I came across this problem and I'm a little stumped?
Question: Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 20. Find P2, which is the IQ separating the bottom 2% from the top 98%
I know I have to utilize z=(x-u)/o At least that's how I think I have to start, I have the table here with me, I'm just not sure how to approach this problem?
On
Its to the left in this case because its says "Bottom 2% of the top 98%" I converted the area of that bell curve to .02 then went on the Normal distribution curve to find the closest value to .02 which is .0202 which that gives you the z-score of -2.05 than I took that and placed it on the equation x = u+z*o. In this case the z was negative so I added the negative.
Let $X$ be a random variable that describes the IQ of a randomly selected adult from the population. Then we are told that $$X \sim \operatorname{Normal}(\mu = 100, \sigma = 20);$$ that is, such scores are normally distributed with a mean of $100$ and a standard deviation of $20$. The question asks you to find a value $x^*$ such that $2\%$ of such adults have IQs below this value (which automatically implies that $98\%$ of adults have IQs above this value).
Your intuition should suggest that this value of $x^*$ must be a fair bit below the mean of $100$, since the symmetry of the normal distribution about its mean implies that exactly $50\%$ of adults have IQs below $100$, and $50\%$ have IQs above $100$. If you were to hazard a guess, you might guess that $x^*$ should be somewhere around $50$. How good is this guess? Let's find out.
In formal mathematical notation, what we want to solve is: $$\Pr[X \le x^*] = 0.02.$$ This rephrases the original question in terms of a probability: what is the value of $x^*$ such that there is only a $2\%$ probability of a randomly chosen adult from the population having an IQ less than $x^*$? Equivalently, we can also write the above as $$\Pr[X > x^*] = 0.98.$$
If you have a (standard) normal distribution table, you know that you can either look up a particular row and column to find a probability. But here, you know the probability and you want to find a score. This means that to find your answer from a table, you will need to look up $0.98$ in the table, and find the row and column values for this probability. Most tables I've seen only show probabilities above $0.5$, which is why I went with $0.98$ rather than $0.02$, but this means we need to do an extra step as you will see shortly.
The other problem with the table is that it corresponds to a standard normal distribution $Z \sim \operatorname{Normal}(\mu = 0, \sigma = 1)$, whereas here, $X$ is not standard normal. So we need to transform $X$ to get $Z$: This is where the formula $$Z = \frac{X - \mu}{\sigma}$$ comes from. So we can write: $$\Pr[X > x^*] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{x^* - 100}{20}\right] = \Pr\left[Z > \frac{x^* - 100}{20}\right] = 0.98.$$ Now we have an expression for the probability in terms of a standard normal. We look up $0.98$ in the normal table and find that $$\Pr[Z \le z^*] = 0.98$$ for $z^* \approx 2.05\color{grey}{(375)}$. (I have included extra digits in grey but these are probably not obtainable from a table lookup). What this means is that $$\frac{x^* - 100}{20} = -z^* = -2.05.$$ Where did the negative sign come from? The reason for this has to do with which probability we looked up: we looked up $0.98$, which will give us a $z$-score to the right of the mean, but remember, the actual score we want should be to the left of the mean. This makes sense when we note that the table gives us $\Pr[Z \le z^*]$, rather than $\Pr[Z > z^*]$. And because many tables only show the upper (right) half of the normal distribution, we have to use this workaround to get the lower (left) half.
Okay, so finally, we just solve: $$x^* \approx 20(-2.05) + 100 = 59.$$ This means that about $2\%$ of adults have IQs less than $59$.