4 Suppose we choose independently \(25\) numbers at random (uniform density) from the interval \([0,20]\). Write the normal densities that approximate the densities of their sum \(S_{25}\), their standardized sum \(S_{25}^*\), and their average \(A_{25}\).
\(a = 0\) and \(b = 20\) in the uniform density function; therefore the mean \(\mu = \frac{1}{2}(a + b) = \frac{1}{2}(0 + 20) = 10\) and the variance \(\sigma^2 = \frac{1}{12}(b - a)^2 = \frac{1}{12}(20 - 0)^2 = \frac{400}{12} = \frac{100}{3}\). The standard deviation is the square root of the variance, so \(\sigma = \sqrt \frac{100}{3} = \frac{10}{\sqrt{3}}\).
Having determined the mean and standard deviation of the uniform distribution, we can conclude that \(S_{25}\) follows a normal distribuion with \(S_\mu = n\mu = 25 \times 10 = 250\) and \(S_\sigma = \sigma \sqrt{n} = \frac{10}{\sqrt{3}}\sqrt{25} = \frac{50}{\sqrt{3}}\). The normal probability distribution is:
\(f(x) = \frac{1}{\sigma \sqrt{2\pi}}e^{{-\frac{1}{2}}(\frac{x - \mu}{\sigma})^2}\)
So, substituting \(S_\mu\) and \(S_\sigma\):
\(f(x) = \frac{\sqrt{3}}{50\sqrt{2\pi}} e^{{-\frac{1}{2}}(\frac{x - 250}{50\sqrt{3}})^2}\)
Having approximated sum \(S_{25}\), we move on to the standardized sum \(S_{25}^*\), which is the same as \(S_{25}\) only with \(\mu = 0\) and \(\sigma = 1\). This means the probability distribution for \(S_{25}^*\) is the standard normal distribution, which is:
\(f(x) = \frac{1}{\sigma \sqrt{2\pi}}e^{{-\frac{1}{2}}x^2}\)
Lastly, \(A_{25}\) follows a normal distribution with a mean that is \(A_\mu = \frac{S_\mu}{25}\) and a standard deviation which is \(A_\sigma = \frac{1}{n}S_\sigma\sqrt{n} = \frac{1}{25}\frac{10}{\sqrt{3}}\sqrt{25} = \frac{2}{\sqrt{3}}\).
Substituting \(A_\mu\) and \(A_\sigma\) into the normal probability distribution, we get:
\(f(x) = \frac{\sqrt{3}}{2\sqrt{2\pi}} e^{{-\frac{1}{2}}(\frac{x - 10}{2\sqrt{3}})^2}\)