To solve this problem, we need to use the properties of the normal distribution. The scores follow a normal distribution with a mean (μ) of 69 and a standard deviation (σ) of 3.

Formula:

To estimate percentages, we calculate the Z-scores using the formula:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the value of interest
• $\mu$ is the mean (69)
• $\sigma$ is the standard deviation (3)

Now, let's break it down for each part:

(a) Between 66 and 72:

We find the Z-scores for 66 and 72.

For $X = 66$:

$Z = \frac{66 - 69}{3} = \frac{-3}{3} = -1$

For $X = 72$:

$Z = \frac{72 - 69}{3} = \frac{3}{3} = 1$

Using the Z-table or a standard normal distribution calculator, the area between $Z = -1$ and $Z = 1$ is approximately 68.27%.

So, approximately 68.27% of the scores are between 66 and 72.

(b) Above 78:

For $X = 78$:

$Z = \frac{78 - 69}{3} = \frac{9}{3} = 3$

The area to the right of $Z = 3$ is about 0.13%.

So, approximately 0.13% of the scores are above 78.

(c) Below 66:

For $X = 66$:

$Z = \frac{66 - 69}{3} = -1$

The area to the left of $Z = -1$ is about 15.87%.

So, approximately 15.87% of the scores are below 66.

(d) Between 60 and 72:

We calculate the Z-scores for 60 and 72.

For $X = 60$:

$Z = \frac{60 - 69}{3} = \frac{-9}{3} = -3$

For $X = 72$:

$Z = \frac{72 - 69}{3} = 1$

The area between $Z = -3$ and $Z = 1$ is the sum of the areas from $Z = -3$ to $Z = 0$ (which is 49.87%) and from $Z = 0$ to $Z = 1$ (34.13%).

So, the total area is 49.87% + 34.13% = 84.0%.

Final answers:

• (a) 68.27%
• (b) 0.13%
• (c) 15.87%
• (d) 84.00%

Would you like any further clarification or details on these calculations?

Related Questions:

1. How does the Z-score help in interpreting the normal distribution?
2. What is the empirical rule and how does it apply to this problem?
3. How can we use a Z-table to calculate probabilities in a normal distribution?
4. How do we calculate probabilities for values outside the mean in a normal distribution?
5. What are the applications of normal distribution in standardized testing?

Tip: Always check if the question asks for approximate percentages or exact ones, as normal distribution tables often give approximate values.