We are given that the dexterity test scores are normally distributed with:

• Mean (expected value), $\mu = 85$
• Variance, $\sigma^2 = 16$, so the standard deviation $\sigma = \sqrt{16} = 4$

We want to find the probability that a participant scores greater than 81, i.e., $P(X > 81)$, where $X$ represents the test scores.

To solve this, we follow these steps:

Step 1: Standardize the variable $X$

We convert $X = 81$ to the standard normal variable $Z$ using the formula:

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

Substituting the values:

$Z = \frac{81 - 85}{4} = \frac{-4}{4} = -1$

Thus, $P(X > 81) = P(Z > -1)$.

Step 2: Use the standard normal distribution table

Next, we look up the cumulative probability for $Z = -1$ in a standard normal distribution table. This gives:

$P(Z \leq -1) \approx 0.1587$

Since we are looking for $P(Z > -1)$, we calculate:

$P(Z > -1) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413$

Final Answer:

The probability that a participant will score greater than 81 is approximately $0.8413$ or 84.13%.

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Do you need further clarification on any part of the explanation? Here are some related questions you might find helpful:

1. What if the score were less than 81 instead of greater?
2. How would the probability change if the mean score was different?
3. What is the probability of scoring between 81 and 85?
4. How do you interpret the meaning of standard deviation in this context?
5. What would be the probability for a score greater than 90?

Tip: When dealing with normal distributions, converting the variable to a standard normal (using the Z-score) simplifies finding probabilities from standard tables or calculators.