The problem asks to compute z-scores and reverse computations for given values of $x$, mean, and standard deviation in a normally distributed variable.

The formula for the z-score is:

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

where:

• $x$ is the value,
• $\mu$ is the mean (here $\mu = 19$),
• $\sigma$ is the standard deviation (here $\sigma = 6$).

Let's solve each part:

a) $x = 21$

Using the z-score formula:

$z = \frac{21 - 19}{6} = \frac{2}{6} = 0.33$

So, z = 0.33.

b) $x = 16$

$z = \frac{16 - 19}{6} = \frac{-3}{6} = -0.50$

So, z = -0.50.

c) $z = 1.67$

We reverse the formula to find $x$:

$x = \mu + z \cdot \sigma = 19 + 1.67 \cdot 6 = 19 + 10.02 = 29.02$

So, x = 29.02.

d) $z = -0.3$

$x = 19 + (-0.3) \cdot 6 = 19 - 1.8 = 17.2$

So, x = 17.2.

e) $z = 0$

When the z-score is 0, $x$ equals the mean:

$x = 19 + 0 \cdot 6 = 19$

So, x = 19.

Would you like further details or have any questions?

Here are five related questions:

1. How is the z-score used in standardizing data?
2. What happens when a z-score is negative?
3. Can z-scores be used to compare data points from different distributions?
4. How do you interpret a z-score of 1.67 in practical terms?
5. What is the empirical rule in relation to z-scores?

Tip: Z-scores can help you understand how far away a value is from the mean, making comparisons across different datasets easier.