Let's go step by step to solve this.

Given:

• Ages: 16, 25, 22, 19, 42, 22
• There are 6 students.

Step 1: Calculate the average age.

The average age is the sum of all the ages divided by the number of students.

$\text{Average age} = \frac{16 + 25 + 22 + 19 + 42 + 22}{6}$

Let's calculate it.

$\text{Sum of ages} = 16 + 25 + 22 + 19 + 42 + 22 = 146$ $\text{Average age} = \frac{146}{6} = 24.33 \, \text{(rounded to two decimal places)}$

So, the average age $\bar{x}$ is approximately 24.33 years.

Step 2: Calculate the standard deviation.

The formula for the standard deviation is:

$\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$

Where:

• $x_i$ are the individual ages,
• $\bar{x}$ is the average age (24.33),
• $n$ is the number of students (6).

Now, let's compute the squared differences for each age:

1. $(16 - 24.33)^2 = (-8.33)^2 = 69.39$
2. $(25 - 24.33)^2 = (0.67)^2 = 0.45$
3. $(22 - 24.33)^2 = (-2.33)^2 = 5.43$
4. $(19 - 24.33)^2 = (-5.33)^2 = 28.41$
5. $(42 - 24.33)^2 = (17.67)^2 = 312.37$
6. $(22 - 24.33)^2 = (-2.33)^2 = 5.43$

Sum of squared differences:

$69.39 + 0.45 + 5.43 + 28.41 + 312.37 + 5.43 = 421.48$

Finally, calculate the standard deviation:

$\sigma = \sqrt{\frac{421.48}{6}} = \sqrt{70.25} \approx 8.38$

So, the standard deviation $\sigma$ is approximately 8.38 years.

Summary:

• Average age: 24.33 years
• Standard deviation: 8.38 years

Would you like to dive deeper into any of the steps or calculations?

Here are 5 related questions for further practice:

1. What would happen to the average and standard deviation if an additional student aged 30 joined the group?
2. How would the standard deviation change if all the students were the same age?
3. What is the variance in the ages of the students?
4. Can you calculate the median age for the group?
5. How does the range (difference between the highest and lowest ages) relate to the standard deviation?

Tip: The standard deviation gives you an idea of how spread out the data points (ages) are from the mean.