Given that IQ tests are standardized to a normal distribution model $N(100, 15)$, this implies the IQ scores follow a normal distribution with a mean $\mu = 100$ and a standard deviation $\sigma = 15$.

a) Cutoff value for the highest 10% of all IQs

We need to find the IQ score that bounds the top 10%, which is the score corresponding to the 90th percentile.

1. Find the z-score corresponding to the 90th percentile using a standard normal distribution table or a calculator.

   • For the 90th percentile, the z-score $z_{0.90} \approx 1.2816$.

2. Use the formula to convert the z-score to the IQ score: $X = \mu + z \cdot \sigma$ Substituting the values: $X = 100 + 1.2816 \cdot 15 \approx 100 + 19.224 = 119.224$ The cutoff for the top 10% is approximately 119.22.

b) Cutoff value for the lowest 20% of all IQs

We now need to find the IQ score corresponding to the 20th percentile.

1. Find the z-score corresponding to the 20th percentile.

   • For the 20th percentile, the z-score $z_{0.20} \approx -0.8416$.

2. Use the same conversion formula: $X = 100 + (-0.8416) \cdot 15 = 100 - 12.624 = 87.376$ The cutoff for the bottom 20% is approximately 87.38.

c) Cutoff values for the middle 70% of the IQs

For the middle 70%, we need the cutoff values that bound the central 70% of the distribution, leaving 15% in the lower tail and 15% in the upper tail.

1. Find the z-scores corresponding to the 15th percentile and the 85th percentile.

   • For the 15th percentile, the z-score $z_{0.15} \approx -1.036$.
   • For the 85th percentile, the z-score $z_{0.85} \approx 1.036$.

2. Convert the z-scores to IQ scores:

   • For the lower bound: $X = 100 + (-1.036) \cdot 15 = 100 - 15.54 = 84.46$
   • For the upper bound: $X = 100 + 1.036 \cdot 15 = 100 + 15.54 = 115.54$ The cutoff values for the middle 70% are approximately 84.46 and 115.54.

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Summary:

• a) The cutoff for the top 10% is 119.22.
• b) The cutoff for the bottom 20% is 87.38.
• c) The middle 70% is bounded by 84.46 and 115.54.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do z-scores relate to percentile ranks in a normal distribution?
2. How would the cutoff values change if the standard deviation were different?
3. Can you explain the relationship between mean and standard deviation in the context of normal distributions?
4. How can we use the empirical rule to estimate cutoff values for the middle 68% instead of 70%?
5. What does the area under a normal distribution curve represent?

Tip: Z-scores allow you to standardize any normal distribution, making it easier to compare scores from different distributions.