An average scanned image occupies 0.5 megabytes of memory with a standard deviation of 0.2 megabytes. If you plan to publish 80 images on your web site, what is the probability that their total size is between 49 megabytes and 53 megabytes?

The question did not specify the exact distribution of the size of those images. However, the sample size [tex]n = 80[/tex] is a rather large number. Assume that:

the sizes of these [tex]80[/tex] images follow the same distribution, with mean [tex]\mu = 0.5[/tex] megabytes and standard deviation [tex]\sigma = 0.2[/tex] megabytes.
the size of each image is independent of one another.

If both assumptions are met, the central limit theorem would apply. This theorem would suggest that the sum of the sizes of these [tex]80[/tex] image (a random variable) will approximately follow a normal distribution. The mean of that sum would be approximately [tex]n\cdot \mu= 80 \times 0.5[/tex]. The standard deviation of that sum would be approximately [tex]\sigma\, \sqrt{n} = 0.2\times \sqrt{80}[/tex].

Let [tex]\displaystyle \Sigma X[/tex] denote the sum of these eighty sizes.

Under these assumption, [tex]\displaystyle \Sigma X \stackrel{\text{app.}}{\sim} \mathrm{N}\left(n\, \mu,\, \sigma\,\sqrt{n}\right)[/tex].

That is: [tex]\displaystyle \Sigma X \stackrel{\text{app.}}{\sim} \mathrm{N}\left({80\times 0.5},\, \left(0.2\times \sqrt{80}\right)^2}\right)[/tex].

The question is asking for the probability [tex]P(49 \le \Sigma X \le 53)[/tex]. Therefore, calculate the [tex]z[/tex]-score that corresponding to [tex]\Sigma X = 49[/tex] and [tex]\Sigma X = 53[/tex]:

For [tex]\Sigma X = 49[/tex], the [tex]z[/tex]-score would be [tex]\displaystyle \frac{\sum X - n\, \mu}{\sigma \sqrt{n}} = \frac{49 - 80 \times 0.5}{0.2\times \sqrt{80}} = 2.25[/tex].
For [tex]\Sigma X = 53[/tex], the [tex]z[/tex]-score would be [tex]\displaystyle \frac{\sum X - n\, \mu}{\sigma \sqrt{n}} = \frac{53 - 80 \times 0.5}{0.2\times \sqrt{80}} = 3.25[/tex].

Make use of a [tex]z[/tex]-table to find these two probabilities:

[tex]P(X \le 49) = P(Z < 2.25) \approx 0.98778[/tex].
[tex]P(X \le 53) = P(Z < 3.25) \approx 0.99942[/tex].

Calculate the probability that this question is asking for:

[tex]\begin{aligned}& P(49 \le \Sigma X \le 53) \\ &= P(\Sigma X < 53) - P(\Sigma X < 49) \\ & \approx 0.99942 - 0.98778 \approx 0.012 = 1.2\%\end{aligned}[/tex].