# Continuous Random Variables

Edited by Paul Ducham

CONTINUOUS PROBABILITY DISTRIBUTIONS

When a random variable may assume any numerical value in one or more intervals on the real number line, then the random variable is called a continuous random variable. For example, the EPAcombined city and highway mileage of a randomly selected midsize car is a continuous random variable. Furthermore, the temperature (in degrees Fahrenheit) of a randomly selected cup of coffee at a fast-food restaurant is also a continuous random variable. We often wish to compute probabilities about the range of values that a continuous random variable x might attain. For example, suppose that marketing research done by a fast-food restaurant indicates that coffee tastes best if its temperature is between 153° F and 167° F. The restaurant might then wish to find the probability that x, the temperature of a randomly selected cup of coffee at the restaurant, will be between 153° and 167°. This probability would represent the proportion of coffee served by the restaurant that has a temperature between 153° and 167°. Moreover, one minus this probability would represent the proportion of coffee served by the restaurant that has a temperature outside the range 153° to 167°.

In general, to compute probabilities concerning a continuous random variable x, we assign probabilities to intervals of values by using what we call a continuous probability distribution. To understand this idea, suppose that f (x) is a continuous function of the numbers on the real line, and consider the continuous curve that results when f (x) is graphed. Such a curve is illustrated in Figure 6.1. Then:

##### Continuous Probability Distributions

The curve f(x) is the continuous probability distribution of the random variable x if the probability that x will be in a specified interval of numbers is the area under the curve f(x) corresponding to the interval. Sometimes we refer to a continuous probability distribution as a probability curve or as a probability density function.

An area under a continuous probability distribution (or probability curve) is a probability. For instance, consider the range of values on the number line from the number a to the number b—that is, the interval of numbers from a to b. If the continuous random variable x is described by the probability curve f(x), then the area under f(x) corresponding to the interval from a to b is the probability that x will attain a value between a and b. Such a probability is illustrated as the shaded area in Figure 6.1. We write this probability as P(a ≤x≤ b). For example, suppose that the continuous probability curve f(x) in Figure 6.1 describes the random variable x = the temperature of a randomly selected cup of coffee at the fast-food restaurant. It then follows that P(153≤ x ≤167)- the probability that the temperature of a randomly selected cup of coffee at the fast-food restaurant will be between 153° and 167°—is the area under the curve f(x) between 153 and 167.

We know that any probability is 0 or positive, and we also know that the probability assigned to all possible values of x must be 1. It follows that, similar to the conditions required for a discrete probability distribution, a probability curve must satisfy the following properties:

Properties of a Continuous Probability Distribution

The continuous probability distribution (or probability curve) f(x) of a random variable x must satisfy the following two conditions: 1 f (x)≥0 for any value of x. 2 The total area under the curve f (x) is equal to 1. Any continuous curve f (x) that satisfies these conditions is a valid continuous probability distribution. Such probability curves can have a variety of shapes—bell-shaped and symmetrical, skewed to the right, skewed to the left, or any other shape. In a practical problem, the shape of a probability curve would be estimated by looking at a frequency (or relative frequency) histogram of observed data. Later in this article, we study probability curves having several different shapes.

We have seen that to calculate a probability concerning a continuous random variable, we must compute an appropriate area under the curve f (x). In theory, such areas are calculated by calculus methods and/or numerical techniques. Because these methods are difficult, needed areas under commonly used probability curves have been compiled in statistical tables. As we need them, we show how to use the required statistical tables. Also, note that since there is no area under a continuous curve at a single point, the probability that a continuous random variable x will attain a single value is always equal to 0. It follows that in Figure 6.1 we have P(x=a) = 0 and P(x = b)=0. Therefore, P(a ≤ x ≤ b) equals P(a < x < b) because each of the interval endpoints a and b has a probability that is equal to 0.

UNIFORM DISTRIBUTION

Suppose that over a period of several days the manager of a large hotel has recorded the waiting times of 1,000 people waiting for an elevator in the lobby at dinnertime (5:00 P.M. to 7:00 P.M.). The observed waiting times range from zero to four minutes. Furthermore, when the waiting times are arranged into a histogram, the bars making up the histogram have approximately equal heights, giving the histogram a rectangular appearance. This implies that the relative frequencies of all waiting times from zero to four minutes are about the same. Therefore, it is reasonable to use The Uniform Distribution to describe the random variable x, the amount of time a randomly selected hotel patron spends waiting for the elevator. In general, the equation that describes the uniform distribution is given in the following box, and this equation is graphed in Figure 6.2(a).

