Sure, let's go through the steps to determine what percentage of the population has an IQ lower than 112, given a mean IQ of 100 and a standard deviation of 15.
1. Calculate the Z-score: The Z-score is a measure of how many standard deviations an individual data point (in this case, an IQ of 112) is from the population mean.
Formula for Z-score:
[tex]\[
Z = \frac{(X - \mu)}{\sigma}
\][/tex]
where [tex]\(X\)[/tex] is the value of the data point (IQ score), [tex]\(\mu\)[/tex] is the population mean, and [tex]\(\sigma\)[/tex] is the standard deviation.
Plug in the values:
[tex]\[
Z = \frac{(112 - 100)}{15} = \frac{12}{15} = 0.8
\][/tex]
2. Find the cumulative probability: The cumulative probability, often denoted as [tex]\(P(Z \leq z)\)[/tex], represents the proportion of the population that falls below the given Z-score.
For [tex]\(Z = 0.8\)[/tex], using standard normal distribution tables or a calculator, we find that the cumulative probability is approximately 0.788.
3. Convert to percentage: To find the percentage of the population with a lower IQ, we convert the cumulative probability to a percentage.
[tex]\[
0.788 \times 100 = 78.8 \%
\][/tex]
Therefore, approximately 78.8% of the population has an IQ lower than 112.
The closest answer choice is:
B. 79%
