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The expected value, often referred to as the mean, is a fundamental concept in probability and statistics. It represents the average or mean value that a random variable is expected to have over many trials or instances. To find the expected value for a continuous random variable, like in our problem, we use the integral of the product of the variable and its probability density function (PDF).
In mathematical terms, the expected value of a continuous random variable \(X\) is given by:
\[ E(X) = \int_{a}^{b} x \, f(x) \, dx \]where ‘\(a\)’ and ‘\(b\)’ are the bounds of \(X\), and \(f(x)\) is the PDF.

• For our specific case, the bounds are from 0 to 2, and \(f(x) = 0.75x(2-x)\).
• This integral calculation gives us \(E(X) = 1\), implying that, on average, the store will take 1 year before making a profit or shutting down.

It is crucial to understand expected value, as it helps in predicting long-term outcomes in uncertain situations.

Variance is a measure of how spread out the values of a random variable are around the mean. It provides insight into the variability or dispersion in a set of values. A high variance means more spread out values, while a low variance indicates values are clustered closely around the mean.
For a continuous random variable, variance is computed using:
\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]where \(E(X^2)\) is the expected value of the square of the variable. For our problem, the variance calculation involves first finding \(E(X^2)\) by integrating:

• The calculation gives \(E(X^2) = 1.2\).
• Substituting \(E(X) = 1\) from earlier, we find \(\text{Var}(X) = 1.2 - 1 = 0.2\).

This variance of 0.2 indicates a relatively small spread, suggesting that the number of years until a store might make a profit is not widely scattered.

Standard deviation is simply the square root of the variance, providing a measure of spread in the same units as the random variable itself. It is a popular statistic because, unlike variance, it presents the spread variability in more interpretable units (years, dollars, etc).
The standard deviation for our random variable \(X\) is:
\[ \text{SD}(X) = \sqrt{\text{Var}(X)} \]Using our earlier result,

• \( \text{SD}(X) = \sqrt{0.2} \approx 0.447 \).

This means in practice, the number of years a store takes to either make a profit or go out of business will likely deviate from the mean by about 0.447 years. It's a practical tool to assess risk or predict variability.

Probability calculations help us determine the likelihood of certain events occurring within a given set of conditions. In the context of a continuous random variable, like our store's time to profit, these calculations involve integrating the probability density function over a specific range.
To find the probability that the store will take at least 1.5 years before making a profit, we calculate:
\[ P(X \geq 1.5) = \int_{1.5}^{2} 0.75x(2-x) \, dx \]

• After evaluating this integral, we find \(P(X \geq 1.5) \approx 0.328\).

This result tells us that there is about a 32.8% chance the store will take 1.5 years or longer to either make a profit or close down. Understanding these probabilities helps businesses to make more informed and strategic decisions based on potential risks.