NORMAL PROBABILITY DISTRIBUTION

The normal curve The bell-shaped appearance of the normal probability distribution is illustrated in Figure 6.3. The equation that defines this normal curve is given in the following box:

The normal probability distribution has several important properties:

1 There is an entire family of normal probability distributions; the specific shape of each normal distribution is determined by its mean µ and its standard deviation σ.

2 The highest point on the normal curve is located at the mean, which is also the median and the mode of the distribution.

3 The normal distribution is symmetrical: The curve’s shape to the left of the mean is the mirror image of its shape to the right of the mean.

4 The tails of the normal curve extend to infinity in both directions and never touch the horizontal axis. However, the tails get close enough to the horizontal axis quickly enough to ensure that the total area under the normal curve equals 1.

5 Since the normal curve is symmetrical, the area under the normal curve to the right of the mean (µ) equals the area under the normal curve to the left of the mean, and each of these areas equals .5 (see Figure 6.3).

Intuitively, the mean m positions the normal curve on the real line. This is illustrated in Figure 6.4(a). This figure shows two normal curves with different means µ1 and µ2 (where µ1 is greater than µ2) and with equal standard deviations. We see that the normal curve with mean  µ1  is centered farther to the right.

The variance σ2 (and the standard deviation σ) measure the spread of the normal curve. This is illustrated in Figure 6.4(b), which shows two normal curves with the same mean and two different standard deviations σ1 and σ2. Because σ1 is greater than σ2, the normal curve with standard deviation σ1 is more spread out (flatter) than the normal curve with standard deviation σ2. In general, larger standard deviations result in normal curves that are flatter and more spread out, while smaller standard deviations result in normal curves that have higher peaks and are less spread out.

Suppose that a random variable x is normally distributed with mean µ and standard deviation σ. If a and b are numbers on the real line, we consider the probability that x will attain a value between a and b. That is, we consider

P(a ≤ x ≤ b)

which equals the area under the normal curve with mean µ and standard deviation σ corresponding to the interval [a, b]. Such an area is depicted in Figure 6.5.We soon explain how to find such areas using a statistical table called a normal table. For now, we emphasize three important areas under a normal curve. These areas form the basis for the Empirical Rule for a normally distributed population. Specifically, if x is normally distributed with mean µ and standard deviation σ, it can be shown (using a normal table) that, as illustrated in Figure 6.6:

Finding normal curve areas        There is a unique normal curve for every combination of µ and σ. Since there are many (theoretically, an unlimited number of) such combinations, we would like to have one table of normal curve areas that applies to all normal curves. There is such a table, and we can use it by thinking in terms of how many standard deviations a value of interest is from the mean. Specifically, consider a random variable x that is normally distributed with mean µ and standard deviation σ. Then the random variable

Table A.3 is a table of cumulative areas under the standard normal curve. This table is called a cumulative normal table, and it is reproduced as Table 6.1. Specifically,

The cumulative normal table gives, for many different values of z, the area under the standard normal curve to the left of z. Two such areas are shown next to Table 6.1—one with a negative z value and one with a positive z value. The values of z in the cumulative normal table range from -3.99 to 3.99 in increments of .01. As can be seen from Table 6.1, values of z accurate to the nearest tenth are given in the far left column (headed z) of the table. Further graduations to the nearest hundredth (.00, .01, .02, . . . , .09) are given across the top of the table. The areas under the normal curve are given in the body of the table, accurate to four (or sometimes five) decimal places.

As an example, suppose that we wish to find the area under the standard normal curve to the left of a z value of 2.00. This area is illustrated in Figure 6.8. To find this area, we start at the top of the leftmost column in Table 6.1 and scan down the column past the negative z values. We then scan through the positive z values until we find the z value 2.0—see the red arrow above. We now scan across the row in the table corresponding to the z value 2.0 until we find the column corresponding to the heading .00. The desired area (which we have shaded blue) is in the row corresponding to the z value 2.0 and in the column headed .00. This area, which equals .9772, is the probability that the random variable z will be less than or equal to 2.00. That is, we have found that P(z ≤ 2) = .9772. Note that, because there is no area under the normal curve at a single value of z, there is no difference between P(z ≤ 2) and P(z < 2). As another example, the area under the standard normal curve to the left of the z value 1.25 is found in the row corresponding to 1.2 and in the column corresponding to .05. We find that this area (also shaded blue) is .8944. That is, P(z ≤ 1.25) = .8944 (see Figure 6.9).

