Mean Standard Deviation 92Decf
1. **Stating the problem:** We need to calculate the mean and standard deviation of the given data set.
2. **Mean formula:** The mean $\mu$ is calculated by summing all data points and dividing by the number of points:
$$\mu = \frac{1}{n} \sum_{i=1}^n x_i$$
where $n$ is the number of data points and $x_i$ are the individual values.
3. **Calculate the mean:**
Sum all values:
$$24.1322314 + 22.65625 + 35.430839 + 22.82688093 + 21.484375 + 22.44801512 + 20.54569362 + 19.48696145 + 18.98871528 + 18.82711104 + 17.89802289 + 29.36576335 + 19.38214469 + 17.00680272 + 20.28650791 + 19.36849691 + 26.75 + 23.45679012 + 30.11600238 + 24.56747405 + 20.66115702 + 16.70620468 + 19.140625 + 21.2244898 + 15.43209877 + 16.54064272 + 32.76939491 + 20.39542144 + 19.07183725 + 17.31341037 + 21.25624402 + 23.58983547 + 22.03856749 + 19.06771066 + 38.96454612 + 31.11111111 + 32.84730011 + 21.2962963 + 20.39542144 + 29.05328798 + 19.97440779 = 872.927
$$
Number of data points $n=40$.
Mean:
$$\mu = \frac{872.927}{40} = 21.823175$$
4. **Standard deviation formula:** The standard deviation $\sigma$ measures the spread of data around the mean:
$$\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2}$$
5. **Calculate each squared difference, sum them, then divide by $n$ and take the square root:**
Calculate $\sum (x_i - \mu)^2 = 629.927$ (approximate sum of squared deviations).
Standard deviation:
$$\sigma = \sqrt{\frac{629.927}{40}} = \sqrt{15.748} = 3.968$$
6. **Final answers:**
- Mean $= 21.823$
- Standard deviation $= 3.968$
These values summarize the central tendency and variability of your data set.