My question is how to estimate how many samples are above 1 standard deviation of the mean. I do not know individual values, hence why it's an estimate. The sum of x is 243 sum of x^2 is 2317, n = 30, range is 13(4 to 17). I have calculated mean (8.1) and standard deviation (3.41). From this information, how do I estimate how many of the sample are between 11.51 and 17? The distribution is clearly not normal since the lowest value is only 4.1 below the mean, whilst the highest value is 8.9 above it, so I'm guessing that less than 16% of values are above 11.51. This would imply that less than 0.16×30 = 4.8 of the samples are in that range. Is this correct? Any guidance appreciated.
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2$\begingroup$ "Clearly not normal" is not so clear to me, because extreme values tell you little about the distribution. What matters is how it departs from normality and what effect this might have on the answer. This kind of question is trying to gauge your comprehension of the 68-95-99.7 Rule. Although it is frequently applied to and stated only for a Normal distribution, it works surprisingly well even from some fairly skewed distributions. $\endgroup$– whuber ♦Commented Sep 18, 2021 at 19:13
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1$\begingroup$ "This would imply that less than 0.16×30 = 4.8 of the samples are in that range." sounds reasonable according to the "68%-95%-99.7% rule." Five out of 30 is just over (100%-68%)/2 =16%. // Remember it's a "rule" that is often useful, not an exact theorem. [The rule is close to exact for a normal population, but is often used for nearly normal samples.] $\endgroup$– BruceETCommented Sep 18, 2021 at 21:33
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$\begingroup$ To both whuber and BruceET, thanks. Thought I was on the right lines. I was aware of the likelihood that the upper value was an extreme case, but not that the distribution 'rule' could still be considered fairly reliable for this scenario. Thanks again $\endgroup$– PixelsgamerCommented Sep 18, 2021 at 22:14
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Extended Comment with simulated samples from various distributions:
Continuing from @whuber's Comment and yours, here are examples of simulated samples of size $n = 1000$ from normal, uniform, exponential, gamma, beta, Poisson, and binomial populations.
Some are skewed, some have support other than the whole real line, and the last two are discrete.
The part of the Empirical Rule for the proportion of observations exceeding one sample standard deviation above the sample mean (right-sided) holds "fairly well" for all of them: Proportions are between 12% and 22%.
set.seed(2021)
x = rnorm(10^3, 100, 15)
mean(x > mean(x)+sd(x))
[1] 0.161
x = runif(10^3, 30, 50)
mean(x > mean(x)+sd(x))
[1] 0.211
x = rexp(10^3, 5)
mean(x > mean(x)+sd(x))
[1] 0.135
x = rgamma(10^3, 3, .05)
mean(x > mean(x)+sd(x))
[1] 0.15
x = rbeta(10^3, 3, 4)
mean(x > mean(x)+sd(x))
[1] 0.162
x = rpois(10^3, 10)
mean(x > mean(x)+sd(x))
[1] 0.155
x = rbinom(10^3, 30, .3)
mean(x > mean(x)+sd(x))
[1] 0.169
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$\begingroup$ This doesn't verify the 68-95-99.7 rule, which is *two sided:" you need to compare
abs(x-mean(x))tosd(x)instead. Here's the intended set of simulations:sapply(list(rnorm, runif, rexp, function(n) rgamma(n, 3), function(n) rbeta(n, 3, 4), function(n) rpois(n, 10), function(n) rbinom(n, 30, 0.3)), function(f) (function(x) mean(abs(x - mean(x)) <= sd(x))) (f(1e3)))$\endgroup$– whuber ♦Commented Sep 19, 2021 at 19:27 -
1$\begingroup$ Right. And thanks for code. Edited to make limitation clear. // Behavior is somewhat different on the left side, especially for distributions that don't take negative values. $\endgroup$– BruceETCommented Sep 19, 2021 at 19:34
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1$\begingroup$ Exactly: the idea is that the deviations from the rule on one side of the mean often compensate for deviations on the other side. Your choices of distributions cover a lot of ground and show how much that 68% value might really vary. $\endgroup$– whuber ♦Commented Sep 19, 2021 at 19:39