Continuous Random Variables

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Course:	MA121: Introduction to Statistics
Book:	Continuous Random Variables

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Date:	Friday, 4 April 2025, 3:55 PM

Description

First, this section talks about how to describe continuous distributions and compute related probabilities, including some basic facts about the normal distribution. Then, it covers how to compute probabilities related to any normal random variable and gives examples of using $z$ -score transformations. Finally, it defines tail probabilities and illustrates how to find them.

Continuous Random Variables
Probability Computations for General Normal Random Variables
- Exercises
- Answers
Areas of Tails of Distributions

Continuous Random Variables

Learning Objectives

To learn the concept of the probability distribution of a continuous random variable, and how it is used to compute probabilities.
To learn basic facts about the family of normally distributed random variables.

The Probability Distribution of a Continuous Random Variable

For a discrete random variable $X$ the probability that $X$ assumes one of its possible values on a single trial of the experiment makes good sense. This is not the case for a continuous random variable. For example, suppose $X$ denotes the length of time a commuter just arriving at a bus stop has to wait for the next bus. If buses run every $30$ minutes without fail, then the set of possible values of $X$ is the interval denoted $[0,30]$ , the set of all decimal numbers between $o$ and $30$ . But although the number $7.211916$ is a possible value of $X$ , there is little or no meaning to the concept of the probability that the commuter will wait precisely $7.211916$ minutes for the next bus. If anything the probability should be zero, since if we could meaningfully measure the waiting time to the nearest millionth of a minute it is practically inconceivable that we would ever get exactly $7.211916$ minutes. More meaningful questions are those of the form: What is the probability that the commuter's waiting time is less than $10$ minutes, or is between $5$ and $10$ minutes? In other words, with continuous random variables one is concerned not with the event that the variable assumes a single particular value, but with the event that the random variable assumes a value in a particular interval.

Definition

The probability distribution of a continuous random variable $X$ is an assignment of probabilities to intervals of decimal numbers using a function $f(x)$ , called $a$ density function, in the following way: the probability that $X$ assumes a value in the interval $[a, b]$ is equal to the area of the region that is bounded above by the graph of the equation $y=f(x)$ , bounded below by the $x$ -axis, and bounded on the left and right by the vertical lines through $a$ and $b$ , as illustrated in Figure 5.1 "Probability Given as Area of a Region under a Curve".

Figure 5.1 Probability Given as Area of a Region under a Curve

This definition can be understood as a natural outgrowth of the discussion in Section 2.1.3 "Relative Frequency Histograms" in Chapter 2 "Descriptive Statistics". There we saw that if we have in view a population (or a very large sample) and make measurements with greater and greater precision, then as the bars in the relative frequency histogram become exceedingly fine their vertical sides merge and disappear, and what is left is just the curve formed by their tops, as shown in Figure 2.5 "Sample Size and Relative Frequency Histograms" in Chapter 2 "Descriptive Statistics". Moreover the total area under the curve is $1$ , and the proportion of the population with measurements between two numbers $a$ and $b$ is the area under the curve and between $a$ and $b$ , as shown in Figure 2.6 "A Very Fine Relative Frequency Histogram" in Chapter 2 "Descriptive Statistics". If we think of $X$ as a measurement to infinite precision arising from the selection of any one member of the population at random, then $P(a < X < b)$ is simply the proportion of the population with measurements between $a$ and $b$ , the curve in the relative frequency histogram is the density function for $X$ , and we arrive at the definition just above.

Every density function $f(x)$ must satisfy the following two conditions:

For all numbers $x, f(x) \geq 0$ , so that the graph of $y=f(x)$ never drops below the $x$ -axis.
The area of the region under the graph of $y=f(x)$ and above the $x$ -axis is $1$ .

Because the area of a line segment is $o$ , the definition of the probability distribution of a continuous random variable implies that for any particular decimal number, say $a$ , the probability that $X$ assumes the exact value $a$ is $o$ . This property implies that whether or not the endpoints of an interval are included makes no difference concerning the probability of the interval.

For any continuous random variable $X$ :

$P(a \leq X \leq b)=P(a < X \leq b)=P(a \leq X < b)=P(a < X < b)$

Example 1

A random variable $X$ has the uniform distribution on the interval $[0,1]$ : the density function is $f(x)=1$ if $x$ is between $0$ and $1$ and $f(x)=0$ for all other values of $x$ , as shown in Figure 5.2 "Uniform Distribution on".

Figure 5.2 Uniform Distribution on $[0,1]$

Find $P(X > 0.75)$ , the probability that $X$ assumes $a$ value greater than $0.75$ .
Find $P(X \leq 0.2)$ , the probability that $X$ assumes $a$ value less than or equal to $0.2$ .
Find $P(0.4 < X < 0.7)$ , the probability that $X$ assumes $a$ value between $0.4$ and $0.7$ .

