Questions: You order some books online and get an estimated delivery date of June 1-June 9. You know you will be our of town June 4th and Sth and are a little concerned abour the package arriving when you are away. Assuming the delivery date follows a discrete uniform distribution, what is the likelihood your package will be delivered while you are out of town? Round your answer to four decimal places, if necessary.

Solution Steps

The delivery date of the package follows a discrete uniform distribution over the interval from June 1 to June 9, which can be represented as \( X \sim U(1, 9) \).

The mean \( E(X) \) of a uniform distribution is calculated using the formula: \[ E(X) = \frac{a + b}{2} = \frac{1 + 9}{2} = 5.0 \]

The variance \( \text{Var}(X) \) is given by: \[ \text{Var}(X) = \frac{(b - a)^2}{12} = \frac{(9 - 1)^2}{12} = \frac{64}{12} = 5.3333 \]

The standard deviation \( \sigma(X) \) is the square root of the variance: \[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{5.3333} \approx 2.3094 \]

To find the probability that the package is delivered while you are out of town (from June 4 to June 5), we calculate: \[ P(4 \leq X \leq 5) = F(5) - F(4) \] where the cumulative distribution function \( F(x; a, b) \) is defined as: \[ F(x; a, b) = \frac{x - a}{b - a}, \quad a \leq x \leq b \] Calculating \( F(5) \) and \( F(4) \): \[ F(5) = \frac{5 - 1}{9 - 1} = \frac{4}{8} = 0.5 \] \[ F(4) = \frac{4 - 1}{9 - 1} = \frac{3}{8} = 0.375 \] Thus, the probability is: \[ P(4 \leq X \leq 5) = 0.5 - 0.375 = 0.125 \]

Final Answer