A set of scores with a normal distribution has a mean of 32 and a standard deviation of 3.7. Which score could be expected to occur the least often?

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To find the score that could be expected to occur least often in a set with a normal distribution, we need to consider the concept of z-scores.

A z-score measures the number of standard deviations a particular data point is away from the mean. It helps us to understand how extreme or unusual a value is relative to the distribution.

In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and roughly 99.7% falls within three standard deviations.

To determine which score could be expected to occur least often, we should find the z-score corresponding to the value. Then, we can see how far that z-score is from the mean.

The formula to calculate a z-score is:

z = (x - μ) / σ

Where:
- z is the z-score
- x is the value
- μ is the mean
- σ is the standard deviation

Substituting the given information into the formula:

z = (x - 32) / 3.7

To find the score that would be expected to occur the least often, we need to identify the z-score that is farthest away from the mean. Since a z-score can be both positive and negative, the absolute value is used to measure the distance.

First, let's calculate the z-scores for the extreme values:

For the highest score:
z_max = (x_max - 32) / 3.7

And for the lowest score:
z_min = (x_min - 32) / 3.7

Now, let's analyze the z-scores:

The score with the least occurrence would be expected to have the largest absolute value of the z-score. Since a normal distribution is symmetric, the z-score at the negative end (i.e., the left tail) would be farthest from the mean.

Therefore, we need to find the z-score closest to negative infinity or the most extreme negative z-score.

Given that a z-table provides information up to a specific z-score, it might be difficult to identify the exact z-score for negative infinity. However, we approximate by selecting the z-score closest to the leftmost end of the z-table, usually denoted as z = -3.

Solving for x using the z-score formula:

-3 = (x_min - 32) / 3.7

Now we can solve for x_min:

x_min = -3 * 3.7 + 32
x_min = -11.1 + 32
x_min = 20.9

Therefore, the score that could be expected to occur the least often in the normal distribution is 20.9.