To find the probability of a random person having an IQ score less than 99, we can use the standard normal distribution (Z-distribution) because IQ scores are assumed to be normally distributed with a mean (\(\mu\)) of 100 and a standard deviation (\(\sigma\)) of 15.
First, calculate the z-score using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where \( X = 99 \), \( \mu = 100 \), and \( \sigma = 15 \). Plug in the values:
\[ Z = \frac{99 - 100}{15} = -0.0667 \]
Next, look up the z-score in a standard normal table or use a calculator to find the cumulative probability corresponding to \( Z = -0.0667 \). The probability \( P(Z < -0.0667) \) represents the probability of a random person having an IQ score less than 99.
Using a standard normal table or a calculator, the probability approximately equals 0.4721 when rounded to 4 decimal places.
So, the probability of a random person on the street having an IQ score of less than 99 is approximately \( 0.4721 \).