution.
Hypothesis testing. One might also say that we reject the null hypothesis $H_0: \mu = \mu_0$ against the alternative $H_a: \mu \ne \mu_0$ at the 10%
level of significance if the 'z-score' 
$$Z = \frac{\bar X - \mu_0}{\sigma_0/\sqrt{n}}$$
has absolute value $|Z| > 1.645.$ This is equivalent to
saying that the hypothetical value $\mu_0$ does not lie
in the 90% confidence interval above.
Large sample test when $\sigma_0$ is unknown. Sometimes, if you don't know $\sigma_0$ and $n$ is big enough
it is safe to say that the sample standard deviation $S$ is
close enough to $\sigma_0$ that you can use the approximate
90% CI $\bar X \pm 1.645S/\sqrt{n}.$ (Note. Unless $n > 30,$ it
would much better to use Student's t distribution here. But that
is another topic.)
(2) Confidence interval in a public opinion poll. A somewhat similar kind of CI is used in public opinion polls to express
one's the 'margin of sampling error'. Suppose 1308 people in a
random sample of 2500 favor Proposition A on an upcoming ballot.
then the estimate of the proportion $p$ of the population in favor
is $\hat p = 1308/2500 = 0.5232.$  In this case, pm can show that
$$Z = \frac{\hat p - p} {\sqrt{\hat p (1- \hat p)/n}}$$
is approximately normally distributed. Then a 90% CI for $p$ is
$$\hat p \pm 1.645\sqrt{\hat p(1-\hat p)/n},$$
which computes to the interval $(0.507, 0.540).$ So we might have
'90% confidence' that Proposition A currently has majority support.
Extensions. Your question was not very specific. These are a few topics related
to it. If the above discussion does not directly answer what you have in mind, I
hope it points the way to relevant topics in an elementary statistics textbook. You chose to ask about the 10% 'error probabilities'.
If it were 5% error probabilities (95% CIs or testing at the 5% level of significance), then simply use 1.960 instead of 1.645.
