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Normal Quantiles

Consider the above problem on height of a male. Assume that height is normally distributed with mean 70" and standard deviation 4". Suppose we want to determine the 90th percentile in height; i.e, the height which is surpassed by only 10% of men. Call this point c.

Well what can we do it? What's that? A z-score of course. Get back to z. Easy. $z = \frac{c-70}{4}$. 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


  
Figure 5.10: P(Z > 1.28) = .10
\begin{figure}
\begin{center}
\epsfig{file=fig24.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

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 player).

How about finding the quartiles  of the general normal distribution with mean $\mu$ and standard deviation $\sigma$? For the first quartile, say, q1, we have $z = \frac{q_1 - \mu}{\sigma}$ and the area to the left is .25. From the CC we get z = -.67, so that $q_1 = (-.67 \times \sigma) + \mu$. 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 $\mu$; hence $q_3 = (+.67 \times \sigma) + \mu$.

Thus the interquartile range for a normal distribution is

\begin{displaymath}iqr = q_3 - q_1 = 2 \times .67 \times \sigma = 1.34 \times \sigma
\end{displaymath}

i.e, the interquartile range for any normal distribution is $1.34 \times \sigma$.

Recall this is the population interquartile range. Hence, suppose I tell you the population is normal, but $\mu$ and $\sigma$ are unknown. Suppose it is extremely important that you obtain an estimate  of $\sigma$. And I give you 1000 data points to use in your estimation of $\sigma$. 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 $\sigma$. Why? Will it be a good estimate? What's that? How much did the estimate miss by? Hey, you catching on.


Exercise 6.4.1  
1.
Consider the height example given above. Determine the first and third population quartiles. (Ans: 67.30204, 72.69796).
2.
In the last problem, a basketball coach remarks that the even shortest professional basketball player, exceeds the 98th percentile in height. Does this make sense? (Help with answer: 98th percentile is 78.215).
3.
Suppose we know that scores on this exam are approximately normally distributed with mean 430 and standard deviation 50. Determine the first and third population quartiles. (Ans: 396.2755, 463.724 5).
4.
Smith College only accepts the upper 20% of people taking the exam in #3. What is the lowest score a person can make on the exam and still be acceptable to Smith College? (Ans. 472.0811).
5.
Verify that for any normal population, the probability that a measurement falls in the interval $\mu - 2\sigma$ to $\mu + 2\sigma$ is .9544997.
6.
Verify that for any normal population, the probability that a measurement falls in the interval $\mu - 3\sigma$ to $\mu + 3\sigma$ is .9973002.


next up previous contents index
Next: Empirical Rule Up: Continuous Probability Models Previous: Normal Distribution

2001-01-01