Next: Problem 3. Up: No Title Previous: Problem 1.

Problem 2.

Suppose the diameter at breast height (1.5 m.) of trees of a certain type is normally distributed with mean 20 cm. and standard deviation 8 cm. Let X be the diameter of a randomly selected tree. Use the table of normal probabilities given below to answer the following questions.

(a) What is the probability that X is between 12 cm. and 24 cm.?

Solution:

$\begin{displaymath} P[ 12 < X < 24 ] \; = \; P \left[ \frac{12-20}{8} < \frac{X-\mu}{\sigma} < \frac{24-20}{8} \right]\end{displaymath}$

$\begin{displaymath} \; = \; P[ -1 < Z < .5 ] \; = \; \Phi(.5) - \Phi(-1) \; = \... ...[ 1 - \Phi(1) ] \; = \; .691 - ( 1 - .841 ) \; = \; 0.532 .\end{displaymath}$

(b) Find a value x so that P[X > x] = 0.200.

Solution: We see from the table that

$\begin{displaymath} P[ Z < 0.8 ] \; = \; 0.788 , \quad P[ Z < 0.9 ] \; = \; 0.816 .\end{displaymath}$

The closest z value is 0.8. Therefore

$\begin{displaymath} P[ Z \gt 0.8 ] \; = \; 1 - .788 \; = \; 0.212 .\end{displaymath}$

(One can use linear interpolation to improve accuracy: this would give z = 0.8 + .1*(12/28) = 0.84.)

Thus, the corresponding x value is

$\begin{displaymath} \mu + z*\sigma \; = \; 20 + (0.8)*8 \; = \; 26.4 .\end{displaymath}$

That is, P[ X > 26.4 ] $\doteq$ .2 (actually .212). (If one uses linear interpolation, the more accurate result is P[ X < 26.72 ] $\doteq$ .2

Table 1: Table of Standard Normal Probabilities. The tabulated values are of $\Phi(z)$ = $P[Z \le z]$ where Z is a normal r.v. with mean and standard deviation 1. So $\Phi(1.2)$ = .885. You may use the symmetry of the normal distribution to compute $\Phi(z)$ for z < 0, e.g. $P[Z \le -1]$ = $P[Z \ge 1]$ .
.0 .1 .2 .3 .4 .5 .6 .7 .8 .9

.500 .540 .579 .618 .655 .691 .726 .758 .788 .816

1 .841 .864 .885 .903 .919 .933 .945 .955 .964 .971

2 .977 .982 .986 .989 .992 .994 .995 .997 .997 .998

**Table 1:** Table of Standard Normal Probabilities. The tabulated values are of $\Phi(z)$ = $P[Z \le z]$ where Z is a normal r.v. with mean and standard deviation 1. So $\Phi(1.2)$ = .885. You may use the symmetry of the normal distribution to compute $\Phi(z)$ for z < 0, e.g. $P[Z \le -1]$ = $P[Z \ge 1]$ .
	.0	.1	.2	.3	.4	.5	.6	.7	.8	.9
	.500	.540	.579	.618	.655	.691	.726	.758	.788	.816
1	.841	.864	.885	.903	.919	.933	.945	.955	.964	.971
2	.977	.982	.986	.989	.992	.994	.995	.997	.997	.998

Next: Problem 3. Up: No Title Previous: Problem 1.

Dennis Cox
3/25/2001