To solve this problem, we can use the concept of the standard normal distribution.
Given that the mean (μ) of IQ scores is 100 and the standard deviation (σ) is 15, we can convert the individual IQ scores into z-scores, which represent the number of standard deviations a particular IQ score is away from the mean.
The z-score can be calculated using the formula:
z = (x - μ) / σ
where x is the IQ score, μ is the mean, and σ is the standard deviation.
In this case, we want to find the probability of a random person having an IQ score less than 96. We need to find the z-score for an IQ score of 96 and then calculate the probability associated with that z-score using the standard normal distribution table.
First, calculate the z-score:
z = (96 - 100) / 15
z = -0.27
Next, we can use a standard normal distribution table or a calculator to find the probability associated with this z-score. The probability of a random person having an IQ score less than 96 is the area under the standard normal curve to the left of the z-score -0.27.
Using a standard normal distribution table or a calculator, the probability associated with a z-score of -0.27 is approximately 0.3944.
Therefore, the probability of a random person on the street having an IQ score of less than 96 is approximately 0.3944 or 39.44%.
I hope this helps! :)