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In a uniform distribution, the probability density function (PDF) indicates that every outcome within a given interval is equally likely. The given PDF, where \( f(x) = 0.1 \) for cable lengths between 1200 mm and 1210 mm, exemplifies this. It is crucial to understand that this 0.1 represents the height of the uniform distribution over the interval. The complete range of values can be represented as a flat, horizontal line in a probability density plot. This height ensures that the total area under the curve within the specified interval equals 1, fulfilling the requirement that the total probability of all outcomes sums up to 1.
An important characteristic of a uniform distribution is that it relies heavily on the range of values. In this case, the range is 10 mm, from 1200 to 1210. However, any value outside this interval would naturally have a probability of zero since they are not included in the specified interval.
Understanding the PDF for a uniform distribution simplifies the calculation of probabilities across intervals, where proportions can be easily calculated by comparing sub-intervals to the total interval length.

Calculating the mean and standard deviation for a uniform distribution is straightforward once you grasp the mathematical formulas used.
For the mean of a uniform distribution, it is always halfway between the upper and lower bounds of the interval. Calculated as \( \mu = \frac{a + b}{2} \), this formula ensures that the mean represents the central point of the distribution. With our example, the interval being 1200 to 1210, the mean is \( 1205 \) mm. This value represents not only the central tendency but also the balance point of the distribution.
In contrast, the standard deviation for a uniform distribution captures the spread of values around the mean. Computed by \( \sigma = \frac{b - a}{\sqrt{12}} \), this equation acknowledges that values are equally dispersed across the entire range. For our cable lengths, the calculation, \( \frac{10}{\sqrt{12}} \approx 2.89 \) mm, suggests a standard deviation that, although not large, underscores the uniform spread of potential cable lengths. Thus, the mean and standard deviation of a uniform distribution give a quick snapshot of the distribution's balance and variability.

Determining the proportion of outcomes within specific limits for a uniform distribution involves examining the overlap between desired specifications and the original distribution's interval.
For this problem, cable specifications are 1195 to 1205 mm, and we are considering the population of cables uniformly distributed between 1200 and 1210 mm. The overlap, or shared range, thus extends from 1200 to 1205 mm.
Computing the proportion within this overlap is simple by recognizing how much of the original interval is contained within the specifications. Calculated as \( \frac{1205 - 1200}{1210 - 1200} = \frac{5}{10} = 0.5 \), we ou find that 50% of cable lengths meet the specifications.
This method highlights the straightforward approach of using the length of overlap divided by the total interval length to easily find the proportion of values within any specification limits.