Your question is a bit vague and it seems the your figure does not quite match the rest of the problem. I think you may have put parts of two similar problems together in your Question.
I'll do my best to give most of the information you requested.
You say the means of the two normal populations are unknown with $\mu_A \le \mu_B,$ and I will assume the two population standard deviations are also unknown.
If it is somehow known that the two population standard deviations are equal, $\sigma_A = \sigma_B,$ then a pooled 2-sample t test of $H_0: \mu_A = \mu_B$ against
$H_1: \mu_A < \mu_B$ is appropriate.
I would use your example with values for the two sample means and standard deviations, but I would need to know the two sample sizes in order to show how to do the test.
So I will use data with somewhat similar sample means and standard deviations, and with sample sizes $n_A = n_B = 40,$ as
sampled in R below:
set.seed(2020)
x.a = rnorm(40, 104, 10)
x.b = rnorm(40, 160, 10)
summary(x.a); length(x.a); sd(x.a)
Min. 1st Qu. Median Mean 3rd Qu. Max.
73.61 100.93 106.45 105.76 113.37 128.35
[1] 40
[1] 12.00162
summary(x.b); length(x.b); sd(x.b)
Min. 1st Qu. Median Mean 3rd Qu. Max.
142.2 154.1 160.7 160.2 165.1 192.0
[1] 40
[1] 9.79959
stripchart(list(x.a, x.b), pch="|", ylim=c(.5, 2.5))
From the summaries and the stripchart, we can see that all values of sample A are
below all values of sample B. There is a complete separation of the two samples.
With such complete separation, there is little doubt that the pooled t test will reject the null hypothesis. [The parameter var.eq=T
calls for the pooled test; without it, R does a Welch two-sample t test when two samples are provided.]
t.test(x.a, x.b, alt="less", var.eq=T)
Two Sample t-test
data: x.a and x.b
t = -22.228, df = 78, p-value < 2.2e-16
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf -50.37798
sample estimates:
mean of x mean of y
105.7579 160.2139
You can find the formulas for doing a pooled two-sample t test in a basic
statistics text. Maybe you should find the formulas and use the sample sizes,
means and standard deviations to compute the pooled variance estimate,
often called $s_p^2$ and then the test statistic $T = 22.228.$
If you choose to do the test at the significance level $\alpha = 1\%$ then
the critical value $c = 2.429$ of the test can be found from a printed table of Student's
t distributions on the row for degrees of freedom $DF = n_A + n_B - 2 = 38$
or by using software as below.
qt(.99, 38)
[1] 2.428568
You asked for a value that separates the two
distributions. Such a value is $c$ and there are probability $0.01$ of rejecting
$H_0$ when it is true. Because the two distributions are so widely separated
the probability of failing to reject $H_0$ when it is false is very small.
This means that we reject the null hypothesis
at the 1% level because $T =22.23 > 2.429.$
[If you know about P-values, the very small P-value (below 1%) is another
indication to reject $H_0.$ Ordinarily, you can't get exact P-values from
printed distribution tables.]
Note: If the distributions were as in the figure you show, then you might
choose the critical value to be $c = 1.5$ Then if you were to rely on a single
observation to decide between A and B, The probability that an observation from A would fall above $c$ is $0.0668,$ which could be found by standardizing and using printed tables of
the standard normal cumulative distribution function. This probability can be found using R (where pnorm
is a normal CDF).
1 - pnorm(1.5, 0, 1)
[1] 0.0668072
Similarly, or by symmetry,
the probability that a single observation from B would fall below $c$ is the same.
pnorm(1.5, 3, 1)
[1] 0.0668072
Addendum, per Comment. Your intuition that it is important to take variability into account is correct. Here is output from a recent
release of Minitab, which explicitly shows the pooled standard deviation. First, I use the summarized data in your Question, and assume both samples are of size 20.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 20 103.72 8.62 1.9
2 20 161.2 13.6 3.0
Difference = μ (1) - μ (2)
Estimate for difference: -57.45
95% upper bound for difference: -51.37
T-Test of difference = 0 (vs <):
T-Value = -15.94 P-Value = 0.000 DF = 38
Both use Pooled StDev = 11.3976
Now, to illustrate the role variability plays, I multiply the sample standard deviations by 10, which amounts to multiplying the variances by 100, and keep the sample sizes the same. [Of course these are no longer real data, but we can pretend.]
The effect is to make the denominator of the $T$-statistic larger, so that the statistic itself is smaller. Now the P-value is $0.06 > 0.05,$ so
the null hypothesis is not rejected at the 5% level.
Two-Sample T-Test and CI
SE
Sample N Mean StDev Mean
1 20 103.7 86.2 19
2 20 161 136 30
Difference = μ (1) - μ (2)
Estimate for difference: -57.4
95% upper bound for difference: 3.3
T-Test of difference = 0 (vs <):
T-Value = -1.59 P-Value = 0.060 DF = 38
Both use Pooled StDev = 113.9756