We now show how to use the cumulative normal table to find several other kinds of normal curve areas. First, suppose that we wish to find the area under the standard normal curve to the right of a z value of 2—that is, we wish to find P(z ≥ 2). This area is illustrated in Figure 6.10 and is called a right-hand tail area. Since the total area under the normal curve equals 1, the area under the curve to the right of 2 equals 1 minus the area under the curve to the left of 2. Because Table 6.1 tells us that the area under the standard normal curve to the left of 2 is .9772, the area under the standard normal curve to the right of 2 is 1 - .9772 = .0228. Said in an equivalent fashion, because P(z ≤ 2) =.9772, it follows that P(z ≥ 2) = 1 -P(z ≤ 2)= 1 - .9772 =.0228.

Next, suppose that we wish to find the area under the standard normal curve to the left of a z value of2. That is, we wish to find P(z≤ -2). This area is illustrated in Figure 6.11 and is called a left-hand tail area. The needed area is found in the row of the cumulative normal table corresponding to2.0 and in the column headed by .00.We find that P(z≤ -2) =.0228. Notice that the area under the standard normal curve to the left of2 is equal to the area under this curve to the right of 2. This is true because of the symmetry of the normal curve.

Figure 6.12 illustrates how to find the area under the standard normal curve to the right of 2. Since the total area under the normal curve equals 1, the area under the curve to the right of 2 equals 1 minus the area under the curve to the left of 2. Because Table 6.1 tells us that the area under the standard normal curve to the left of -2 is .0228, the area under the standard normal curve to the right of -2 is 1 - .0228 = .9772. That is, because P(z ≤ -2) = .0228, it follows that P(z ≥ -2) - 1 - P(z ≤ -2) = 1 - .0228 = .9772.

The smallest z value in Table 6.1 is 3.99, and the table tells us that the area under the standard normal curve to the left of 3.99 is .00003 (see Figure 6.13). Therefore, if we wish to find the area under the standard normal curve to the left of any z value less than 3.99, the most we can say (without using a computer) is that this area is less than .00003. Similarly, the area under the standard normal curve to the right of any z value greater than 3.99 is also less than .00003 (see Figure 6.13).

Figure 6.14 illustrates how to find the area under the standard normal curve between 1 and 2. This area equals the area under the curve to the left of 2, which the normal table tells us is .9772, minus the area under the curve to the left of 1, which the normal table tells us is .8413. Therefore, P(1 ≤ z ≤ 2) = .9772 - .8413 = .1359.

To conclude our introduction to using the normal table, we will use this table to justify the empirical rule. Figure 6.15(a) illustrates the area under the standard normal curve between - 1 and 1. This area equals the area under the curve to the left of 1, which the normal table tells us is .8413, minus the area under the curve to the left of - 1, which the normal table tells us is .1587. Therefore, P(-1 ≤ z ≤ 1) =.8413 -.1587 =.6826. Now, suppose that a random variable x is normally distributed with mean m and standard deviation s, and remember that z is the number of standard deviations σ that x is from µ. It follows that when we say that P(- 1 ≤ z ≤ 1) equals .6826, we are saying that 68.26 percent of all possible observed values of x are between a point that is one standard deviation below µ (where z equals - 1) and a point that is one standard deviation above µ (where z equals 1). That is, 68.26 percent of all possible observed values of x are within (plus or minus) one standard deviation of the mean µ.

Figure 6.15(b) illustrates the area under the standard normal curve between - 2 and 2. This area equals the area under the curve to the left of 2, which the normal table tells us is .9772, minus the area under the curve to the left of -2, which the normal table tells us is .0228. Therefore, P(- 2 ≤ z ≤ 2) = .9772 - .0228 = .9544. That is, 95.44 percent of all possible observed values of x are within (plus or minus) two standard deviations of the mean µ.

