Normal Distribution

An individuals’ heights are normally distributed with mean 70 and standard deviation 3 (inches).
a) What proportion of the population is over 6 feet tall?
For this question I used the formula z = X - mean / standard deviation.
So, I got (X-70)/3. Then I multiplied 6 *12 to get 72 to match the inches, and then plugged it into the equation. 72-70/3= 2/3. Now, I am stuck as to what to do on how to find the proportion? Is the answer 2/3?

b) If 75 percent of the population is taller than Mike, what is Mike’s height?
I thought to use the same equation in a). So, it would be 1- 0.75= 0.25. 0.25 = (X-70)/3
So, X = 70.75. Is that the correct height?
Thanks for your time.

Mean Joe

TS Contributor
Do you know a table (in your book?) such as this:

Use this table to get your answer. What you have calculated for part (a) is z = 2/3. You want to find P[Z > 2/3] (">" because the problem says "over 6 feet tall"). Note that this table gives you P[Z < z].
It's a column *and* a row ! The columns give the probabilities for the second decimal place of z.

By way of example, P(Z<0.45) = 0.6736

Note the comment by Mean Joe about ">" and also note that P(Z<0)=0.5