This is an exercise illustrating the Empirical Rule:
To start, let's get the correct mean $\bar X = 3466.667$ and standard
deviation $S = 961.7692.$ (Please do some checking here. Did one of us type the observations incorrectly,
or did you incorrectly compute $\bar X$ and $S$.)
Then observations 3000, 3300, 3300, 3600, 3900, and 4350 (six of the nine)
are within one standard deviation of the mean. (Please verify this: Find the
endpoints of the interval $\bar X \pm S$. Then check which observations
fall within the interval.)
So the desired proportion
is $6/9 \approx 66.7\%$ (The Empirical Rule suggests this proportion
is often 'about 68%`, so your percentage is pretty close. The ER is an approximate rule;
don't necessarily expect exact results.)
For reference, computations from R statistical software are as follows:
x = c(2250,2250,3000,3300,3300,3600,3900,4350,5250)
a = mean(x); s = sd(x)
a; s
## 3466.667 # average
## 961.7692 # SD
x[abs(a-x)<=s] # list of observations within one SD of average
## 3000 3300 3300 3600 3900 4350
pm = c(-1,1); a + pm*s
## 2504.897 4428.436
mean(abs(x-a)<s) # proportion of obs within one SD of avg
## 0.6666667
One more thing: Now, to consolidate you understanding, please try this: The ER suggests that about 95% of
the observations fall within two standard deviations of the mean. How
many of your observations fall within the interval $\bar X \pm 2S ?$
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 $\mu$, $\sigma$.
Best Answer
Let $X$ be normally distributed with mean $\mu$ and standard deviation $\sigma$. Then $$\Pr(\mu-3\sigma\le X\le \mu+3\sigma)=\Pr\left(\left|\frac{X-\mu}{\sigma}\right|\le 3\right)=\Pr(|Z|\le 3),$$ where $Z$ is standard normal. From tables, we find that $\Pr(Z\le 3)\approx 0.99865$. Thus $$\Pr(\mu-3\sigma\le X\le \mu+3\sigma)\approx 2(0.99865-0.5)=0.9973.$$ Informally, the probability that $X$ is more than $3$ standard deviation units away from the mean is $\approx 2.7\times 10^{-3}$, quite small. Thus only a small proportion of observations will lie outside the interval $\mu\pm 3\sigma$, which has width $6\sigma$.