The Stanford-Binet IQ Test: Normal Curve with a Mean of 100 and a Standard deviation of 20

To estimate the probability of a person having an IQ below 90 using the Stanford-Binet IQ Test, we can use the Z-score formula. The Z-score represents how many standard deviations away from the mean a particular score is. Here's the Z-score formula: Z = (X - Mean) / Standard Deviation where X = the IQ score, Mean = 100, and Standard Deviation = 20. Now to calculate the Z-score for an IQ of 90: Z = (90 - 100) / 20 Z = -10 / 20 Z = -0.5 A Z-score of -0.5 means that 90 is 0.5 standard deviations below the mean. To find the probability, we can refer to a standard normal distribution table, which gives us the area under the curve (i.e., the probability) up to a given Z-score. Looking up a Z-score of -0.5 in the standard normal distribution table gives us a value of 0.3085. Therefore, the probability of a randomly selected person having an IQ below 90 is approximately 30.85%.