Solution:

$P(X > 0.75)$ is the area of the rectangle of height $1$ and base length $1-0.75=0.25$ , hence is $base\ \times height\ =(0.25) \cdot(1)=0.25$ . See Figure 5.3 "Probabilities from the Uniform Distribution on "(a).
$P(X \leq 0.2)$ is the area of the rectangle of height $1$ and base length $0.2-0=0.2$ , hence is $base\ \times height\ =(0.2) \cdot(1)=0.2$ . See Figure 5.3 "Probabilities from the Uniform Distribution on "(b).
$P(0.4 < X < 0.7)$ is the area of the rectangle of height $1$ and length $0.7-0.4=0.3$ , hence is $base\ \times height\ =(0.3) \cdot(1)=0.3$ . See Figure 5.3 "Probabilities from the Uniform Distribution on "(c).

Figure 5.3 Probabilities from the Uniform Distribution on $[0,1]$

Example 2

A man arrives at a bus stop at a random time (that is, with no regard for the scheduled service) to catch the next bus. Buses run every 30 minutes without fail, hence the next bus will come any time during the next 30 minutes with evenly distributed probability (a uniform distribution). Find the probability that a bus will come within the next 10 minutes.

Solution:

The graph of the density function is a horizontal line above the interval from $0$ to $30$ and is the $x$ -axis everywhere else. Since the total area under the curve must be $1$ , the height of the horizontal line is $1/30$ . See Figure 5.4 "Probability of Waiting At Most $10$ Minutes for a Bus". The probability sought is $P(0 \leq X \leq 10)$ . By definition, this probability is the area of the rectangular region bounded above by the horizontal line $f(x)=1 / 30$ , bounded below by the $x$ -axis, bounded on the left by the vertical line at $0$ (the $y$ -axis), and bounded on the right by the vertical line at $10$ . This is the shaded region in Figure 5.4 "Probability of Waiting At Most $10$ Minutes for a Bus". Its area is the base of the rectangle times its height, $10 \cdot(1 / 30)=1 / 3$ . Thus $P(0 \leq X \leq 10)=1 / 3$ .

Figure 5.4 Probability of Waiting At Most $10$ Minutes for a Bus

Normal Distributions

Most people have heard of the "bell curve". It is the graph of a specific density function $f(x)$ that describes the behavior of continuous random variables as different as the heights of human beings, the amount of a product in a container that was filled by a high-speed packing machine, or the velocities of molecules in a gas. The formula for $f(x)$ contains two parameters $\mu$ and $\sigma$ that can be assigned any specific numerical values, so long as $\sigma$ is positive. We will not need to know the formula for $f(x)$ , but for those who are interested it is

$f(x)=\frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{1}{2}(\mu-x)^{2} / \sigma^{2}}$

where $\pi \approx 3.14159$ and $e \approx 2.71828$ is the base of the natural logarithms.

Each different choice of specific numerical values for the pair $\mu$ and $\sigma$ gives a different bell curve. The value of $\mu$ determines the location of the curve, as shown in Figure 5:5 "Bell Curves with ". In each case the curve is symmetric about $\mu$ .

Figure 5.5 Bell Curves with $\sigma=0.25$ and Different Values of $\mu$

The value of $\sigma$ determines whether the bell curve is tall and thin or short and squat, subject always to the condition that the total area under the curve be equal to $1$ . This is shown in Figure 5.6 "Bell Curves with ", where we have arbitrarily chosen to center the curves at $\mu=6$ .

Figure 5.6 Bell Curves with $\mu=6$ and Different Values of $\sigma$

Definition

The probability distribution corresponding to the density function for the bell curve with parameters $\mu$ and $\sigma$ is called the normal distribution with mean $\mu$ and standard deviation $\sigma$ .

Definition

A continuous random variable whose probabilities are described by the normal distribution with mean $\mu$ and standard deviation $\sigma$ is called a normally distributed random variable, or a normal random variable for short, with mean $\mu$ and standard deviation $\sigma$ .

Figure 5.7 "Density Function for a Normally Distributed Random Variable with Mean" shows the density function that determines the normal distribution with mean $\mu$ and standard deviation $\sigma$ . We repeat an important fact about this curve:

The density curve for the normal distribution is symmetric about the mean.

Figure 5.7 Density Function for a Normally Distributed Random Variable with Mean $\mu$ and Standard Deviation $\sigma$

Example 3

Heights of 25 -year-old men in a certain region have mean $69.75$ inches and standard deviation $2.59$ inches. These heights are approximately normally distributed. Thus the height $X$ of a randomly selected 25 -year-old man is a normal random variable with mean $\mu=69.75$ and standard deviation $\sigma=2.59$ . Sketch a qualitatively accurate graph of the density function for $X$ . Find the probability that a randomly selected 25-year-old man is more than $69.75$ inches tall.