Figure 6.15(c) illustrates the area under the standard normal curve between 3 and 3. This area equals the area under the curve to the left of 3, which the normal table tells us is .99865, minus the area under the curve to the left of -3, which the normal table tells us is .00135. Therefore, P( - 3 ≤ z ≤ 3) = . 99865 - .00135 = .9973. That is, 99.73 percent of all possible observed values of x are within (plus or minus) three standard deviations of the mean µ.

Although the empirical rule gives the percentages of all possible values of a normally distributed random variable x that are within one, two, and three standard deviations of the mean µ, we can use the normal table to find the percentage of all possible values of x that are within any particular number of standard deviations of µ. For example, later we will need to know the percentage of all possible values of x that are within plus or minus 1.96 standard deviations of µ. Figure 6.15(d) illustrates the area under the standard normal curve between - 1.96 and 1.96. This area equals the area under the curve to the left of 1.96, which the normal table tells us is .9750, minus the area under the curve to the left of 1.96, which the table tells us is .0250. Therefore, P(- 1.96 ≤ z ≤ 1.96) = .9750 - .0250 = .9500. That is, 95 percent of all possible values of x are within plus or minus 1.96 standard deviations of the mean µ.

Some practical applications     We have seen how to use z values and the normal table to find areas under the standard normal curve. However, most practical problems are not stated in such terms. We now consider an example in which we must restate the problem in terms of the standard normal random variable z before using the normal table.

#### EXAMPLE 6.2 The Car Mileage Case

Recall that an automaker has recently introduced a new midsize model and that we have used the sample of 50 mileages to estimate that the population of mileages of all cars of this type is normally distributed with a mean mileage equal to 31.56 mpg and a standard deviation equal to .798 mpg. Suppose that a competing automaker produces a midsize model that is somewhat smaller and less powerful than the new midsize model. The competitor claims, however, that its midsize model gets better mileages. Specifically, the competitor claims that the mileages of all its midsize cars are normally distributed with a mean mileage µ equal to 33 mpg and a standard deviation σ equal to .7 mpg. In the next example we consider one way to investigate the validity of this claim. In this example we assume that the claim is true, and we calculate the probability that the mileage, x, of a randomly selected competing midsize car will be between 32 mpg and 35 mpg. That is, we wish to find P(32≤ x ≤ 35). As illustrated in Figure 6.16, this probability is the area between 32 and 35 under the normal curve having mean µ = 33 and standard deviation σ =.7. In order to use the normal table, we must restate the problem in terms of the standard normal random variable z. The z value corresponding to 32 is

Example 6.2 illustrates the general procedure for finding a probability about a normally distributed random variable x. We summarize this procedure in the following box:

#### Finding Normal Probabilities

1 Formulate the problem in terms of the random variable x.

2 Calculate relevant z values and restate the problem in terms of the standard normal random variable

z = x - µ / σ

3 Find the required area under the standard normal curve by using the normal table.

4 Note that it is always useful to draw a picture illustrating the needed area before using the normal table.

#### EXAMPLE 6.3 The Car Mileage

Case Recall from Example 6.2 that the competing automaker claims that the population of mileages of all its midsize cars is normally distributed with mean m33 and standard deviation σ = .7. Suppose that an independent testing agency randomly selects one of these cars and finds that it gets a mileage of 31.2 mpg when tested as prescribed by the EPA. Because the sample mileage of 31.2 mpg is less than the claimed mean µ = 33, we have some evidence that contradicts the competing automaker’s claim. To evaluate the strength of this evidence, we will calculate the probability that the mileage, x, of a randomly selected midsize car would be less than or equal to 31.2 if, in fact, the competing automaker’s claim is true. To calculate P(x ≤ 31.2) under the assumption that the claim is true, we find the area to the left of 31.2 under the normal curve with mean µ = 33 and standard deviation σ =.7 (see Figure 6.17). In order to use the normal table, we must find the z value corresponding to 31.2. This z value is

