Well what can we do it? What's that? A z-score of course. Get
back to z. Easy.
We can't determine z,
but the area to the left of z on the standard normal distribution
curve is .90. See Figure 5.10
Now we need to find the value of z which corresponds to
.90. It's easy. Just ask CC. This time when you click go to the Normal
Percentage Point part of the page and enter .90 in the p-window. Then click submit and read the answer, which is 1.28. Try it.
Hence, by the above equation,
1.28 = (c-70)/4; i.e., c = 75.12.
Therefore only 10% of men are taller than 75", (a very short basketball
How about finding the quartiles of the general normal distribution with
and standard deviation ? For the first quartile,
say, q1, we have
and the area to the left is .25. From the CC we get z = -.67, so that
Now you try it for the
third quartile, q3. Another way is to use symmetry : the quartiles have got to be symmetric with respect to ;
Thus the interquartile range for a normal distribution is
Recall this is the population interquartile range. Hence, suppose I tell you the population is normal, but and are unknown. Suppose it is extremely important that you obtain an estimate of . And I give you 1000 data points to use in your estimation of . But then the batteries in your calculator died. And then your friend the computer whiz went to the movies. So what do you do? You don't want to compute the sample standard deviation . This is 1000 numbers! Alas! But then you notice that on the last sheet of the pages containing the data, someone has sketched a boxplot . You obtain the length of the box and divide it by 1.34. This is an estimate of . Why? Will it be a good estimate? What's that? How much did the estimate miss by? Hey, you catching on.