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In the world of statistics, a **probability density function (pdf)** plays a crucial role in defining how probabilities are distributed over different outcomes in a continuous setting. The pdf for a uniform distribution is particularly straightforward. It assigns equal likelihood to all values within a given range. This is why it is often pictorially represented as a flat, horizontal line when graphed.

For a uniform distribution defined over the interval \[A, B\], the pdf is given by the formula:

• \( f(x) = \frac{1}{B - A} \) for \( A \leq x \leq B \)

In our case, where the values of the diameter of a weld are uniformly distributed from \(0.20\) mm to \(4.25\) mm, we substitute these into the formula:
The resulting pdf is \( f(x) = \frac{1}{4.05} \), applicable to any x-value within our specified range.

Understanding the pdf of a uniform distribution helps us to calculate probabilities for different intervals by simply measuring the length of the interval and multiplying by this constant density value.

The **mean of a uniform distribution** is another important concept that represents the 'center' or average of the distribution. Since all outcomes within the interval are equally likely, the mean becomes the midpoint of the distribution.

For a uniform distribution ranging from \[A, B\], the mean can be calculated as:

• \( \text{Mean} = \frac{A + B}{2} \)

Applying this to our specific case with \(A = 0.20\) and \(B = 4.25\), the mean works out to be \(2.225\) mm. This tells us that on average, the diameter of a weld is most likely around \(2.225\) mm.

Knowing the mean is beneficial when calculating the probability of values falling within certain intervals near this central point. For example, to find the probability that a weld diameter is within \(1\) mm of the mean, we consider values from \(1.225\) mm to \(3.225\) mm. This range gives us a broader sense of where most of our data is likely concentrated.

**Calculating interval probabilities** in a uniform distribution is a straightforward task. Since each segment of a continuous distribution is equally probable, we just need to determine the length of the interval of interest.

For example, if we want to determine the probability that the weld diameter exceeds \(3\) mm, we focus on the segment from \(3\) mm to the upper bound, \(4.25\) mm. This probability is calculated by:

• \( P(X > 3) = \frac{\text{length of interval}}{\text{total range}} = \frac{1.25}{4.05} \approx 0.3086 \)

Similarly, when examining the probability of the weld diameter being exactly \(1\) mm long anywhere within the acceptable range, for any \(a\) from \(0.20 < a < a+1 < 4.25\):

• The probability simplifies to \( \frac{1}{4.05} \)

This simplicity and predictability make uniform distributions very accessible, turning interval probability calculations into a matter of straightforward application of the pdf and basic arithmetic calculations.