Here's the question. Need to understand how to solve as opposed to just the answer that way I can be able to figure out how to answer the other questions myself! Thanks!
- A person must score in the upper 2% of the population in an IV test to qualify for membership in Mensa. If the ID scores are normally distribute with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa?
Let $X$ denote the score on the IQ test. Then in the population $X \sim N(100,15^2)$.
You then have to solve equation $$Pr(X>a)=0.02$$ where $a$ is the cut-off for the score.
Now, $$Pr(X>a)=Pr\left(\frac{X-100}{15}>\frac{a-100}{15}\right) \\ =1-Pr\left(\frac{X-100}{15}<\frac{a-100}{15}\right) \\ = 1 - \Phi\left(\frac{a-100}{15}\right)$$
where $\Phi$ is the CDF of a standard normal distribution.
We want $$ 1 - \Phi\left(\frac{a-100}{15}\right)=0.02 $$ So we solve the above equation for $a$: $$ \Phi\left(\frac{a-100}{15}\right) = 0.98 \\ \frac{a-100}{15} = \Phi^{-1}(0.98 ) \\ a = 15 \Phi^{-1}(0.98 ) +100 $$ where $\Phi^{-1}$ is the inverse CDF of the standard normal distribution.
Numerically, $a \approx 130.8$
Let me know if any of the above steps is unclear.