Finding a point on the horizontal axis under a normal curve   In order to use many of the formulas, we must be able to find the z value so that the tail area to the right of z under the standard normal curve is a particular value. For instance, we might need to find the z value so that the tail area to the right of z under the standard normal curve is .025. This z value is denoted z.025, and we illustrate z.025 in Figure 6.19(a). We refer to z.025 as the point on the horizontal axis under the standard normal curve that gives a right-hand tail area equal to .025. It is easy to use the cumulative normal table to find such a point. For instance, in order to find z.025, we note from Figure 6.19(b) that the area under the standard normal curve to the left of z.025 equals .975. Remembering that areas under the standard normal curve to the left of z are the four-digit (or five-digit) numbers given in the body of Table 6.1, we scan the body of the table and find the area .9750. We have shaded this area in Table 6.1, and we note that the area .9750 is in the row corresponding to a z of 1.9 and in the column headed by .06. It follows that the z value corresponding to .9750 is 1.96. Because the z value 1.96 gives an area under the standard normal curve to its left that equals .975, it also gives a right-hand tail area equal to .025. Therefore, z.025 = 1.96.

#### EXAMPLE 6.5

A large discount store sells 50 packs of HX-150 blank DVDs and receives a shipment every Monday. Historical sales records indicate that the weekly demand, x, for HX-150 DVD 50 packs is normally distributed with a mean of µ = 100 and a standard deviation of σ = 10. How many 50 packs should be stocked at the beginning of a week so that there is only a 5 percent chance that the store will run short during the week?

If we let st equal the number of 50 packs that will be stocked, then st must be chosen to allow only a .05 probability that weekly demand, x, will exceed st. That is, st must be chosen so that

P(x > st) = .05

Figure 6.20(a) shows that the number stocked, st, is located under the right-hand tail of the normal curve having mean µ = 100 and standard deviation σ = 10. In order to find st, we need to determine how many standard deviations st must be above the mean in order to give a right-hand tail area that is equal to .05.

The z value corresponding to st is

z = st - µ/ σ = st - 100/ 10

and this z value is the number of standard deviations that st is from m. This z value is illustrated in Figure 6.20(b), and it is the point on the horizontal axis under the standard normal curve that gives a right-hand tail area equal to .05. That is, the z value corresponding to st is z.05. Since the area under the standard normal curve to the left of z.05 is 1 - .05 = .95—see Figure 6.20(b)—we look for .95 in the body of the normal table. In Table 6.1, we see that the areas closest to .95 are .9495, which has a corresponding z value of 1.64, and .9505, which has a corresponding z value of 1.65. Although it would probably be sufficient to use either of these z values, we will (because it is easy to do so) interpolate halfway between them and assume that z.05 equals 1.645. To find st, we solve the equation

#### EXAMPLE 6.6

Extensive testing indicates that the lifetime of the Everlast automobile battery is normally distributed with a mean of µ = 60 months and a standard deviation of σ = 6 months. The Everlast’s manufacturer has decided to offer a free replacement battery to any purchaser whose Everlast battery does not last at least as long as the minimum lifetime specified in its guarantee. How can the manufacturer establish the guarantee period so that only 1 percent of the batteries will need to be replaced free of charge?

If the battery will be guaranteed to last l months, l must be chosen to allow only a .01 probability that the lifetime, x, of an Everlast battery will be less than l. That is, we must choose l so that

P(x < l ) = .01

Figure 6.22(a) shows that the guarantee period, l, is located under the left-hand tail of the normal curve having mean µ = 60 and standard deviation σ = 6. In order to find l, we need to determine how many standard deviations l must be below the mean in order to give a left-hand tail area that equals .01. The z value corresponding to l is

#### EXAMPLE 6.7

Consider computing a tolerance interval [µ ±kσ] that contains 99 percent of the measurements in a normally distributed population having mean m and standard deviation s. As illustrated in Figure 6.23, we must find the value k so that the area under the normal curve having mean m and standard deviation s between (µ - kσ) and (µ + kσ) is .99. Because the total area under this normal curve is 1, the area under the normal curve that is not between (µ - kσ) and (µ + kσ) is 1 .99 .01. This implies, as illustrated in Figure 6.23, that the area under the normal curve to the left of (µ + kσ) is .01/2 =.005, and the area under the normal curve to the right of (µ + kσ) is also .012 .005. This further implies, as illustrated in Figure 6.23, that the area under the normal curve to the left of (µ + kσ) is .995.Because the z value corresponding to a value of x tells us how many standard deviations x is from µ, the z value corresponding to (µ + kσ) is obviously k. It follows that k is the point on the horizontal axis under the standard normal curve so that the area to the left of k is .995. Looking up .995 in the body of the normal table, we find that the values closest to .995 are .9949, which has a corresponding z value of 2.57, and .9951, which has a corresponding z value of 2.58. Although it would be sufficient to use either of these z values, we will interpolate halfway between them, and we will assume that k equals 2.575. It follows that the interval [µ ± 2.575 σ] contains 99 percent of the measurements in a normally distributed population having mean µ and standard deviation σ.