Solution:

The distribution of heights looks like the bell curve in Figure 5.8 "Density Function for Heights of 25 Year-Old Men". The important point is that it is centered at its mean, $69.75$ , and is symmetric about the mean.

Figure 5.8 Density Function for Heights of 25-Year-Old Men

Since the total area under the curve is $1$ , by symmetry the area to the right of $69.75$ is half the total, or $0.5$ . But this area is precisely the probability $P(X > 69.75)$ , the probability that a randomly selected 25 year-old man is more than $69.75$ inches tall.

We will learn how to compute other probabilities in the next two sections.

This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work's original creator or licensor.

Key Takeaways

For a continuous random variable $X$ the only probabilities that are computed are those of $X$ taking a value in a specified interval.
The probability that $X$ take a value in a particular interval is the same whether or not the endpoints of the interval are included.
The probability $P(a < X < b)$ , that $X$ take a value in the interval from $a$ to $b$ , is the area of the region between the vertical lines through $a$ and $b$ , above the $x$ -axis, and below the graph of a function $f(x)$ called the density function.
A normally distributed random variable is one whose density function is a bell curve.
Every bell curve is symmetric about its mean and lies everywhere above the $x$ -axis, which it approaches asymptotically (arbitrarily closely without touching).

Exercises

Basic

1. A continuous random variable $X$ has a uniform distribution on the interval $[5,12]$ . Sketch the graph of its density function.

3. A continuous random variable $X$ has a normal distribution with mean $100$ and standard deviation $10$ . Sketch a qualitatively accurate graph of its density function.

5. A continuous random variable $X$ has a normal distribution with mean $73$ . The probability that $X$ takes a value greater than $80$ is $0.212$ . Use this information and the symmetry of the density function to find the probability that $X$ takes a value less than $66$ . Sketch the density curve with relevant regions shaded to illustrate the computation.

7. A continuous random variable $X$ has a normal distribution with mean $50.5$ . The probability that $X$ takes a value less than $54$ is $0.76$ . Use this information and the symmetry of the density function to find the probability that $X$ takes a value greater than $47$ . Sketch the density curve with relevant regions shaded to illustrate the computation.

9. The figure provided shows the density curves of three normally distributed random variables $X_{A}$ , $X_{B}$ , and $X_{C}$ . Their standard deviations (in no particular order) are $15$ , $7$ , and $20$ . Use the figure to identify the values of the means $\mu_{A}$ , $\mu_{B}$ , and $\mu_{C}$ and standard deviations $\sigma_{A}$ , $\sigma_{B}$ , and $\sigma_{C}$ of the three random variables.

Applications

11. Dogberry's alarm clock is battery operated. The battery could fail with equal probability at any time of the day or night. Every day Dogberry sets his alarm for 6:30 a.m. and goes to bed at 10:00 p.m. Find the probability that when the clock battery finally dies, it will do so at the most inconvenient time, between 10:00 p.m. and 6:30 a.m.

13. The amount $X$ of orange juice in a randomly selected half-gallon container varies according to a normal distribution with mean $64$ ounces and standard deviation $0.25$ ounce.

a. Sketch the graph of the density function for

$X$ .
b. What proportion of all containers contain less than a half gallon (

$64$ ounces)? Explain.

c. What is the median amount of orange juice in such containers? Explain.

Answers

1. The graph is a horizontal line with height $1 / 7$ from $x=5$ to $x=12$

3. The graph is a bell-shaped curve centered at $100$ and extending from about $70$ to $130$ .

5. $0.212$

7. $0.76$

9. $\mu_{A}=100$ , $\mu_{B}=200$ , $\mu_{C}=300$ , $\sigma_{A}=7$ , $\sigma_{B}=20$ , $\sigma_{C}=15$

11. $0.3542$

13. a. The graph is a bell-shaped curve centered at

$64$ and extending from about

$63.25$ to

$64.75$ .
b.

$0.5$
c.

$64$

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.

Exercises

Basic

1. $X$ is a normally distributed random variable with mean $57$ and standard deviation $6$ . Find the probability indicated.

$P(X < 59.5)$
b.

$P(X < 46.2)$
c.

