To find the standard deviation of the ages, we need to follow several steps which include finding the mean (expected value) of the distribution, the variance, and finally, the standard deviation. Here's a detailed step-by-step solution:
1. Calculate the Mean (Expected Value) of the Distribution:
The mean ([tex]\(\mu\)[/tex]) of a probability distribution is found by multiplying each possible value by its probability and summing all these products.
The formula is:
[tex]\[
\mu = \sum (x_i \times P(x_i))
\][/tex]
Where [tex]\(x_i\)[/tex] represents the age values and [tex]\(P(x_i)\)[/tex] represents their respective probabilities.
[tex]\[
\mu = (13 \times 0.08) + (14 \times 0.23) + (15 \times 0.23) + (16 \times 0.28) + (17 \times 0.15) + (18 \times 0.03)
\][/tex]
Calculating this sum, we have:
[tex]\[
\mu = 13 \times 0.08 + 14 \times 0.23 + 15 \times 0.23 + 16 \times 0.28 + 17 \times 0.15 + 18 \times 0.03 = 1.04 + 3.22 + 3.45 + 4.48 + 2.55 + 0.54 = 15.28
\][/tex]
So, the mean age is 15.28.
2. Calculate the Variance of the Distribution:
The variance ([tex]\(\sigma^2\)[/tex]) measures the spread of the distribution and is calculated using the following formula:
[tex]\[
\sigma^2 = \sum ((x_i - \mu)^2 \times P(x_i))
\][/tex]
Substituting our values:
[tex]\[
\sigma^2 = (13 - 15.28)^2 \times 0.08 + (14 - 15.28)^2 \times 0.23 + (15 - 15.28)^2 \times 0.23 + (16 - 15.28)^2 \times 0.28 + (17 - 15.28)^2 \times 0.15 + (18 - 15.28)^2 \times 0.03
\][/tex]
Calculating each term:
[tex]\[
= (2.28)^2 \times 0.08 + (1.28)^2 \times 0.23 + (0.28)^2 \times 0.23 + (0.72)^2 \times 0.28 + (1.72)^2 \times 0.15 + (2.72)^2 \times 0.03
\][/tex]
Then:
[tex]\[
= 5.1984 \times 0.08 + 1.6384 \times 0.23 + 0.0784 \times 0.23 + 0.5184 \times 0.28 + 2.9584 \times 0.15 + 7.3984 \times 0.03
\][/tex]
Then:
[tex]\[
= 0.4159 + 0.3768 + 0.0180 + 0.1452 + 0.4438 + 0.2219 = 1.6216
\][/tex]
So, the variance is 1.6216.
3. Calculate the Standard Deviation:
The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance:
[tex]\[
\sigma = \sqrt{\sigma^2} = \sqrt{1.6216} \approx 1.2734
\][/tex]
Therefore, the standard deviation of the ages, rounded to four decimal places, is:
[tex]\[
\boxed{1.2734}
\][/tex]
