To determine the percentage of IQ scores greater than 116 from a normal distribution with a mean of 100 and a standard deviation of 15, we'll follow these steps:

Step 1: Convert the IQ score to a Z-score

The Z-score represents how many standard deviations the value is from the mean. The formula for calculating the Z-score is:

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

Where:

• $X$ = 116 (the IQ score),
• $\mu$ = 100 (the mean),
• $\sigma$ = 15 (the standard deviation).

Substituting the values:

$Z = \frac{116 - 100}{15} = \frac{16}{15} \approx 1.07$

So, the Z-score for an IQ of 116 is approximately 1.07.

Step 2: Use the Z-table to find the cumulative probability

Now, we look up the Z-score of 1.07 in the standard normal distribution table (Z-table) to find the cumulative probability (i.e., the probability of a score being less than 116).

From the Z-table, the cumulative probability for a Z-score of 1.07 is approximately 0.8577. This means that about 85.77% of the scores fall below 116.

Step 3: Find the percentage greater than 116

Since the total area under the curve of a normal distribution is 100%, the percentage of scores greater than 116 is:

$100\% - 85.77\% = 14.23\%$

Final Answer:

The percentage of IQ scores greater than 116 is approximately 14.23%.

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Would you like more detailed steps or have any questions on this process? Here are some related questions for you to explore further:

1. How do you calculate the Z-score for a different IQ value?
2. What is the role of the standard deviation in a normal distribution?
3. How do you find the probability for a range of values in a normal distribution?
4. What is the concept of the 68-95-99.7 rule in normal distribution?
5. How can you calculate the percentile rank of a score in a normal distribution?

Tip: The Z-table gives the cumulative probability from the left (less than the given value), so for values greater than the Z, always subtract the cumulative probability from 100%.