To find the probability that the sample mean IQ of 300 randomly selected adults is greater than 115, we can use the Central Limit Theorem. Here are the steps:

1. Identify the parameters:

   • Population mean ($\mu$) = 112
   • Population standard deviation ($\sigma$) = 25
   • Sample size ($n$) = 300

2. Calculate the standard error of the mean (SEM): $\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{300}}$ Calculating that: $\text{SEM} = \frac{25}{\sqrt{300}} \approx 1.443$

3. Find the Z-score for a sample mean of 115: The Z-score formula is: $Z = \frac{X - \mu}{\text{SEM}}$ where $X = 115$.

   Substituting the values: $Z = \frac{115 - 112}{1.443} \approx 2.08$

4. Find the probability for this Z-score: Using standard normal distribution tables or a calculator, we can find the probability of $Z > 2.08$.

   For $Z = 2.08$, the cumulative probability to the left is approximately 0.9812. Therefore, the probability of a Z-score greater than 2.08 is: $1 - 0.9812 = 0.0188$

So, the probability that the sample mean IQ is greater than 115 is approximately 0.019.

Answer:

The correct answer is:

• 0.019

Would you like more details, or have any questions?

Here are some related questions to expand your understanding:

1. How does the Central Limit Theorem help in determining probabilities for sample means?
2. What would happen to the probability if the sample size increased to 500?
3. How would the probability change if the population standard deviation were higher?
4. Why do we use the standard error instead of the population standard deviation for sample means?
5. What other types of distributions can be used for probabilities if the sample size is small?

Tip: Always check if the sample size is large enough to use the Central Limit Theorem for normal approximation, typically $n > 30$ is considered sufficient.