![]() That estimate is within 2% of the actual sample standard deviation. ![]() The estimated variance is the weighted average of the squared difference from the mean: In : var = np.average((mids - mean)**2, weights=n) In this case, it is pretty close to the mean of the original data. The estimate of the mean is the weighted average of mids: In : mean = np.average(mids, weights=n) Mids is the midpoints of the bins it has the same length as n: In : mids = 0.5*(bins + bins) This means that: Pr (mu - sigma le X le mu + sigma) approx 0.68 Pr(. First, the Empirical Rule says that the probability within 1 standard deviation from the mean is approximately 68. I'll use numpy.histogram to compute the histogram: In : n, bins = np.histogram(x) The Empirical Rule tells us about the approximate probability that is found within a certain number of standard deviations from the population mean. ![]() We'll compute the sample mean, variance and standard deviation of the input before computing the histogram. Change the increment of t to t += n*(bins - mean)**2īy the way, you can simplify (and speed up) your calculation by using numpy.average with the weights argument. You haven't weighted the contribution of each bin with n.
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