Certainly! We are dealing with a problem involving a uniform distribution. Let’s break down the problem and solve it step-by-step.
### Step 1: Understanding the Uniform Distribution
For a uniform distribution between two values [tex]\( a \)[/tex] and [tex]\( b \)[/tex], every interval of the same length within [tex]\([a,b]\)[/tex] is equally likely. The probability density function (PDF), [tex]\( P(x) \)[/tex], for a uniform distribution over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ P(x) = \frac{1}{b - a} \][/tex]
where [tex]\( a \)[/tex] is the minimum time and [tex]\( b \)[/tex] is the maximum time.
### Step 2: Define the Parameters
Given:
- Minimum time ([tex]\( a \)[/tex]): 12 minutes
- Maximum time ([tex]\( b \)[/tex]): 19 minutes
### Step 3: Probability Density Calculation
We can calculate the probability density [tex]\( P(x) \)[/tex]:
[tex]\[ P(x) = \frac{1}{19 - 12} = \frac{1}{7} \][/tex]
### Step 4: Define the Interval of Interest
We are interested in the time interval between 13 minutes and 18 minutes. So:
- Lower bound of the interval: 13 minutes
- Upper bound of the interval: 18 minutes
### Step 5: Calculate the Probability
The probability that the time taken to wash the dishes will fall between any two points in a uniform distribution is equal to the area under the probability density function over that interval. Since the distribution is uniform, this corresponds to the length of the interval multiplied by the uniform density.
[tex]\[ \text{Probability} = (\text{Upper bound} - \text{Lower bound}) \times P(x) \][/tex]
Substituting the given values:
[tex]\[ \text{Probability} = (18 - 13) \times \frac{1}{7} = 5 \times \frac{1}{7} \approx 0.71 \][/tex]
### Step 6: Result
Therefore, the probability that washing dishes will take between 13 and 18 minutes is approximately 0.71.
So, to two decimal places, the answer is:
[tex]\[ \boxed{0.71} \][/tex]
