# Self-study Order Statistics

So I got this exercise from a book and I'm confused by a statement they made.

Example: In a 100-meter Olympic race, the running times can be considered to be $$U$$~$$(9.6, 10.0)$$-distributed. Suppose that there are eight competitors in the finals. We wish to find the probability that the winner breaks the world record of 9.69 seconds. All units are seconds.

Solution

We wish to determine $$P(X(1) < 9.69)$$.

Since $$f_X(x) = 2.5$$ for $$9.6 < x < 10.0$$ and zero otherwise, it follows that in

the interval $$9.8 < x < 10.2$$ we have $$F_X(x) = 2.5x − 24$$ and hence $$F_{X_{(1)}} (x) = 1 − (25 − 2.5x)^8$$

(since we assume that the running times are independent). Since the desired

probability equals $$F_{X_{(1)}}(9.69)$$, the answer to our problem is $$1 − (25 − 2.5·9.69)^8 \approx 0.8699.$$

My question is that I don't see why to look for the CDF on $$9.8 < x < 10.2$$ since that interval hasn't been mentioned in the question

It is a typo: they meant $$9.6 < x < 10.0$$. The CDF is still $$F_X(x) = \frac{x - 9.6}{10.0-9.6} = 2.5 x - 24$$.