The standard-Binet IQ test is designed so that scores are normally distributed with a mean of 100 and a standard deviation of 16 A person got an IQ score of 135 What IQ score corresponds to the 95th p

To find the IQ score corresponding to the 95th percentile, we need to find the z-score that corresponds to the 95th percentile and then use that z-score to solve for the IQ score.

First, we need to find the z-score corresponding to the 95th percentile. We can use a standard normal distribution table or calculator to find that the z-score corresponding to the 95th percentile is approximately 1.645.

Next, we can use the formula for converting a z-score to a raw score:

z = (x - μ) / σ

where z is the z-score, x is the raw score (IQ score in this case), μ is the mean, and σ is the standard deviation.

Plugging in the values we know:

1.645 = (x - 100) / 16

Solving for x:

x = 1.645 * 16 + 100

x = 126.32

Therefore, an IQ score of 126.32 corresponds to the 95th percentile.