Assume that men have IQ scores that are normally distributed with a mean of 99 and a standard deviation of 13. Find the IQ score that separated the lower 30% of the others. Round up to the nearest integer.
To find the IQ score that separates the lower 30% of others, we need to determine the corresponding z-score and then convert it back to an IQ score.
Here's how we can do it:
Step 1: Calculate the z-score corresponding to the lower 30%.
To find the z-score, we use the cumulative distribution function (CDF) of a standard normal distribution.
The formula for z-score is:
z = (x - μ) / σ,
where
x is the IQ score,
μ is the mean of the distribution (99 in this case), and
σ is the standard deviation of the distribution (13 in this case).
The CDF gives us the probability of getting a value less than or equal to a given value in a standard normal distribution. In this case, we want to find the IQ score that corresponds to the lower 30% (0.3).
Using a standard normal distribution table or a statistical software, we find that the z-score corresponding to the lower 30% is approximately -0.524.
Step 2: Convert the z-score back to an IQ score.
We use the formula for z-score to convert the z-score back to the IQ score:
x = z * σ + μ.
Substituting the values, we have:
x = -0.524 * 13 + 99,
x ≈ 92.876.
Now, we can round 92.876 up to the nearest integer, which gives us an IQ score of 93.
Therefore, the IQ score that separates the lower 30% of others is 93.