$P(X > 52.2)$

d. $P(X > 70)$

3. $X$ is a normally distributed random variable with mean $112$ and standard deviation $15$ . Find the probability indicated.

a. $P(100 < X < 125)$
b. $P(91 < X < 107)$
c. $P(118 < X < 160)$

5. $X$ is a normally distributed random variable with mean $500$ and standard deviation $25$ . Find the probability indicated.

a. $P(X < 400)$
b. $P(466 < X < 625)$

7. $X$ is a normally distributed random variable with mean $15$ and standard deviation $1$ . Use Figure 12.2 "Cumulative Normal Probability" to find the first probability listed. Find the second probability using the symmetry of the density curve. Sketch the density curve with relevant regions shaded to illustrate the computation.

a. $P(X < 12), P(X > 18)$
b. $P(X < 14), P(X > 16)$
c. $P(X < 11.25), P(X > 18.75)$
d. $P(X < 12.67), P(X > 17.33)$

9. $X$ is a normally distributed random variable with mean $67$ and standard deviation $13$ . The probability that $X$ takes a value in the union of intervals $(-\infty, 67-a] \cup[67+a, \infty)$ will be denoted $P(X \leq 67-a$ or $X \geq 67+a)$ . Use Figure 12.2 "Cumulative Normal Probability" to find the following probabilities of this type. Sketch the density curve with relevant regions shaded to illustrate the computation. Because of the symmetry of the density curve you need to use Figure 12.2 "Cumulative Normal Probability" only one time for each part.

a. $P(X < 57$ or $X > 77)$
b. $P(X < 47$ or $X > 87)$
c. $P(X < 49$ or $X > 85)$
d. $P(X < 37$ or $X > 97)$

Applications

11. The amount $X$ of beverage in a can labeled $12$ ounces is normally distributed with mean $12.1$ ounces and standard deviation $0.05$ ounce. A can is selected at random.

a. Find the probability that the can contains at least

$12$ ounces.
b. Find the probability that the can contains between

$11.9$ and

$12.1$ ounces.

13. The systolic blood pressure $X$ of adults in a region is normally distributed with mean $112 \mathrm{~mm} \mathrm{Hg}$ and standard deviation $15 \mathrm{~mm} \mathrm{Hg}$ . A person is considered "prehypertensive" if his systolic blood pressure is between $120$ and $130 \mathrm{~mm}\mathrm{Hg}$ . Find the probability that the blood pressure of a randomly selected person is prehypertensive.

15. Heights $X$ of adult men are normally distributed with mean $69.1$ inches and standard deviation $2.92$ inches. Juliet, who is $63.25$ inches tall, wishes to date only men who are taller than she but within $6$ inches of her height. Find the probability that the next man she meets will have such a height.

17. A regulation golf ball may not weigh more than $1.620$ ounces. The weights $X$ of golf balls made by a particular process are normally distributed with mean $1.361$ ounces and standard deviation $0.09$ ounce. Find the probability that a golf ball made by this process will meet the weight standard.

19. The amount of non-mortgage debt per household for households in a particular income bracket in one part of the country is normally distributed with mean $\$ 28,350$ and standard deviation $\$ 3,425$ . Find the probability that a randomly selected such household has between $\$ 20,000$ and $\$ 30,000$ in non-mortgage debt.

21. The distance from the seat back to the front of the knees of seated adult males is normally distributed with mean $23.8$ inches and standard deviation $1.22$ inches. The distance from the seat back to the back of the next seat forward in all seats on aircraft flown by a budget airline is $26$ inches. Find the proportion of adult men flying with this airline whose knees will touch the back of the seat in front of them.

23. The useful life of a particular make and type of automotive tire is normally distributed with mean $57,500$ , miles and standard deviation $950$ miles.

a. Find the probability that such a tire will have a useful life of between $57,000$ and $58,000$ miles.
b. Hamlet buys four such tires. Assuming that their lifetimes are independent, find the probability that all four will last between $57,000$ and $58,000$ miles. (If so, the best tire will have no more than $1,000$ miles left on it when the first tire fails.) Hint: There is a binomial random variable here, whose value of $p$ comes from part (a).

25. The lengths of time taken by students on an algebra proficiency exam (if not forced to stop before completing it) are normally distributed with mean $28$ minutes and standard deviation $1.5$ minutes.

a. Find the proportion of students who will finish the exam if a $30$ -minute time limit is set.
b. Six students are taking the exam today. Find the probability that all six will finish the exam within the $30$ -minute limit, assuming that times taken by students are independent. Hint: There is a binomial random variable here, whose value of $p$ comes from part (a).

27. A regulation hockey puck must weigh between $5.5$ and $6$ ounces. In an alternative manufacturing process the mean weight of pucks produced is $5.75$ ounce. The weights of pucks have a normal distribution whose standard deviation can be decreased by increasingly stringent (and expensive) controls on the manufacturing process. Find the maximum allowable standard deviation so that at most $0.005$ of all pucks will fail to meet the weight standard. (Hint: The distribution is symmetric and is centered at the middle of the interval of acceptable weights).

Answers

1. a.

$0.6628$
b.

$0.0359$
c.

$0.7881$

d. $0.0150$

3. a.

$0.5959$
b.

$0.2899$

C. $0.3439$

5. a.

$0.0000$

b. $0.9131$

7. a.

$0.0013,0.0013$
b.

$0.1587,0.1587$
c.

