@Semoi's Answer (+1) in terms of a 95% confidence interval is a correct
approach for this two-sided test at the 5% level (but it should be centered at the class average 535). Here I show the test in
what I hope is a familiar format.
You need to think of $\mu = 525$ as a population mean and $\sigma = 80$ as the population standard deviation (for the 'region'). You need to think of
$\bar X = 535$ as the mean of a sample of size $n = 90$ (this one class) as
explained in @NuclearHoagie's Comment.
"Different from" implies a 2-sided alterative for your z-test. The test statistic is
$z = \frac{535 - 525}{80/\sqrt{90}}= 1.18585.$ The two-sided P-value is
$$P(|Z| > 1.18585) = 0.2357 > 0.05 = 5\%,$$ where $Z$ is standard normal.
So do not reject
$H_0:\mu = 525$ against the two-sided alternative at the 5% level.
In R:
z = (535-525)/(80/sqrt(90)); z
[1] 1.185854
2*pnorm(-z)
[1] 0.2356799
From a recent release of Minitab statistical software (which has a one-sample z-test procedure for summarized data):
One-Sample Z
Test of μ = 525 vs ≠ 525
The assumed standard deviation = 80
N Mean SE Mean 95% CI Z P
90 535.00 8.43 (518.47, 551.53) 1.19 0.236
Below is a simulation of class averages a
from a thousand
classes of size $n = 90.$ The summary shows class averages were
as small as 479 and as large as 557. Half of the 1000 classes
had scores between 518 and 532. So an average score of 535 in a class of
90 students is not surprising. [Such a class with an average score of 555 would be surprising.]
set.seed(2021)
a = replicate(1000, mean(rnorm(80, 525, 90)))
summary(a)
Min. 1st Qu. Median Mean 3rd Qu. Max.
479.1 518.0 525.3 525.4 532.3 557.7
A histogram of the 1000 simulated class averages is shown below. The vertical dashed line is at 535.