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Mean Standard Deviation 92Decf

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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.