$0.0001,0.0001$

d. $0.0099,0.0099$

9. a.

$0.4412$
b.

$0.1236$
C.

$0.1676$
d.

$0.0208$

11. a. $0.9772$
b. $0.5000$

13. $0.1830$

15. $0.4971$

17. $0.9980$

19. $0.6771$

21. $0.0359$

23. a. $0.4038$
b. $0.0266$

25. a. $0.9082$
b. $0.5612$

27. $0.089$

Areas of Tails of Distributions

Learning Objective

To learn how to find, for a normal random variable $X$ and an area $a$ , the value $\mathrm{x}^{*}$ of $X$ so that $P\left(X < \mathrm{x}^{*}\right)=a$ or that $P\left(X > \mathrm{x}^{*}\right)=a$ , whichever is required.

Definition

The left tail of a density curve $y=f(x)$ of a continuous random variable $X$ cut off by a value $\mathrm{x}^{*}$ of $X$ is the region under the curve that is to the left of $\mathrm{x}^{*}$ , as shown by the shading in Figure 5.19 "Right and Left Tails of a Distribution" (a). The right tail cut off by $\mathrm{x}^{*}$ is defined similarly, as indicated by the shading in Figure 5.19 "Right and Left Tails of a Distribution"(b).

Figure 5.19 Right and Left Tails of a Distribution

The probabilities tabulated in Figure 12.2 "Cumulative Normal Probability" are areas of left tails in the standard normal distribution.

Tails of the Standard Normal Distribution

At times it is important to be able to solve the kind of problem illustrated by Figure 5.20. We have a certain specific area in mind, in this case the area $0.0125$ of the shaded region in the figure, and we want to find the value $\mathrm{z}^{*}$ of $Z$ that produces it. This is exactly the reverse of the kind of problems encountered so far. Instead of knowing a value $\mathrm{z}^{*}$ of $Z$ and finding a corresponding area, we know the area and want to find $\mathrm{z}^{*}$ . In the case at hand, in the terminology of the definition just above, we wish to find the value $\mathrm{z}^{*}$ that cuts off a left tail of area $0.0125$ in the standard normal distribution.

The idea for solving such a problem is fairly simple, although sometimes its implementation can be a bit complicated. In a nutshell, one reads the cumulative probability table for $Z$ in reverse, looking up the relevant area in the interior of the table and reading off the value of $Z$ from the margins.

Figure 5.20 $Z$ Value that Produces a Known Area

Example 12

Find the value $\mathrm{z}^{*}$ of $Z$ as determined by Figure 5.20: the value $\mathrm{z}^{*}$ that cuts off a left tail of area $0.0125$ in the standard normal distribution. In symbols, find the number $\mathrm{z}^{*}$ such that $P\left(Z < \mathrm{z}^{*}\right)=0.0125$ .

Solution:

The number that is known, $0.0125$ , is the area of a left tail, and as already mentioned the probabilities tabulated in Figure 12.2 "Cumulative Normal Probability" are areas of left tails. Thus to solve this problem we need only search in the interior of Figure 12.2 "Cumulative Normal Probability" for the number $0.0125$ . It lies in the row with the heading $-2.2$ and in the column with the heading $0.04$ . This means that $P(Z < -2.24)=0.0125$ , hence $\mathrm{z}^{*}=-2.24$ .

Example 13

Find the value $\mathrm{z}^{*}$ of $Z$ as determined by Figure 5.21: the value $\mathrm{z}^{*}$ that cuts off a right tail of area $0.0250$ in the standard normal distribution. In symbols, find the number $\mathrm{z}^{*}$ such that $P\left(Z > \mathrm{z}^{*}\right)=0.0250$ .

Figure 5.21 $Z$ Value that Produces a Known Area

Solution:

The important distinction between this example and the previous one is that here it is the area of a right tail that is known. In order to be able to use Figure 12.2 "Cumulative Normal Probability" we must first find that area of the left tail cut off by the unknown number $\mathrm{z}^{*}$ . Since the total area under the density curve is $1$ , that area is $1-0.0250=0.9750$ . This is the number we look for in the interior of Figure 12.2 "Cumulative Normal Probability". It lies in the row with the heading $1.9$ and in the column with the heading $0.06$ . Therefore $\mathrm{z}^{*}=1.96$ .

Definition

The value of the standard normal random variable $Z$ that cuts off a right tail of area $c$ is denoted $z_{c}$ . By symmetry, value of $Z$ that cuts off a left tail of area $c$ is $-z_{c}$ . See Figure 5.22 "The Numbers".

Figure 5.22 The Numbers $z_{c}$ and $-z_{c}$ .

The previous two examples were atypical because the areas we were looking for in the interior of Figure 12.2 "Cumulative Normal Probability" were actually there. The following example illustrates the situation that is more common.

