To solve this, we use the empirical rule, which states that for a bell-shaped (normal) distribution:

1. Approximately 68% of the data falls within 1 standard deviation of the mean.
2. Approximately 95% of the data falls within 2 standard deviations of the mean.
3. Approximately 99.7% of the data falls within 3 standard deviations of the mean.

Step 1: Compute the number of standard deviations $z$ from the mean

We are asked to find the percentage of IQ scores greater than 133. First, calculate how many standard deviations $133$ is above the mean:

$z = \frac{133 - 99}{17} = \frac{34}{17} = 2$

So, $133$ is 2 standard deviations above the mean.

Step 2: Apply the empirical rule

From the empirical rule:

• 95% of the data lies within 2 standard deviations of the mean (i.e., between $99 - 34 = 65$ and $99 + 34 = 133$).
• This means 5% of the data lies outside this range (2.5% below 65 and 2.5% above 133).

Step 3: Interpret the result

The percentage of IQ scores greater than 133 is therefore:

$\text{Percentage greater than 133} = 2.5\%$

Final Answer:

The percentage of IQ scores greater than 133 is 2.5% (without rounding).

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Would you like further details or have any questions? Here are 5 related questions to expand your understanding:

1. What is the percentage of IQ scores less than 65 based on the empirical rule?
2. How would the calculation change if the standard deviation were 20 instead of 17?
3. What percentage of IQ scores fall between 82 and 116?
4. How does the empirical rule compare to using a z-score table for such calculations?
5. How can the empirical rule be applied to a distribution that is not symmetric?

Tip: The empirical rule is a quick way to approximate probabilities, but z-score tables or statistical software give more precise values for non-standard situations.