To find the probability that a randomly selected adult has an IQ between 83 and 117, given that IQ scores are normally distributed with a mean $\mu = 100$ and standard deviation $\sigma = 20$, we can use the standard normal distribution (Z-distribution).

Step 1: Convert the IQ scores to Z-scores

The Z-score formula is:

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

where:

• $X$ is the value we want to convert,
• $\mu$ is the mean,
• $\sigma$ is the standard deviation.

For $X = 83$:

$Z_1 = \frac{83 - 100}{20} = \frac{-17}{20} = -0.85$

For $X = 117$:

$Z_2 = \frac{117 - 100}{20} = \frac{17}{20} = 0.85$

Step 2: Find the probabilities corresponding to these Z-scores

Using the standard normal distribution table or a calculator:

• The probability corresponding to $Z_1 = -0.85$ is approximately 0.1977.
• The probability corresponding to $Z_2 = 0.85$ is approximately 0.8023.

Step 3: Find the probability between the two Z-scores

The probability that the IQ is between 83 and 117 is the difference between the two probabilities:

$P(83 < X < 117) = P(Z_2) - P(Z_1) = 0.8023 - 0.1977 = 0.6046$

Final Answer:

The probability that a randomly selected adult has an IQ between 83 and 117 is approximately 0.6046 or 60.46%.

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Would you like further details on how to use the Z-tables or any clarifications?

Here are 5 related questions to explore further:

1. How do you find the probability for IQ scores above a certain value?
2. What is the probability of having an IQ below 83?
3. How does changing the standard deviation affect the probability distribution?
4. How would you calculate the probability for IQ scores between two non-standard values, like 90 and 110?
5. What is the Z-score interpretation in real-world terms?

Tip: Always remember that a Z-score tells you how many standard deviations away a particular value is from the mean.