ch 3* the scores of adults on an iq test are approximately normal with mean 100 and standard deviation 15…

ch 3* the scores of adults on an iq test are approximately normal with mean 100 and standard deviation 15. the organization mensa, which calls itself \the high iq society,\ requires an iq score of 130 or higher for membership. what percent of adults would qualify for membership?\n5%\n2.5%\n95%\nquestion 6\n1 pts\nch 3 consider a normal model that describes student scores in a chemistry class. faye has a standardized score (z - score) of - 1.75. this means that faye\nis 1.75 standard deviations below average for the class.\nis 1.75 points below average for the class.\nhas a standard deviation of 1.75.\nhas a score that is 1.75 times the average for the class.

ch 3* the scores of adults on an iq test are approximately normal with mean 100 and standard deviation 15. the organization mensa, which calls itself \the high iq society,\ requires an iq score of 130 or higher for membership. what percent of adults would qualify for membership?\n5%\n2.5%\n95%\nquestion 6\n1 pts\nch 3 consider a normal model that describes student scores in a chemistry class. faye has a standardized score (z - score) of - 1.75. this means that faye\nis 1.75 standard deviations below average for the class.\nis 1.75 points below average for the class.\nhas a standard deviation of 1.75.\nhas a score that is 1.75 times the average for the class.

Answer

Explanation:

Step1: Calculate the z - score for MENSA IQ requirement

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 130$ (MENSA IQ score requirement), $\mu = 100$ (mean IQ score), and $\sigma=15$ (standard deviation of IQ scores). $z=\frac{130 - 100}{15}=\frac{30}{15}=2$

Step2: Find the proportion of the area to the right of the z - score

We know that the total area under the normal curve is 1. Using the standard normal distribution table, the area to the left of $z = 2$ is approximately 0.9772. The area to the right of $z=2$ (the proportion of adults with IQ 130 or higher) is $1 - 0.9772=0.0228\approx2.5%$

For the second question about Faye's z - score: The z - score represents the number of standard deviations a data - point is from the mean. A negative z - score of $z=-1.75$ means that Faye is 1.75 standard deviations below the mean (average) of the class.

Answer:

For the first question: 2.5% For the second question: is 1.75 standard deviations below average for the class.