Since the mean and standard deviation of the lifetime of a battery in a portable computer are 3.5 and 1.0 hours respectively, the mean and standard deviation of the lifetime of 25 batteries in a portable computer are 3.5 and 1.0/5 hours respectively. Thus, z = (3.25 - 3.5)/0.2 .
Actually, it is not obvious that the average speed is $L/t_m$. In fact, this is not mathematically true, since in general $$\operatorname{E}[1/X] \ne 1/\operatorname{E}[X].$$
To see why, suppose you realize a sample of race times among $n$ riders: $$\boldsymbol T = (t_1, t_2, \ldots, t_n).$$ Now, their average time is $\bar T = \frac{1}{n} \sum_{i=1}^n t_i$. For a track $L = 1$ (the units are irrelevant), the race speeds observed are $$\boldsymbol S = (s_1, s_2, \ldots s_n) = (1/t_1, 1/t_2, \ldots, 1/t_n).$$ But this has the average value $$\bar S = \frac{1}{n}\sum_{i=1}^n s_i = \frac{1}{n}\sum_{i=1}^n \frac{1}{t_i} \ne \frac{n}{\sum_{i=1}^n t_i} = \frac{1}{\bar T}.$$ In fact, this inequality is especially apparent if the number of riders is small; e.g., if $n = 2$, it is quite apparent that $$\frac{1}{2}\left(\frac{1}{t_1} + \frac{1}{t_2}\right) \ne \frac{2}{t_1 + t_2}.$$ Equality is attained if and only if $t_1 = t_2$. Suffice it to say that the correct way to calculate average speed is to take the observed sample of the race times, compute $s_i = L/t_i$ for each rider $i$, and take the arithmetic mean.
For the sample standard deviation, you'd do the same thing.
However, from a theoretical standpoint, it is immediately clear that since $$\operatorname{E}[1/Z] = \int_{z=-\infty}^\infty \frac{1}{z} e^{-z^2/2} \, dz$$ is unbounded for a standard normal random variable $Z$, you have no hope of obtaining a closed-form expression for the expectation (let alone standard deviation) of the reciprocal of a normal random variable with general parameters $\mu$, $\sigma$.
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I had the very same issue. As @mihaild mentions in the comment, your solution is the inverse of the true one. The real reason for this is in the text though.
What you (and I) calculated was the probability that Yuki's time is greater than Zana's. However, the question is about the probability that Yuki's time is faster, not greater. Because, of course, she is faster when her time is lower.