To solve these questions, we need to convert between $x$-values (raw scores) and $z$-values (standard scores) for a normal distribution with a mean $\mu = 4.9$ and a standard deviation $\sigma = 0.2$.

The conversion formula between $x$ and $z$ is: $z = \frac{x - \mu}{\sigma}$ and to convert from $z$ back to $x$: $x = z \cdot \sigma + \mu$

Let's go through each part:

Part (a) Convert $4.5 < x$ to a $z$ interval.

Given $x = 4.5$: $z = \frac{4.5 - 4.9}{0.2} = \frac{-0.4}{0.2} = -2$

So, the interval $4.5 < x$ converts to: $z > -2$

Part (b) Convert $x < 4.2$ to a $z$ interval.

Given $x = 4.2$: $z = \frac{4.2 - 4.9}{0.2} = \frac{-0.7}{0.2} = -3.5$

So, the interval $x < 4.2$ converts to: $z < -3.5$

Part (c) Convert $4.0 < x < 5.5$ to a $z$ interval.

1. For $x = 4.0$: $z = \frac{4.0 - 4.9}{0.2} = \frac{-0.9}{0.2} = -4.5$

2. For $x = 5.5$: $z = \frac{5.5 - 4.9}{0.2} = \frac{0.6}{0.2} = 3$

So, the interval $4.0 < x < 5.5$ converts to: $-4.5 < z < 3$

Part (d) Convert $z < -1.44$ to an $x$ interval. (Round your answer to one decimal place.)

Given $z = -1.44$: $x = (-1.44 \cdot 0.2) + 4.9 = -0.288 + 4.9 = 4.6$

So, $z < -1.44$ converts to: $x < 4.6$

Part (e) Convert $z < 1.28$ to an $x$ interval. (Round your answer to one decimal place.)

Given $z = 1.28$: $x = (1.28 \cdot 0.2) + 4.9 = 0.256 + 4.9 = 5.2$

So, $z < 1.28$ converts to: $x < 5.2$

Part (f) Convert $-2.25 < z < -1.00$ to an $x$ interval. (Round your answers to one decimal place.)

1. For $z = -2.25$: $x = (-2.25 \cdot 0.2) + 4.9 = -0.45 + 4.9 = 4.5$

2. For $z = -1.00$: $x = (-1.00 \cdot 0.2) + 4.9 = -0.2 + 4.9 = 4.7$

So, $-2.25 < z < -1.00$ converts to: $4.5 < x < 4.7$

Part (g) If a female had an RBC count of 5.9 or higher, would that be considered unusually high? Explain using $z$-values.

To determine if 5.9 is unusually high, we convert it to a $z$-score and see if it falls far from the mean:

$z = \frac{5.9 - 4.9}{0.2} = \frac{1.0}{0.2} = 5$

A $z$-score of 5 is extremely high, indicating that an RBC count of 5.9 or more would be very unusual, as it lies far outside the typical range of $z$-scores (usually within $\pm3$).

So, the correct answer is: Yes. A $z$-score of 5.00 implies that this RBC is unusually high.

Would you like more details on any part, or have additional questions?

Additional Questions

1. How do we interpret the meaning of a $z$-score in a normal distribution?
2. What are some real-world applications of converting between $x$-values and $z$-scores?
3. How can we determine if a value is "usual" or "unusual" based on the $z$-score?
4. Why do we round $x$-values to one decimal place in certain parts of the question?
5. What other measures of spread, besides standard deviation, can affect how we interpret data in a normal distribution?

Tip

To quickly identify if a value is unusually high or low, check if its $z$-score is beyond $\pm2$ (for "unusual" values) or $\pm3$ (for "extreme" values) in a normal distribution.