In a continuous uniform distribution, the probability of an event occurring between two values $a$ and $b$ is given by the formula:

$P(a < X < b) = \frac{b - a}{d - c}$

Where:

• $a$ and $b$ are the lower and upper bounds of the range you are interested in.
• $c$ and $d$ are the bounds of the entire distribution.

In this case:

• The class lengths are uniformly distributed between 50.0 min and 52.0 min, so $c = 50.0$ and $d = 52.0$.
• We are interested in the probability that the class length is between 51.2 min and 51.4 min, so $a = 51.2$ and $b = 51.4$.

Using the formula:

$P(51.2 < X < 51.4) = \frac{51.4 - 51.2}{52.0 - 50.0}$

Now, calculating the values:

$P(51.2 < X < 51.4) = \frac{0.2}{2.0} = 0.1$

So, the probability that the class length is between 51.2 and 51.4 minutes is 0.1 or 10%.

Would you like any further explanation on this, or do you have additional questions?

Related questions:

1. How do you find the probability for other intervals in a uniform distribution?
2. What is the mean and variance of a continuous uniform distribution?
3. How would the probability change if the class times were distributed between 51.5 and 52.0 minutes?
4. Can you calculate probabilities for a normal distribution in a similar way?
5. How does the concept of continuous distributions differ from discrete ones?

Tip: For uniform distributions, always remember that the probability density is constant across the entire interval, making the calculation of probabilities straightforward using the formula above.