Let's solve the problem step by step.

a. Finding the Mean

The mean (average) of a set of numbers is calculated by summing all the values and dividing by the number of values.

Given dataLet's solve the problem step by step.

a. Finding the Mean

The mean (average) of a set of numbers is calculated by summing all the values and dividing by the number of values.

Given $8.8, 7.1, 7.2, 8.2, 7.6, 8.6$

The mean $\bar{x}$ is calculated as: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$ Where $n = 6$ is the number of samples, and $x_i$ are the observed values.

$\bar{x} = \frac{8.8 + 7.1 + 7.2 + 8.2 + 7.6 + 8.6}{6}$

$\bar{x} = \frac{47.5}{6} \approx 7.92 \, \text{mg/L}$

b. Finding the Variance

The variance measures how much the values differ from the mean.

The formula for variance $\sigma^2$ is: $\sigma^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n}$

1. Calculate the differences from the mean for each sample:

   • $8.8 - 7.92 \approx 0.88$
   • $7.1 - 7.92 \approx -0.82$
   • $7.2 - 7.92 \approx -0.72$
   • $8.2 - 7.92 \approx 0.28$
   • $7.6 - 7.92 \approx -0.32$
   • $8.6 - 7.92 \approx 0.68$

2. Square each difference:

   • $0.88^2 \approx 0.7744$
   • $(-0.82)^2 \approx 0.6724$
   • $(-0.72)^2 \approx 0.5184$
   • $0.28^2 \approx 0.0784$
   • $(-0.32)^2 \approx 0.1024$
   • $0.68^2 \approx 0.4624$

3. Sum these squared differences:

   • $0.7744 + 0.6724 + 0.5184 + 0.0784 + 0.1024 + 0.4624 \approx 2.6084$

4. Finally, divide by the number of samples: $\sigma^2 = \frac{2.6084}{6} \approx 0.4347 \, \text{mg}^2/\text{L}^2$

Final Answers:

• Mean: $7.92 \, \text{mg/L}$
• Variance: $0.4347 \, \text{mg}^2/\text{L}^2$

Would you like more details or have any questions?

Related Questions:

1. How do you calculate the standard deviation from the variance?
2. What is the importance of variance in analyzing data?
3. How would the variance change if an additional sample were added?
4. Can variance be negative? Why or why not?
5. How does the sample size affect the variance and standard deviation?

Tip:

Always ensure that you square the differences when calculating variance to avoid cancellation of positive and negative deviations from the mean.