What Is the Probability That a Randomly Selected Person Has an IQ Below 90?

To estimate the probability that a randomly selected person has an IQ below 90 using the Stanford-Binet IQ Test distribution, you can use the properties of the normal distribution. Here are the steps involved: 1. **Find the Z-score**: The Z-score indicates how many standard deviations away a particular score is from the mean. You can calculate it using the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \(X\) is the score (90 in this case), \(\mu\) is the mean (100), and \(\sigma\) is the standard deviation (20). 2. **Calculate the Z-score for 90**: \[ Z = \frac{(90 - 100)}{20} = \frac{-10}{20} = -0.5 \] 3. **Consult the Z-table**: The Z-table (also known as the standard normal distribution table) provides the probability that a standard normal variable is less than or equal to a given value. For a Z-score of -0.5, you can look it up in the Z-table. The value corresponding to -0.5 is approximately 0.3085. This means that the probability of a randomly selected person having an IQ below 90 is approximately **0.3085** or **30.85%**. So, there is about a 30.85% chance that a randomly selected person will have an IQ below 90.

To estimate the probability that a randomly selected person has an IQ below 90, we can use the z-score formula and a standard normal distribution table. Here's how: 1. Calculate the z-score: z = (X - μ) / σ, where X is the IQ score (90 in this case), μ is the mean (100), and σ is the standard deviation (20). z = (90 - 100) / 20 z = -10 / 20 z = -0.5 2. Look up the z-score (-0.5) in a standard normal distribution table or use a calculator that provides this information. The table will give you the area to the left of the z-score, which represents the probability of obtaining a value lower than the given z-score. The probability you find in the table for a z-score of -0.5 is approximately 0.3085, which means there is a 30.85% chance that a randomly selected person has an IQ below 90. Please keep in mind that the values presented above are only estimates and should not be taken as exact probabilities.