How to use Standard Deviation and Mean

Question

I have the following population

| Amounts |
|---------|
| 3022.03 |
| 3870.91 |
| 2796.78 |
| 5104.81 |
| 3413.48 |
| 4409.38 |
| 7265.16 |
| 2396    |
| 827.73  |
| 653.83  |
| 1072.73 |
| 231.28  |
| 112.47  |
| 672.16  |
| 264.52  |
| 213.97  |
| 2034.23 |
| 2660.1  |
| 879.87  |
| 1234.62 |
| 1334.81 |
| 729.68  |
| 507.21  |
| 312.49  |
| 412.72  |
| 317.11  |
| 143.34  |
| 79.34   |
| 20.14   |
| 36.39   |
| 29.25   |
| 209.39  |
| 385.88  |
| 273.69  |
| 360.87  |
| 85.24   |
| 33.09   |
| 20.26   |
| 3.7     |
| 5.51    |
| 10.12   |
| 5.08    |

The mean is 1153.60 and the Standard Deviation 1649.15

The numbers within one Standard Deviation (around 70% of the cases) should be between 1153.60+1649.15=2802.75 (max) and 1153.60-1649.15=-495.55 (min) but -495.55 is a negative number.

What am I missing out?

Answer 1

You could find $a$ and $b$ where $$ \hat{F}(a) \approx \frac{1 - 0.70}{2} = 0.15 \qquad \text{and} \qquad \hat{F}(b) \approx 1 - \frac{1 - 0.70}{2} = 0.85 $$ and $\hat{F}$ is the empirical distribution. Note that this does not make any guarantees about the data generating distribution. Here is some Python code that computes it for your data, stored in the variable values:

>>> import numpy as np
>>> np.quantile(values, 0.15), np.quantile(values, 0.85)
(29.825999999999997, 2776.2780000000002)

Alternatively, you could assume a parametric model for the data generating distribution. For example, if you assume it is a log-normal distribution, then you can estimate the interval by a method of your choosing. Indeed, this is what you did above but you made the assumption of a normal distribution (arguably a bad assumption).