To transform the test scores from their current normal distribution to a new normal distribution, we use a linear transformation in the form:

$\text{new score} = a + b \times (\text{old score})$

Given Information:

• Old mean $\mu_{\text{old}} = 22$
• Old standard deviation $\sigma_{\text{old}} = 4$
• New mean $\mu_{\text{new}} = 80$
• New standard deviation $\sigma_{\text{new}} = 10$

Step 1: Solve for $b$

The standard deviation of the transformed scores will be $b \times \sigma_{\text{old}}$. Since the teacher wants the new distribution to have a standard deviation of 10, we can set up the equation:

$b \times \sigma_{\text{old}} = \sigma_{\text{new}}$

Substitute the values:

$b \times 4 = 10$

Solve for $b$:

$b = \frac{10}{4} = 2.5$

Step 2: Solve for $a$

The mean of the transformed scores will be $a + b \times \mu_{\text{old}}$. Since the teacher wants the new distribution to have a mean of 80, we can set up the equation:

$a + b \times \mu_{\text{old}} = \mu_{\text{new}}$

Substitute the values:

$a + 2.5 \times 22 = 80$

Simplify and solve for $a$:

$a + 55 = 80$ $a = 80 - 55 = 25$

Final Answer:

• $a = 25$
• $b = 2.5$

Thus, the transformation formula is:

$\text{new score} = 25 + 2.5 \times (\text{old score})$

Would you like a detailed explanation of how the transformation preserves the normal distribution? Here are some related questions:

1. How does scaling by $b$ affect the variance of a distribution?
2. Can this transformation be applied to any normal distribution?
3. What happens if the old scores are not normally distributed?
4. How would the formula change if the desired new standard deviation were 15 instead of 10?
5. What is the inverse transformation to return to the old scores?

Tip: The linear transformation $a + b(x)$ preserves the normal distribution's shape; it only changes the mean and standard deviation.