The average score of a special test of knowledge of wood refinishing has a mean of 53 and standard deviation of 6. Find the range of values in which at least 75% of the scores will lie.

To find the range of values in which at least 75% of the scores will lie, we need to find the z-scores corresponding to the 25th and 75th percentiles.

First, we find the z-score corresponding to the 25th percentile:
z = (X - μ) / σ
z = (X - 53) / 6
Using a z-table or calculator, we find that the z-score corresponding to the 25th percentile is approximately -0.6745.

Next, we find the z-score corresponding to the 75th percentile:
z = (X - μ) / σ
z = (X - 53) / 6
Using a z-table or calculator, we find that the z-score corresponding to the 75th percentile is approximately 0.6745.

Now, we can find the corresponding scores:
For the 25th percentile:
-0.6745 = (X - 53) / 6
-4.047 = X - 53
X = 48.953

For the 75th percentile:
0.6745 = (X - 53) / 6
4.047 = X - 53
X = 57.047

Therefore, at least 75% of the scores will lie in the range of 48.953 to 57.047.

We can use the empirical rule, also known as the 68-95-99.7 rule, to find the range of values in which at least 75% of the scores will lie.

The empirical rule states that:

- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.

In this case, the mean is 53 and the standard deviation is 6.

So, 75% of the scores will lie within two standard deviations of the mean, which is:

53 - 2(6) = 41 to 53 + 2(6) = 65

Therefore, at least 75% of the scores will lie between 41 and 65.