To solve for the $z$-scores, we use the formula:

$z = \frac{x - \mu}{\sigma}$

where:

• $x$ is the value of the random variable,
• $\mu = 26$ is the mean,
• $\sigma = 8.9$ is the standard deviation.

Step-by-step calculations:

(a) $x = 11$

$z = \frac{11 - 26}{8.9} = \frac{-15}{8.9} \approx -1.69$

(b) $x = 20$

$z = \frac{20 - 26}{8.9} = \frac{-6}{8.9} \approx -0.67$

(c) $x = 15$

$z = \frac{15 - 26}{8.9} = \frac{-11}{8.9} \approx -1.24$

(d) $x = 18$

$z = \frac{18 - 26}{8.9} = \frac{-8}{8.9} \approx -0.90$

(e) $x = 14$

$z = \frac{14 - 26}{8.9} = \frac{-12}{8.9} \approx -1.35$

(f) $x = 26$

$z = \frac{26 - 26}{8.9} = \frac{0}{8.9} = 0$

Final Results:

• (a) $z \approx -1.69$
• (b) $z \approx -0.67$
• (c) $z \approx -1.24$
• (d) $z \approx -0.90$
• (e) $z \approx -1.35$
• (f) $z = 0$

Would you like further clarification or detailed explanations?

Follow-Up Questions:

1. How are $z$-scores used in determining probabilities under the normal curve?
2. Can you explain how standard deviation impacts the spread of $z$-scores?
3. What happens if $\sigma$ were smaller? How would it change $z$-scores?
4. How would you find $x$ given a $z$-score and the parameters $\mu$ and $\sigma$?
5. How are $z$-scores related to percentiles in the normal distribution?

Tip:

Always double-check your subtraction and division when computing $z$-scores to ensure accuracy, as small errors can affect results significantly!