- The assembly time \(X\) follows a uniform distribution on the interval \([6, 10]\).
- The probability of the time falling within the interval \([7, 9]\) is calculated as the length of the target interval divided by the total interval: \(\frac{9-7}{10-6} = \frac{2}{4}\).
- Simplifying this gives the probability: \(0.50\).
- The correct option is \(\boxed{a}\).
Explanation
- Understand the distribution and parameters
The assembly time \(X\) is modeled as a continuous random variable following a uniform distribution on the interval \([a, b] = [6, 10]\). In a uniform distribution, the probability of the variable falling within any sub-interval \([c, d]\) contained within \([a, b]\) is given by the ratio of the length of the sub-interval to the length of the total interval. The formula is:
\[P(c \leq X \leq d) = \frac{d - c}{b - a}\]
Here, we have \(a = 6\), \(b = 10\), \(c = 7\), and \(d = 9\).
- Calculate the probability
To find the probability that the assembly time \(X\) falls between \(7\) and \(9\) minutes, we apply the formula:
\[P(7 \leq X \leq 9) = \frac{9 - 7}{10 - 6}\]
Performing the subtractions in the numerator and the denominator:
\[P(7 \leq X \leq 9) = \frac{2}{4}\]
Simplifying the fraction:
\[P(7 \leq X \leq 9) = 0.50\]
This represents the proportion of the total possible assembly time interval that is covered by the range \(7\) to \(9\) minutes.
- Compare with options and conclude
Comparing our result \(0.50\) with the provided options:
A\(0.50\)
B\(1.00\)
C\(0\)
D\(0.75\)
We can see that the calculated probability matches option a.
Examples
Uniform distributions are frequently used in project management to model tasks where there is a minimum and maximum possible time for completion, but any value within that range is equally likely. For instance, if you are estimating the time to complete a specific task in a software project, like writing a module, you might estimate it will take between 2 and 5 hours. Knowing this, you can calculate the probability of finishing in a specific sub-window, helping with scheduling and resource management. If the time is uniformly distributed over \([2, 5]\), the probability of finishing between 3 and 4 hours is \(\frac{4-3}{5-2} = \frac{1}{3}\).