Example 14

Find $z_{.01}$ and $-z_{.01}$ , the values of $Z$ that cut off right and left tails of area $0.01$ in the standard normal distribution.

Solution:

Since $-z_{.01}$ cuts off a left tail of area $0.01$ and Figure 12.2 "Cumulative Normal Probability" is a table of left tails, we look for the number $0.0100$ in the interior of the table. It is not there, but falls between the two numbers $0.0102$ and $0.0099$ in the row with heading $-2.3$ . The number $0.0099$ is closer to $0.0100$ than $0.0102$ is, so for the hundredths place in $-z_{.01}$ we use the heading of the column that contains $0.0099$ , namely, $0.03$ , and write $-z_{.01} \approx-2.33$ .

The answer to the second half of the problem is automatic: since $-z_{.01}=-2.33$ , we conclude immediately that $z_{.01}=2.33$ .

We could just as well have solved this problem by looking for $z_{.01}$ first, and it is instructive to rework the problem this way. To begin with, we must first subtract $0.01$ from $1$ to find the area $1-0.0100=0.9900$ of the left tail cut off by the unknown number $z_{.01}$ . See Figure 5.23
"Computation of the Number ". Then we search for the area $0.9900$ in Figure 12.2 "Cumulative Normal Probability". It is not there, but falls between the numbers $0.9898$ and $0.9901$ in the row with heading $2.3$ . Since $0.9901$ is closer to $0.9900$ than $0.9898$ is, we use the column heading above it, $0.03$ , to obtain the approximation $z_{.01} \approx 2.33$ . Then finally $-z_{.01} \approx-2.33$ .

Figure 5.23 Computation of the Number $z_{.01}$

Tails of General Normal Distributions

The problem of finding the value $\mathrm{x}^{*}$ of a general normally distributed random variable $X$ that cuts off a tail of a specified area also arises. This problem may be solved in two steps.

Suppose $X$ is a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$ . To find the value $\mathrm{x}^{*}$ of $X$ that cuts off a left or right tail of area $c$ in the distribution of $X$ :

1. find the value $\mathrm{z}^{*}$ of $Z$ that cuts off a left or right tail of area $c$ in the standard normal distribution;
2. $\mathrm{z}^{*}$ is the $z$ -score of $\mathrm{x}^{*}$ ; compute $\mathrm{x}^{*}$ using the destandardization formula

$\mathrm{x}^{*}=\mu+\mathrm{z}^{*} \sigma$

In short, solve the corresponding problem for the standard normal distribution, thereby obtaining the $z$ -score of $x^{*}$ , then destandardize to obtain $x^{*}$ .

Example 15

Find $\mathrm{x}^{*}$ such that $P\left(X < \mathrm{x}^{*}\right)=0.9332$ , where $X$ is a normal random variable with mean $\mu=10$ and standard deviation $\sigma=2.5$ .

Solution:

All the ideas for the solution are illustrated in Figure 5.24 "Tail of a Normally Distributed Random Variable". Since $0.9332$ is the area of a left tail, we can find $\mathrm{z}^{*}$ simply by looking for $0.9332$ in the interior of Figure 12.2 "Cumulative Normal Probability". It is in the row and column with headings $1.5$ and $0.00$ , hence $\mathrm{z}^{*}=1.50$ . Thus $\mathrm{x}^{*}$ is $1.50$ standard deviations above the mean, so

$\mathrm{x}^{*}=\mu+\mathrm{z}^{*} \sigma=10+1.50 \cdot 2.5=13.75$

Figure 5.24 Tail of a Normally Distributed Random Variable

Example 16

Find $\mathrm{x}^{*}$ such that $P\left(X > \mathrm{x}^{*}\right)=0.65$ , where $X$ is a normal random variable with mean $\mu=175$ and standard deviation $\sigma=12$ .

Solution:

The situation is illustrated in Figure 5.25 "Tail of a Normally Distributed Random Variable". Since $0.65$ is the area of a right tail, we first subtract it from $1$ to obtain $1-0.65=0.35$ , the area of the complementary left tail. We find $\mathrm{z}^{*}$ by looking for $0.3500$ in the interior of Figure 12.2 "Cumulative Normal Probability". It is not present, but lies between table entries $0.3520$ and $0.3483$ . The entry $0.3483$ with row and column headings $-0.3$ and $0.09$ is closer to $0.3500$ than the other entry is, so $\mathrm{z}^{*} \approx-0.39$ . Thus $\mathrm{x}^{*}$ is $0.39$ standard deviations below the mean, so

$\mathrm{x}^{*}=\mu+\mathrm{z}^{*} \sigma=175+(-0.39) \cdot 12=170.32$

Figure 5.25 Tail of a Normally Distributed Random Variable

Example 17

Scores on a standardized college entrance examination (CEE) are normally distributed with mean $510$ and standard deviation $60$ . A selective university decides to give serious consideration for admission to applicants whose CEE scores are in the top $5 \%$ of all CEE scores. Find the minimum score that meets this criterion for serious 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 minimum score that is in the top $5 \%$ of all CEE is the score $\mathrm{x}^{*}$ that cuts off a right tail in the distribution of $X$ of area $0.05$ ( $5 \%$ expressed as a proportion). See Figure 5.26 "Tail of a Normally Distributed Random Variable".

