Question

ch 3* the scores of adults on an iq test are approximately normal with mean 100 and standard deviation 15. clara scores 132 on such a test. she scores higher than what percent of all adults?
about 98%
about 10%
about 90%

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 132$ (Clara's score), $\mu = 100$ (mean), and $\sigma=15$ (standard deviation). So $z=\frac{132 - 100}{15}=\frac{32}{15}\approx2.13$.

Step2: Find the proportion of scores below Clara's

We use the standard normal distribution table (z - table). Looking up the value of $z = 2.13$ in the z - table, the proportion of scores to the left of $z = 2.13$ is approximately $0.9834$.

Step3: Find the proportion of scores above Clara's

The proportion of scores above Clara's is $1 - 0.9834=0.0166$. So Clara scores higher than approximately $100\times(1 - 0.0166)=98.34\%\approx98\%$ of all adults.

Answer:

About 98%