Genius society requires an IQ that is in the top 2% of the population in order to join. If an IQ test has a mean of 100 and a standard deviation of 15, what is the minimum qualifying score to join the genius society?

For number 1 we need Pr(X > x) = 0.02

which is pr(X<x) = 0.98

We will determine z such that pr(Z<z) = 0.98

and in place of the table, I will use Microsoft Excel

=norminv(0.98,0,1)
which gives the number z satisfying Pr(Z<z) = 0.98

Result is 2.053748176

so

x - 100
▬▬▬..=...2.053748176

15

so

x = 2.053748176(15) + 100 = 130.8

Answer Link

xcute40 xcute40 · Answer 2 · 2019-08-23T20:02:55+02:00

Answer:

130.75

Step-by-step explanation:

To be in the top 2%, you have to have an IQ higher than 98% of the population (100% - 2%). This is also known as the 98th percentile. For this problem, you are given the probability and need to determine the raw score.

Since you have the probability, go to the z–table and find the probability that is closest to 0.98. Then look across the row and up the column to determine the z-score.

The closest probability is 0.97982 and that corresponds to a z-score of 2.05.

With a z-score of 2.05, you can use the z formula to compute the raw score.

[tex]z=\frac{x-mean}{standard |deviation}[/tex]

[tex]2.05=\frac{x-100}{15}[/tex]

[tex]2.05(15)=\frac{x-100}{15} (15)\\\\30.75+100=x-100+100\\\\130.75=x[/tex]