Figure 5.26 Tail of a Normally Distributed Random Variable

Since $0.0500$ is the area of a right tail, we first subtract it from $1$ to obtain $1-0.0500=0.9500$ , the area of the complementary left tail. We find $\mathrm{z}^{*}=z_{.05}$ by looking for $0.9500$ in the interior of Figure 12.2 "Cumulative Normal Probability". It is not present, and lies exactly half-way between the two nearest entries that are, $0.9495$ and $0.9505$ . In the case of a tie like this, we will always average the values of $Z$ corresponding to the two table entries, obtaining here the value $\mathrm{z}^{*}=1.645$ . Using this value, we conclude that $\mathrm{x}^{*}$ is $1.645$ standard deviations above the mean, so

$\mathrm{x}^{*}=\mu+\mathrm{z}^{*} \sigma=510+1.645 \cdot 60=608.7$

Example 18

All boys at a military school must run a fixed course as fast as they can as part of a physical examination. Finishing times are normally distributed with mean $29$ minutes and standard deviation $2$ minutes. The middle $75 \%$ of all finishing times are classified as "average". Find the range of times that are average finishing times by this definition.

Solution:

Let $X$ denote the finish time of a randomly selected boy. Then $X$ is normally distributed with mean $29$ and standard deviation $2$ . The probability that $X$ lie in a particular interval is the same as the proportion of all finish times that lie in that interval. Thus the situation is as shown in Figure 5.27 "Distribution of Times to Run a Course". Because the area in the middle corresponding to "average" times is $0.75$ , the areas of the two tails add up to $1-0.75=0.25$ in all. By the symmetry of the density curve each tail must have half of this total, or area $0.125$ each. Thus the fastest time that is "average" has $z$ -score $-z_{.125}$ , which by Figure 12.2 "Cumulative Normal Probability" is $-1.15$ , and the slowest time that is "average" has $z$ -score $z_{.125}=1.15$ . The fastest and slowest times that are still considered average are

$x_{\text {fast }}=\mu+\left(-z_{.125}\right) \sigma=29+(-1.15) \cdot 2=26.7$

and

$x_{\text {slow }}=\mu+z_{.125} \sigma=29+(1.15) \cdot 2=31.3$

Figure 5.27 Distribution of Times to Run a Course

A boy has an average finishing time if he runs the course with a time between $26.7$ and $31.3$ minutes, or equivalently between $26$ minutes $42$ seconds and $31$ minutes $18$ seconds.

Key Takeaways

The problem of finding the number $\mathrm{z}^{*}$ so that the probability $P\left(Z < \mathrm{z}^{*}\right)$ is a specified value $c$ is solved by looking for the number $c$ in the interior of Figure 12.2 "Cumulative Normal Probability" and reading $\mathrm{z}^{*}$ from the margins.
The problem of finding the number $\mathrm{z}^{*}$ so that the probability $P\left(Z > \mathrm{z}^{*}\right)$ is a specified value $c$ is solved by looking for the complementary probability $1-c$ in the interior of Figure 12.2 "Cumulative Normal Probability" and reading $z^*$ from the margins.
For a normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$ , the problem of finding the number $\mathrm{x}^{*}$ so that $P\left(X < \mathrm{x}^{*}\right)$ is a specified value $c$ (or so that $P\left(X > \mathrm{x}^{*}\right)$ is a specified value $c$ ) is solved in two steps: (1) solve the corresponding problem for $Z$ with the same value of $c$ , thereby obtaining the $z$ -score, $\mathrm{z}^{*}$ , of $\mathrm{x}^{*}$ ; (2) find $\mathrm{x}^{*}$ using $\mathrm{x}^{*}=\mu+\mathrm{z}^{*} \cdot \sigma$ .
The value of $Z$ that cuts off a right tail of area $c$ in the standard normal distribution is denoted $z_{c}$ .

Answers

1. a.

$-2.43$
b.

$2.17$
c.

$-1.28$

d. $2.29$

3. a.

$-1.04$
b.

$0.67$
c.

$0.43$

d. $-0.84$

5. a.

$1.96$

b. $0.84$

7. a.

$87.52$

b. $89.58$

9. $15.32$

11. a.

$382$

b. $415$

13. $7030.14, 7069, 7107.86$

15. $85.90$

17. $65,054$

19.

