Let's break down this problem step by step.

Part (a): Calculate $\bar{x}$ (the sample mean) and the deviations from the mean

The sample data points for Young's modulus (in GPa) are: $116.8, 115.5, 114.9, 115.1, 115.6$

1. Calculate the mean ($\bar{x}$):

$\bar{x} = \frac{116.8 + 115.5 + 114.9 + 115.1 + 115.6}{5}$

Let me compute the mean.

$\bar{x} = \frac{577.9}{5} = 115.58$

So, $\bar{x} = 115.58$.

2. Calculate the deviations from the mean:

The deviations for each data point are calculated as: $\text{deviation} = x_i - \bar{x}$ Where $x_i$ are the individual data points.

Let me calculate these deviations for each data point:

• For $116.8$, deviation = $116.8 - 115.58 = 1.22$
• For $115.5$, deviation = $115.5 - 115.58 = -0.08$
• For $114.9$, deviation = $114.9 - 115.58 = -0.68$
• For $115.1$, deviation = $115.1 - 115.58 = -0.48$
• For $115.6$, deviation = $115.6 - 115.58 = 0.02$

So, the deviations are: $1.22, -0.08, -0.68, -0.48, 0.02$

Part (b): Calculate the sample variance ($s^2$) and sample standard deviation ($s$)

The formulas are:

• Sample variance ($s^2$):

$s^2 = \frac{1}{n - 1} \sum_{i=1}^{n} (x_i - \bar{x})^2$

• Sample standard deviation ($s$):

$s = \sqrt{s^2}$

Where $n = 5$ is the number of observations, and the deviations are squared before summing.

Now, let's compute the squared deviations and the sample variance:

• $(1.22)^2 = 1.4884$
• $(-0.08)^2 = 0.0064$
• $(-0.68)^2 = 0.4624$
• $(-0.48)^2 = 0.2304$
• $(0.02)^2 = 0.0004$

The sum of squared deviations is: $1.4884 + 0.0064 + 0.4624 + 0.2304 + 0.0004 = 2.188$

Now calculate the sample variance $s^2$:

$s^2 = \frac{2.188}{5 - 1} = \frac{2.188}{4} = 0.547$

Finally, calculate the sample standard deviation $s$:

$s = \sqrt{0.547} = 0.740$

Final answers:

• Sample variance ($s^2$): $0.547 \, \text{GPa}^2$
• Sample standard deviation ($s$): $0.740 \, \text{GPa}$

Would you like more details or have any other questions?

Related Questions:

1. How is sample variance different from population variance?
2. What is the significance of using the sample standard deviation?
3. Can you explain when to use degrees of freedom in variance calculation?
4. How does Young's modulus relate to the physical properties of materials?
5. What are other methods of measuring material stiffness?

Tip:

In small samples, the variance and standard deviation help account for potential variation in the data, providing a measure of how spread out the values are from the mean.