$t$-distribution and finding probability of sample mean There is a question from a statistics book:
If four observations are taken from a normal distribution what is the probability that the difference between the observed mean $\bar x$ and the true mean $\mu$ will be less than five times the observed standard deviation, $s$?
The book gives an answer "About 0.002 since the chance that a $t$-variate with 3 d.f. will exceed 10.21 is 0.001". The reason they use 10.21 is because thats just what the nearest discretisation of the t exceeding probabilities table in the book is. I cannot figure out though, why they arrive at around the 10 to look up the probability exceeding and why they give the answer 0.002, because the question states that "...will be less than five times the observed standard deviation, s?" so surely the probability should be substantial because it will cover the bulk of the mass in the center of the t-distribution curve. Is the book wrong or am I severely overlooking something?
Also, first time posting on this forum, so I may not have the text editing right. Apologies in advance.
 A: There seems a problem with this answer. If you're sure you've looked up everything correctly, I would email the author. 
How the problem should be addressed:
As it is written here, I'm going to assume that we are interested in $P(\frac{\bar x - \mu}{s} < 5)$, (not $P(\frac{|\bar x -\mu|}{s } > 5)$ as the book apparently answers). Note that this isn't even quite right, as $s$, the square root of the unbiased estimator of the variance, is not the observed standard deviation. See the bottom for more details. 
As such, we use the fact that we know that $\frac{\bar x - \mu}{s/\sqrt{n}} \sim t_{n-1}$. Because $n = 4$, we then know that 
$P(\frac{\bar x - \mu}{s} < 5) = $
$P(\frac{\bar x - \mu}{s } \times 2 < 5 \times 2) = $
$P(\frac{\bar x - \mu}{s/2} < 10) = $
Since $n = 4$, we then know that the above is equal to 
$P(t_{3}  < 10)$
I think you've got it from there. 
Now that we know how the problem should have been addressed, what was my complaint about $s$ and observed standard deviations? The observed standard deviation is, by definition, $\hat \sigma = \sqrt{\sum_{i = 1}^n \frac{(x_i - \bar x)^2}{n} }$
However, since $\hat \sigma^2$ is a biased estimator of $\sigma$, we often use 
$s = \sqrt{\sum_{i = 1}^n \frac{(x_i - \bar x)^2}{n-1} }$ 
as $s^2$ is unbiased for $\sigma^2$. But note that this is not the observed standard deviation!
