Final answer:
The amount dispensed by the beverage machine is a continuous random variable. Its mean and standard deviation can be calculated using formulas for a uniform distribution. Probabilities for specific intervals or values are determined based on the uniform distribution's properties.
Step-by-step explanation:
The amount dispensed by the beverage machine is a continuous random variable because it can take on any value within a continuous range, specifically between 6.5 and 7.5 ounces. A discrete random variable, on the other hand, consists of separate, indivisible values with no possible intermediate values.
To graph the frequency function, one must plot a rectangle with a base on the x-axis from 6.5 to 7.5 and a height that ensures the total area is 1. Since this is a uniform distribution, the height will be constant across the entire range. The rectangle's area represents the probability distribution.
The mean (μ) of a uniform distribution on [a, b] is μ = (a + b) / 2, and the standard deviation (σ) is σ = √((b - a)² / 12). Inserting the values 6.5 and 7.5 for a and b, respectively, we get a mean of 7 ounces and a standard deviation of approximately 0.29 ounces. On the graph, μ ± 2σ would span the interval from the mean plus and minus twice the standard deviation.
To find P(x ≥ 7), note that since 7 is the midpoint of the distribution and it is uniform, P(x ≥ 7) = 0.5. P(x < 6) is 0 because 6 ounces is not within the range of the distribution. P(6.5 ≤ x ≤ 7.25) is calculated as the proportion of the range that falls between 6.5 and 7.25, which is 0.75. The probability that each of two bottles dispenses more than 7.25 ounces of beverage, assuming independence, is P(x > 7.25)², since each bottle situation is independent and has a probability of P(x > 7.25).