This solution explains the step-by-step process to calculate the standard deviation for a given set of ages: 25, 30, 35, 40, 45, and 50 years. Standard deviation measures the amount of variation or dispersion of a set of values.
First, we find the mean ($\mu$) of the given ages. The mean is the sum of all values divided by the total number of values.
The ages are: 25, 30, 35, 40, 45, 50.
Number of observations ($n$) = 6.
Sum of ages = $25 + 30 + 35 + 40 + 45 + 50 = 225$ years.
The formula for the mean is:
$$ \mu = \frac{\sum x_i}{n} $$
Calculating the mean:
$$ \mu = \frac{225}{6} = 37.5 \text{ years} $$
Next, we find the difference between each age and the mean ($\mu$).
| Age ($x_i$) | Deviation ($x_i - \mu$) |
|---|---|
| 25 | $25 - 37.5 = -12.5$ |
| 30 | $30 - 37.5 = -7.5$ |
| 35 | $35 - 37.5 = -2.5$ |
| 40 | $40 - 37.5 = 2.5$ |
| 45 | $45 - 37.5 = 7.5$ |
| 50 | $50 - 37.5 = 12.5$ |
Now, we square each of the deviations calculated in the previous step.
| Deviation ($x_i - \mu$) | Squared Deviation $(x_i - \mu)^2$ |
|---|---|
| $-12.5$ | $(-12.5)^2 = 156.25$ |
| $-7.5$ | $(-7.5)^2 = 56.25$ |
| $-2.5$ | $(-2.5)^2 = 6.25$ |
| $2.5$ | $(2.5)^2 = 6.25$ |
| $7.5$ | $(7.5)^2 = 56.25$ |
| $12.5$ | $(12.5)^2 = 156.25$ |
The variance ($\sigma^2$) is the average of the squared deviations. Since we are given a specific group of 6 people and asked for the standard deviation of *their* ages, we treat this as a population. Therefore, we divide the sum of squared deviations by the number of observations ($n$).
Sum of Squared Deviations = $156.25 + 56.25 + 6.25 + 6.25 + 56.25 + 156.25 = 437.5$
The formula for population variance is:
$$ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n} $$
Calculating the variance:
$$ \sigma^2 = \frac{437.5}{6} \approx 72.9167 $$
The standard deviation ($\sigma$) is the square root of the variance.
The formula for population standard deviation is:
$$ \sigma = \sqrt{\sigma^2} $$
Calculating the standard deviation:
$$ \sigma = \sqrt{72.9167} \approx 8.5404 $$
We need to round the standard deviation to two decimal places.
$$ \sigma \approx 8.54 $$
Therefore, the standard deviation of the ages is approximately 8.54 years.
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