IQ scores are normally distributed with mean 100 and standard deviation 15 . Ten percent of IQ scores are above what value? 119 75 125 81

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1. Let $$ X $$ be the IQ score, where $$ X \sim N(100, 15^2) $$.

2. We are given that 10% of IQ scores are above a value $$ x $$, which means:
   $$
   P(X > x) = 0.10.
   $$
   Therefore, $$ P(X \leq x) = 0.90 $$. Thus, $$ x $$ is the 90th percentile.

3. To find $$ x $$, we first find the corresponding $$ z $$-score for the 90th percentile. We have:
   $$
   P(Z \leq z_{0.90}) = 0.90.
   $$
   Using a standard normal distribution table or calculator, we find:
   $$
   z_{0.90} \approx 1.282.
   $$

4. The formula relating $$ x $$ and $$ z $$ in a normal distribution is:
   $$
   x = \mu + z \sigma,
   $$
   where $$\mu = 100$$ and $$\sigma = 15$$. Substituting the values, we have:
   $$
   x = 100 + (1.282)(15) \approx 100 + 19.23 \approx 119.23.
   $$

5. Rounding $$ x $$ to the nearest whole number gives $$ x \approx 119 $$.

Thus, 10% of IQ scores are above $$\boxed{119}$$.
10% of IQ scores are above 119.

IQ scores are normally distributed with mean 100 and standard deviation 15 . Ten percent of IQ scores are above what value? 119 75 125 81

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