Assume that adults have IQ score that are
normally distributed with a mean of 105
and a standard deviation 20. Find P, which
is the IQ score separating the bottom 10%
from the top 90%. Round to the nearest
hundredth as needed.
To find the IQ score that separates the bottom 10% from the top 90% in a normally distributed set with a mean of 105 and a standard deviation of 20, we can use the concept of z-scores.
A z-score measures the number of standard deviations a given value is from the mean. We can calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation.
First, let's find the z-score that corresponds to the bottom 10% of the distribution.
To do this, we need to find the z-score that corresponds to a cumulative area of 0.10 in the left tail of the standard normal distribution. We can use a standard normal distribution table or a calculator to find this value. In this case, the z-score corresponding to the 10th percentile is approximately -1.28.
Next, we can use the z-score formula to find the corresponding IQ score:
-1.28 = (P - 105) / 20
Solving for P, we get:
P - 105 = -1.28 * 20
P - 105 = -25.6
P = -25.6 + 105
P ≈ 79.4
Therefore, the IQ score that separates the bottom 10% from the top 90% is approximately 79.4.