Continuous Random Variables

Probability Computations for General Normal Random Variables

Learning Objective

To learn how to compute probabilities related to any normal random variable.

If $X$ is any normally distributed normal random variable then Figure 12.2 "Cumulative Normal Probability" can also be used to compute a probability of the form $P(a < X < b)$ by means of the following equality.

If $X$ is a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$ , then

$P(a < X < b)=P\left(\frac{a-\mu}{\sigma} < Z < \frac{b-\mu}{\sigma}\right)$

where $Z$ denotes a standard normal random variable. $a$ can be any decimal number or $-\infty ; b$ can be any decimal number or $\infty$ .

The new endpoints $(a-\mu) / \sigma$ and $(b-\mu) / \sigma$ are the $z$ -scores of $a$ and $b$ as defined in Section 2.4.2 in Chapter 2 "Descriptive Statistics".

Figure 5.14 "Probability for an Interval of Finite Length" illustrates the meaning of the equality geometrically: the two shaded regions, one under the density curve for $X$ and the other under the density curve for $Z$ , have the same area. Instead of drawing both bell curves, though, we will always draw a single generic bell-shaped curve with both an $x$ -axis and a $z$ -axis below it.

Figure 5.14 Probability for an Interval of Finite Length

Example 9

Let $X$ be a normal random variable with mean $\mu=10$ and standard deviation $\sigma=2.5$ . Compute the following probabilities.

$P(X < 14)$ .

b. $P(8 < X < 14)$ .

Solution:

a. See Figure 5.15 "Probability Computation for a General Normal Random Variable".

$\begin{aligned}P(X < 14) &=P\left(Z < \frac{14-\mu}{\sigma}\right) \\&=P\left(Z < \frac{14-10}{2.5}\right) \\&=P(Z < 1.60) \\&=0.9452 \end{aligned}$

Figure 5.15 Probability Computation for a General Normal Random Variable

a. See Figure 5.16 "Probability Computation for a General Normal Random Variable".

$\begin{aligned}P(8 < X < 14) &=P\left(\frac{8-10}{2.5} < Z < \frac{14-10}{2.5}\right) \\&=P(-0.80 < Z < 1.60) \\&=0.9452-0.2119 \\&=0.7333\end{aligned}$

Figure 5.16 Probability Computation for a General Normal Random Variable

Example 10

The lifetimes of the tread of a certain automobile tire are normally distributed with mean 37,500 miles and standard deviation 4,500 miles. Find the probability that the tread life of a randomly selected tire will be between 30,000 and 40,000 miles.

Solution:

Let $X$ denote the tread life of a randomly selected tire. To make the numbers easier to work with we will choose thousands of miles as the units. Thus $\mu=37.5, \sigma=4.5$ , and the problem is to compute $P(30 < X < 40)$ . Figure 5.17 "Probability Computation for Tire Tread Wear" illustrates the following computation:

Figure 5.17 Probability Computation for Tire Tread Wear

$\begin{aligned}P(30 < X < 40) &=P\left(\frac{30-\mu}{\sigma} < Z < \frac{40-\mu}{\sigma}\right) \\&=P\left(\frac{30-37.5}{4.5} < Z < \frac{40-37.5}{4.5}\right) \\&=P(-1.67 < Z < 0.56) \\&=0.7123-0.0475 \\&=0.6648\end{aligned}$

Note that the two $z$ -scores were rounded to two decimal places in order to use Figure 12.2 "Cumulative Normal Probability".

Example 11

Scores on a standardized college entrance examination (CEE) are normally distributed with mean 510 and standard deviation 60. A selective university considers for admission only applicants with CEE scores over 650. Find percentage of all individuals who took the CEE who meet the university's CEE requirement for consideration for admission.

Solution:

Let $X$ denote the score made on the CEE by a randomly selected individual. Then $X$ is normally distributed with mean 510 and standard deviation 60. The probability that $X$ lie in a particular interval is the same as the proportion of all exam scores that lie in that interval. Thus the solution to the problem is $P(X > 650)$ , expressed as a percentage. Figure 5.18 "Probability Computation for Exam Scores" illustrates the following computation:

Figure 5.18 Probability Computation for Exam Scores

$\begin{aligned}P(X > 650) &=P\left(Z > \frac{650-\mu}{\sigma}\right) \\&=P\left(Z > \frac{650-510}{60}\right) \\&=P(Z > 2.33) \\&=1-0.9901 \\&=0.0099\end{aligned}$

The proportion of all CEE scores that exceed 650 is $0.0099$ , hence $0.99 \%$ or about $1 \%$ do.

Key Takeaway

Probabilities for a general normal random variable are computed using Figure 12.2 "Cumulative Normal Probability" after converting $x$ -values to z-scores.