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In a discrete uniform distribution, the calculation of the mean offers an average value of the possible outcomes. This helps us understand the tendency or the central value of the distribution.
For a discrete uniform distribution between integers 'a' and 'b', the mean is calculated using the formula \[E(X) = \frac{a+b}{2} \]This formula is derived from the idea that each value between 'a' and 'b' is equally likely, and the average of these values gives the central tendency.
In the original problem, since the random variable \(X\) ranges from 0 to 9, the mean, or expected value, \(E(X)\) is:\[E(X) = \frac{0+9}{2} = 4.5\]This result indicates that, on average, the possible outcomes for \(X\) tend to cluster around 4.5.
When scaling \(X\) by a factor of 5 to get \(Y = 5X\), the mean gets scaled too, given by:\[E(Y) = 5 \cdot E(X) = 5 \cdot 4.5 = 22.5\]This demonstrates that multiplying the variable increases the mean by the same factor, showing a consistent shift in expected outcome.

Variance provides a measure of how much the values of a distribution are spread out from the mean. For a discrete uniform distribution, variance indicates the extent to which the values vary about their mean value.
The variance of \(X\), having a discrete uniform distribution, is calculated using the formula:\[Var(X) = \frac{(b-a+1)^2 - 1}{12}\]This is because the squared deviations from the mean are averaged over all the values from 'a' to 'b'. In the example where \(X\) ranges from 0 to 9, we find:\[Var(X) = \frac{(9-0+1)^2 - 1}{12} = 8.25\]This value (8.25) shows how much the possible outcomes of \(X\) typically differ from their average value.
When scaling the variable to \(Y = 5X\), the variance changes in a squared manner, calculated as:\[Var(Y) = 5^2 \cdot Var(X) = 25 \cdot 8.25 = 206.25\]This illustrates that when a variable is multiplied by a constant, its variance increases by the square of that constant, leading to an increased spread of the distribution.

Standard deviation represents the average amount of variability in your dataset. It is the square root of the variance and provides insight into the clustering of data points around the mean.
The formula for standard deviation is:\[\sigma(X) = \sqrt{Var(X)}\]By substituting the variance of \(X\) obtained earlier:\[\sigma(X) = \sqrt{8.25} \approx 2.87\]What this tells us is that most of the values of \(X\) lie within 2.87 units from the mean (4.5) on either side.
When the variable is modified to \(Y = 5X\), the standard deviation also adjusts, calculated as:\[\sigma(Y) = \sqrt{Var(Y)} = \sqrt{206.25} \approx 14.36\]Hence, we see that scaling a random variable by a factor similarly scales the standard deviation by the same amount. This reflects in how spread out the variable \(Y\) is compared to \(X\), further emphasized by the larger standard deviation.