Data. You have $n = 103$ 'grades' with values $v = (55. 60, 65, \dots, 90)$
occurring with respective frequencies $f = (2, 6, 18, 27, 25, 22, 3).$
Treating this as a sample, we use $A =\bar X = \frac 1n \sum_{i=1}^7 f_iv_i,$
where $n = \sum_{i=1}^7 f_i = 103,$ so that the sample mean $A = \bar X = 77.039$ estimates the mean $\mu$ of a population.
v = c(60, 65, 70, 75, 80, 85, 90)
f = c( 2, 6, 18, 27, 25, 22, 3)
a = sum(v*f)/sum(f); a
[1] 77.03883
Similarly, the sample variance is
$S^2 = \frac{1}{n-1}\sum_{i=1}^7 f_i(v_i - a)^2 = 44.822$ so that
the population variance $\sigma^2$ is estimated by $44.822.$ Also,
the population standard deviation is estimated by $S = \sqrt{44.822}
= 6.695.$
n = sum(f); sum(f*(v-a)^2)/(n-1)
[1] 44.82201
Shapiro-Wilk test of normality. The 103 observations are not normally distributed, because they have been rounded to only seven different values. For data not so severely rounded,
an excellent test of normality would be the Shapiro-Wilk test. But your
103 observations fail this test with a P-value $0.00012$ far below $0.05.$
x = rep(v, times=f)
shapiro.test(x)
Shapiro-Wilk normality test
data: x
W = 0.9384, p-value = 0.0001217
Chi-squared GOF test. However, one can still ask whether the unrounded values might be
consistent with the normal distribution $\mathsf{Norm}(\mu =77.039, \sigma=6.695).$ Here is a histogram where the
bins are chosen to put the observed values at the bin centers. The fit isn't excellent, but a chi-squared goodness-of-fit (GOF) test may not
reject this normal distribution as a possible fit. [See Addendum for R code.]
A difficulty with the GOF test is that it requires expected count larger than $5$ in each of the intervals, and this normal distribution doesn't
assign enough probability to the first and last intervals. Thus we are
left with five intervals: $(-\infty, 67.5],$ $(67.5, 72.5],$
$(72.5, 77,5],$ $(77.5, 82.5],$ $(82.5, \infty).$
We need to know the probability that $\mathsf{Norm}(77.039, 6.695)$ assigns to each of the five intervals. That can be found using the normal CDF, called pnorm in R. (This is where you had trouble in your Question.)
int.end = c(-Inf, 67.5, 72.5, 77.5, 82.5, Inf)
int.prb = diff(pnorm(int.end, 77.04, 6.695))
sum(int.prb)
[1] 1
Then, according to the normal model we can find the expected counts $E_i$ in
each interval by multiplying by $n = 103.$ We combine observed counts $O_i$ for the first and last intervals.
int.exp = n*int.prb
int.obs = c(8, 18, 27, 25, 25)
Here is a table of the observed and expected counts, which are used
to find the chi-squared GOF statistic.
cbind(int.obs, int.exp)
int.obs int.exp
[1,] 8 7.93993
[2,] 18 17.69146
[3,] 27 28.68967
[4,] 25 27.31845
[5,] 25 21.36049
The chi-squared GOF statistic
$$Q = \sum_{i=1}^5 \frac{(O_i - E_i)^2}{E_i} = 0.922$$
has approximately a chi-squared distribution with $\nu = 5 - 3$
degrees of freedom. The P-value 0.922 is far above 5% so we do
not reject the normal model. Rejection would have required $Q$ to
exceed the critical value $c = 5.99.$
q = sum((int.obs - int.exp)^2/int.exp); q
[1] 0.9222296
1 - pchisq(q, 2)
[1] 0.6305803
qchisq(.95,2)
[1] 5.991465
If you have many of these GOF fit tests to do, you will want to find
convenient R code for doing the whole process at once. Some of the code suggested in @Joe's Answer (+1) may be helpful. Just remember
that you need to check that expected counts exceed 5, and combine
bins of they don't.
Approximate methods. Finally, here is an approximate method for using the Shapiro-Wilk test.
As we saw at the beginning of this Answer, this test rejected your normal model, likely because the observations are rounded to only a few integers.
If you add random noise to these observations, roughly speaking, to
'unround' them, then they pass the Shapiro-Wilk test. In your case
random noise from the distribution $\mathsf{Unif}(-2.5, 2.5)$ should
suffice. Then we have approximate P-value 0.318 and the normal model
is not rejected. (You will get a somewhat different answer each time
you try this random approximation: subsequent runs gave P-values 0.1856 and 0.4348.)
set.seed(1234)
shapiro.test(x + runif(n, -2.5, 2.5))
Shapiro-Wilk normality test
data: x + runif(n, -2.5, 2.5)
W = 0.98538, p-value = 0.318
If you have many of these tests to do, you might try this approximate
procedure as a screening device, and then check cases where normality
is rejected with the chi-squared GOF test.
Looking at histograms and
normal probability plots might also give quick view. Here is a normal probability plot of your data.
qqnorm(x); qqline(x, col="red")
If the 'clusters' touch the red line (except possibly for the smaller
first and last ones), you might take that as some evidence of a fit to normal.
Addendum per Comments: Here is R code for the figure above with histogram of data and density curve. The first few lines make sure the data is in memory.
v = c(60, 65, 70, 75, 80, 85, 90)
f = c( 2, 6, 18, 27, 25, 22, 3)
x = rep(v, times=f)
cutp = seq(60-2.5, 90+2.5, by = 5)
hist(x, prob=T, br=cutp, col="skyblue2", xlim=c(40,110), ylim=c(0,.06))
curve(dnorm(x, 77.03, 6.695), add=T, lwd=2, col="red")
