# 5.3: Probability Computations for General Normal Random Variables

Learning Objectives

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

If $$X$$ is any normally distributed normal random variable then Figure $$\PageIndex{1}$$ can also be used to compute a probability of the form $$P(a<X<b)$$ by means of the following equality.

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 $$\infty$$. Figure $$\PageIndex{1}$$: Cumulative Normal Probability

The new endpoints $$\frac{(a-\mu )}{\sigma }$$ and $$\frac{(b-\mu )}{\sigma }$$ are the $$z$$-scores of $$a$$ and $$b$$ as defined in Chapter 2.

Figure $$\PageIndex{2}$$ 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 $$\PageIndex{2}$$: Probability for an Interval of Finite Length

Example $$\PageIndex{1}$$

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

1. $$P(X<14)$$.
2. $$P(8<X<14)$$.

Solution:

1. See Figure $$\PageIndex{3}$$ "Probability Computation for a General Normal Random Variable". \begin{align*} 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{align*} Figure $$\PageIndex{3}$$: Probability Computation for a General Normal Random Variable
1. See Figure $$\PageIndex{4}$$ "Probability Computation for a General Normal Random Variable". \begin{align*} P(8<X<14) &= P\left ( \frac{8-10}{2.5}<Z<\frac{14-10}{2.5} \right )\\ &= P\left ( -0.80<Z<1.60 \right )\\ &= 0.9452-0.2119\\ &= 0.7333 \end{align*} Figure $$\PageIndex{4}$$: Probability Computation for a General Normal Random Variable

Example $$\PageIndex{2}$$

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 $$\PageIndex{5}$$ "Probability Computation for Tire Tread Wear" illustrates the following computation:

\begin{align*} 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\left ( -1.67<Z<0.56\right )\\ &= 0.7123-0.0475\\ &= 0.6648 \end{align*} Figure $$\PageIndex{5}$$: Probability Computation for Tire Tread Wear

Note that the two $$z$$-scores were rounded to two decimal places in order to use Figure $$\PageIndex{1}$$ "Cumulative Normal Probability".

Example $$\PageIndex{3}$$

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 $$\PageIndex{6}$$ "Probability Computation for Exam Scores" illustrates the following computation:

\begin{align*} 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{align*} Figure $$\PageIndex{6}$$: Probability Computation for Exam Scores

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 $$\PageIndex{1}$$ after converting $$x$$-values to $$z$$-scores.

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