Computing average on normal distribution problem

A school makes a test and its scores has average $22.8$ and standard deviation $4.8$. The distribution of scores is approximately normal.

a) Choose one person randomly. What is the probability that her score is at least 25?

b) Choose 25 persons randomly. What is the probability that the average of their scores is at least 25?

c) Which one of your computations gives more accurate probability?

I managed to do a) and b) but in c) I don't know the answer. I just assumed that the distribution of scores is exactly normal so exact probabilities are

$$1-\tfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{11/24}e^{-t^2/2}dt$$

and

$$1-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{55/24}e^{-t^2/2}dt.$$

So is the answer that computations has the same accuracy? Or how can I evaluate the error as the distribution is not exactly normal? Or if I evaluate the expressions by Sage:

sage: 1-N(1/sqrt(2*pi)*integrate(e^(-x^2/2),x,-infinity,11/24))
0.323356489807950
sage: 1-N(1/sqrt(2*pi)*integrate(e^(-x^2/2),x,-infinity,55/24))
0.0109624426491708


do I have to find out how accurate methods Sage uses to evaluate the integrals?

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For (c) you are probably supposed to say something like "the distribution of a average of several independent observations will have a closer approximation to a normal distribution due to the Central Limit Theorem"

As well as other issues, more "accurate" answers may depend on whether a score by an individual should be an integer or not. If yes, then you could make a continuity correction, so for (a) it might be better to look at $\gt 24.5$ and for (b) $\gt 24.98$.

The arithmetic accuracy of Sage doesn't really come into it. For example in R you could have the same results

> pnorm(25, mean=22.8, sd=4.8, lower.tail=FALSE)
[1] 0.3233565
> pnorm(25, mean=22.8, sd=4.8/sqrt(25), lower.tail=FALSE)
[1] 0.01096244


or with the continuity correction

> pnorm(25-1/2, mean=22.8, sd=4.8, lower.tail=FALSE)
[1] 0.361607
> pnorm(25-(1/2)*(1/25), mean=22.8, sd=4.8/sqrt(25), lower.tail=FALSE)
[1] 0.01157853

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