Let's break down the problem.

Part (a)

You are asked to compute the standard deviation, $s$, for the data set: $5, 15, 13, 7, 11$ The standard deviation formula is: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$ where:

• $x_i$ are the data points,
• $\bar{x}$ is the mean of the data points,
• $n$ is the number of data points.

Let me compute this step by step:

1. Data points: 5, 15, 13, 7, 11

2. Mean ($\bar{x}$): $\bar{x} = \frac{5 + 15 + 13 + 7 + 11}{5} = \frac{51}{5} = 10.2$

3. Deviation from mean: Subtract the mean from each data point:

   • $5 - 10.2 = -5.2$
   • $15 - 10.2 = 4.8$
   • $13 - 10.2 = 2.8$
   • $7 - 10.2 = -3.2$
   • $11 - 10.2 = 0.8$

4. Square the deviations:

   • $(-5.2)^2 = 27.04$
   • $(4.8)^2 = 23.04$
   • $(2.8)^2 = 7.84$
   • $(-3.2)^2 = 10.24$
   • $(0.8)^2 = 0.64$

5. Sum of squares: $27.04 + 23.04 + 7.84 + 10.24 + 0.64 = 68.8$

6. Divide by $n - 1$ (degrees of freedom): $\frac{68.8}{4} = 17.2$

7. Standard deviation: $s = \sqrt{17.2} \approx 4.147$

So the correct standard deviation for part (a) is approximately 4.147.

Part (b)

You are asked to add 8 to each data value, giving a new data set: $13, 23, 21, 15, 19$ Let's compute the new standard deviation.

1. New data points: 13, 23, 21, 15, 19

2. New mean: $\bar{x} = \frac{13 + 23 + 21 + 15 + 19}{5} = \frac{91}{5} = 18.2$

3. Deviation from the new mean:

   • $13 - 18.2 = -5.2$
   • $23 - 18.2 = 4.8$
   • $21 - 18.2 = 2.8$
   • $15 - 18.2 = -3.2$
   • $19 - 18.2 = 0.8$

4. Square the deviations (same as before, because adding a constant doesn't change deviations):

   • $(-5.2)^2 = 27.04$
   • $(4.8)^2 = 23.04$
   • $(2.8)^2 = 7.84$
   • $(-3.2)^2 = 10.24$
   • $(0.8)^2 = 0.64$

5. Sum of squares: The same, $68.8$.

6. Divide by $n - 1$: $\frac{68.8}{4} = 17.2$

7. New standard deviation: $s = \sqrt{17.2} \approx 4.147$

So the correct standard deviation for part (b) is also 4.147.

Part (c)

The correct conclusion is that adding the same constant to each data value does not change the standard deviation. Therefore, the correct answer to the question is:

• "Adding the same constant $c$ to each data value results in the standard deviation remaining the same."

Let me know if you want more details or if you have questions!

Here are 5 related questions to expand your understanding:

1. How does multiplying each data point by a constant affect the standard deviation?
2. Why does adding a constant to each data point leave the standard deviation unchanged?
3. Can standard deviation ever be negative? Why or why not?
4. What is the relationship between variance and standard deviation?
5. How does standard deviation differ from range as a measure of variability?

Tip: Always remember that standard deviation measures the spread of data around the mean. Adding a constant shifts the entire dataset without affecting the spread.