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How would you interpret the standard deviation of a discrete random variable and state it simply for someone with limited statistical background? For example,

Assume I am betting on an event with a 21% chance of occurring. If I bet it will occur and am correct, I win \$50. If I lose, I pay \$20. Let $X=\text{profit from bet}$.

$P(x=\$50) =0.21$

$P(x=-\$20)=0.79$

I logically understand the expected value $\mu$, which in this case $\mu\approx-\$5$. If I was to explain what this value meant to someone, I would say:

If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5 per bet.

The standard deviation ($\sigma$) is equal to approximately \$29. But I struggle for some reason with how to explain this conceptually. Does this refer to the average deviation of each individual bet, just like $\mu$? Would it be valid to say:

If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5 per bet, plus/minus \$29.

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    $\begingroup$ Standard deviation is not the average amount by which an observation differs from the mean. That’s called mean absolute deviation. $MAD=\mathbb{E}[\vert X-\mu_X\vert]\le \sqrt{\mathbb{E}[(X-\mu_X)^2]}=sd$ (usually strict inequality, $<$). // Getting to your question, how would you explain standard deviation in the case of a continuous or even Gaussian random variable? $\endgroup$
    – Dave
    Oct 1 '20 at 9:56
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    $\begingroup$ Your final conclusion is definitely wrong. The uncertainty in the average long term outcome depends on how many bets you made and, for a sufficiently large number of bets, is arbitrarily small. You appear to confuse the standard deviation with the standard error. Good intro stats books do an excellent job of explaining this to people with limited (and no) statistical backgrounds. I am fond of Freedman, Pisani, & Purves, but many others of similar expository quality have appeared in the last generation, too. Consider consulting one of them. $\endgroup$
    – whuber
    Oct 1 '20 at 13:21
  • $\begingroup$ I have consulted a textbook and multiple websites and have yet to find a clear explanation, which is why I've posted my question. $\endgroup$ Oct 1 '20 at 17:08
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Since the technical definition of a standard deviation is non-trivial, any attempt to explain it back to its definition is going to be difficult for a layperson to understand. If you really want to impart the technical meaning then you are going to have to grind through the formula and explain it a bit at a time. If you prefer to avoid this, I recommend that you explain the quantity it in a purely heuristic way ---e.g., "Standard deviation is a measure of spread of an uncertain outcome; the bigger the standard deviation the more variable the outcome." If you want to elaborate, you can also note that this measure has the property that shifting the outcome by a fixed amount does not affect the standard deviation, and re-scaling it commensurately rescales the standard deviation.

Would it be valid to say: "If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5 per bet, plus/minus \$29."

No, that does not make sense. A long-term average of values is a single number (albeit an uncertain one); so how can it be plus/minus something? If you are referring to some kind of interval estimate of the uncertain value, this begs the question of how you measure the uncertainty of falling in this interval (which depends on aspects of the underlying distribution beyond its standard deviation).

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