Whenever we use a normal table to find a z point corresponding to a particular normal curve area, we will use the halfway interpolation procedure illustrated in Examples 6.5 and 6.7 if the area we are looking for is exactly halfway between two areas in the table. Otherwise, as illustrated in Example 6.6, we will use the z value corresponding to the area in the table that is closest to the desired area.

BINOMIAL DISTRIBUTION USING NORMAL DISTRIBUTION

Figure 6.24 illustrates several binomial distributions. In general, we can see that as n gets larger and as p gets closer to .5, the graph of a binomial distribution tends to have the symmetrical, bell-shaped appearance of a normal curve. It follows that, under conditions given in the following box, we can approximate the binomial distribution by using a normal distribution.

This approximation is often useful because binomial tables for large values of n are often unavailable. The conditions np≥5 and n(1-p)≥5 must be met in order for the approximation to be appropriate. Note that if p is near 0 or near 1, then n must be larger for a good approximation, while if p is near .5, then n need not be as large.

When we say that we can approximate the binomial distribution by using a normal distribution, we are saying that we can compute binomial probabilities by finding corresponding areas under a normal curve (rather than by using the binomial formula). We illustrate how to do this in the following example.

Making the proper continuity correction can sometimes be tricky. A good way to approach this is to list the numbers of successes that are included in the event for which the binomial probability is being calculated. Then assign the appropriate area under the normal curve to each number of successes in the list. Putting these areas together gives the normal curve area that must be calculated. For example, again consider the binomial random variable x with n=50 and p = .5. If we wish to find P(27≤x≤29), then the event 27≤x≤29 includes 27, 28, and 29 successes. Because we assign the areas under the normal curve corresponding to the intervals [26.5, 27.5], [27.5, 28.5], and [28.5, 29.5] to the values 27, 28, and 29, respectively, then the area to be found under the normal curve is P(26.5≤x≤29.5). Table 6.2 gives several other examples.

#### EXAMPLE 6.9 The Cheese Spread Case

A food processing company markets a soft cheese spread that is sold in a plastic container with an “easy pour” spout. Although this spout works extremely well and is popular with consumers, it is expensive to produce. Because of the spout’s high cost, the company has developed a new, less expensive spout. While the new, cheaper spout may alienate some purchasers, a company study shows that its introduction will increase profits if fewer than 10 percent of the cheese spread’s current purchasers are lost. That is, if we let p be the true proportion of all current purchasers who would stop buying the cheese spread if the new spout were used, profits will increase as long as p is less than .10.

Suppose that (after trying the new spout) 63 of 1,000 randomly selected purchasers say that they would stop buying the cheese spread if the new spout were used. To assess whether p is less than .10, we will assume for the sake of argument that p equals .10, and we will use the sample information to weigh the evidence against this assumption and in favor of the conclusion that p is less than .10. Let the random variable x represent the number of the 1,000 purchasers who say they would stop buying the cheese spread. Assuming that p equals .10, then x is a binomial random variable with n = 1,000 and p =.10. Since the sample result of 63 is less than µ=np= 1,000(.1) = 100, the expected value of x when p equals .10, we have some evidence to contradict the assumption that p equals .10. To evaluate the strength of this evidence, we calculate the probability that 63 or fewer of the 1,000 randomly selected purchasers would say that they would stop buying the cheese spread if the new spout were used if, in fact, p equals .10.

we find that

P(x ≤ 63.5) = P(z≤ - 3.85)

Using the normal table, we find that the area under the standard normal curve to the left of - 3.85 is .00006. This says that, if p equals .10, then in only 6 in 100,000 of all possible random samples of 1,000 purchasers would 63 or fewer say they would stop buying the cheese spread if the new spout were used. Since it is very difficult to believe that such a small chance (a .00006 chance) has occurred, we have very strong evidence that p does not equal .10 and is, in fact, less than .10. Therefore, it seems that using the new spout will be profitable