$1.58, 1.86$

21. $66.5, 78.5$

23. a. $0.5$
b. $2.005$

Exercises

Basic

1. Find the value of $\mathrm{z}^{*}$ that yields the probability shown.

$P\left(Z < \mathrm{z}^{*}\right)=0.0075$
b.

$P\left(Z < \mathrm{z}^{*}\right)=0.9850$
c.

$P\left(Z > \mathrm{z}^{*}\right)=0.8997$
d.

$P\left(Z > \mathrm{z}^{*}\right)=0.0110$

3. Find the value of $\mathrm{z}^{*}$ that yields the probability shown.

a. $P\left(Z < \mathrm{z}^{*}\right)=0.1500$
b. $P\left(Z < \mathrm{z}^{*}\right)=0.7500$
c. $P\left(Z > \mathrm{z}^{*}\right)=0.3333$
d. $P\left(Z > \mathrm{z}^{*}\right)=0.8000$

5. Find the indicated value of $Z$ . (It is easier to find $-z_{c}$ and negate it.)

a. $z 0.025$
b. $z 0.20$

7. Find the value of $x^{*}$ that yields the probability shown, where $X$ is a normally distributed random variable $X$ with mean $83$ and standard deviation $4$ .

a. $P\left(X < \mathrm{x}^{*}\right)=0.8700$
b. $P\left(X > \mathrm{x}^{*}\right)=0.0500$

9. $X$ is a normally distributed random variable $X$ with mean $15$ and standard deviation $0.25$ . Find the values $x_{L}$ and $x_{R}$ of $X$ that are symmetrically located with respect to the mean of $X$ and satisfy $P\left(x_{L} < X < x_{R}\right) = 0.80$ . (Hint. First solve the corresponding problem for $Z$ .)

Applications

11. Scores on a national exam are normally distributed with mean $382$ and standard deviation $26$ .

a. Find the score that is the

$50$ th percentile.
b. Find the score that is the

$90$ th percentile.

13. The monthly amount of water used per household in a small community is normally distributed with mean $7,069$ gallons and standard deviation $58$ gallons. Find the three quartiles for the amount of water used.

15. Scores on the common final exam given in a large enrollment multiple section course were normally distributed with mean $69.35$ and standard deviation $12.93$ . The department has the rule that in order to receive an A in the course his score must be in the top $10 \%$ of all exam scores. Find the minimum exam score that meets this requirement.

17. Tests of a new tire developed by a tire manufacturer led to an estimated mean tread life of $67,350$ miles and standard deviation of $1,120$ miles. The manufacturer will advertise the lifetime of the tire (for example, a " $50,000$ mile tire") using the largest value for which it is expected that $98 \%$ of the tires will last at least that long. Assuming tire life is normally distributed, find that advertised value.

19. The weights $X$ of eggs produced at a particular farm are normally distributed with mean $1.72$ ounces and standard deviation $0.12$ ounce. Eggs whose weights lie in the middle $75 \%$ of the distribution of weights of all eggs are classified as "medium". Find the maximum and minimum weights of such eggs. (These weights are endpoints of an interval that is symmetric about the mean and in which the weights of $75 \%$ of the eggs produced at this farm lie.)

21. All students in a large enrollment multiple section course take common in-class exams and a common final, and submit common homework assignments. Course grades are assigned based on students' final overall scores, which are approximately normally distributed. The department assigns a $C$ to students whose scores constitute the middle $2 / 3$ of all scores. If scores this semester had mean $72.5$ and standard deviation $6.14$ , find the interval of scores that will be assigned a $C$ .

Additional Exercises

23. A machine for filling $2$ -liter bottles of soft drink delivers an amount to each bottle that varies from bottle to bottle according to a normal distribution with standard deviation $0.002$ liter and mean whatever amount the machine is set to deliver.

a. If the machine is set to deliver

$2$ liters (so the mean amount delivered is

$2$ liters) what proportion of the bottles will contain at least

$2$ liters of soft drink?
b. Find the minimum setting of the mean amount delivered by the machine so that at least

$99 \%$ of all bottles will contain at least

$2$ liters.

Continuous Random Variables

Description

Table of contents

Continuous Random Variables

Learning Objectives

The Probability Distribution of a Continuous Random Variable

Definition

Example 1

Example 2

Normal Distributions

Definition

Definition

Example 3

Key Takeaways

Exercises

Basic

Applications

Answers

Probability Computations for General Normal Random Variables

Learning Objective

Example 9

Example 10

Example 11

Key Takeaway

Exercises

Basic

Applications

Answers

Areas of Tails of Distributions

Learning Objective

Definition

Tails of the Standard Normal Distribution

Example 12

Example 13

Definition

Example 14

Tails of General Normal Distributions

Example 15

Example 16

Example 17

Example 18

Key Takeaways

Answers

Exercises

Basic

Applications

Additional Exercises