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Ch5 Continuous Random Variables

Homework:1,11,15,18,19,21,31,39,

40,72,76,79,81,95,101

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Sec 5.1: Continuous Distributions

We will discuss three popular continuous distributions, the uniform distribution, the normal distribution and the exponential distribution in this chapter. We will not discuss the topic of approximating a binomial distribution with a normal distribution because it is no longer an important issue.

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Sec 5.2: The Uniform Distribution

A continuous random variable that has equal likely outcomes over it entire range of possible values possesses an uniform distribution. Suppose that the random variable x can assume values only in the interval c < x < d. Then the probability density function of x is

p xd c

ifc x d( ) .

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The expectation, the variance, and the standard deviation of x are

and

respectively.

d c d c

2 122

2,

d c 2

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<Example 5.1>: (Basic)

Suppose x is a uniform random variable with c = 1 and d = 9.

(a) Find the probability density function of x.

(b) Find the mean, variance, and standard deviation of x.

(c) Compute P(x > 2).

(d) Compute the probability P( - x < (e) Compute the probability P( -2< x < +2

<Solutions>:

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<Example 5.2>:

X is a uniformly distributed random variable in the interval (1 , 11).

(a) Find the probability density function of X.

(b) Find mean and standard deviation of X.

(c) Find s such that P(X > s) = 0.7.

(d) Find t such that P(X < t) = 0.25.

<Solutions>:

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Sec 5.3: The Normal Distribution

Normal distribution is the most popular distribution function. Some reasons for the popularity of the normal distribution are as follows:

(1) The distributions of many random variables, such as the height of a group of students, the length of ears of corns, the errors made in measuring a person's blood pressure, are approximately normally distributed.

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(2) The normal distribution is relatively easy to work with mathematically, and is well tabulated.

(3) Some measurements do not have a normal distribution, but a simple transformation of the original scale of the measurement may induce approximate normality.

(4) Even if the distribution of the original population is far from normal, the distribution of sample means tends to become normally distributed if the sample sizes are sufficiently large.

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The probability density function of a normal random variable with mean and standard deviation is

Normal distribution has several very good properties. First, the normal distribution are symmetric. Second, the normal distribution has an unique mode. Third, the population mean, the population median, and the population mode of a normal random are overlap.

p x e if x

x

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2

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The standard normal random variable is a normal random variable with mean zero and standard deviation one. If x is a normal random variable with meanand standard deviation of then z = (x -) / is a standard normal random variable.

It is very important to find a probability corresponding to a normal random variable. Although it is very easy to use statistical packages to find the probability corresponding to a normal random variable, I still want you learn to find this probability with normal probability table.

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We can use the following steps to find the probabilities associated with a normal random variable from a normal probability table.

(1). Draw the normal curve and shade the area corresponding to the probability for which you want to find.

(2). Convert the x values to the standard normal random variable values z using the formula

z = (x - u) / s.

(3). Use the table IV in appendix A to find the area corresponding to the z values.

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<Example 5.3>: (Basic)

Find the following probabilities for the standard normal random variable z.

(a) P(z > 1.64).

(b) P(z < -1.28).

(c) P(-2.58 < z < 1.28).

(d) P(1.96 > z > 1.28).

(e) P(-1.00 > z > -3.00).

<Solutions>:

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<Example 5.4>: (Intermediate)

Suppose x is the random variable of the height of students attending University of Central Florida. If x has a normal distribution with mean 67.56 inches and standard deviation 2.57 inches.

(a) Find P(x > 62 inches).

(b) Find P(65 inches < x < 70 inches).

(c) Is x a symmetric random variable?

<Solutions>:

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<Example 5.5>: (Intermediate)

The life of a certain type of battery is normally distributed with a mean of 90 hours and a standard deviation of 7 hours.

(a) Find the probability of this type of batteries that will have a life of less than 79 hours.

(b) Find the probability of this type of batteries that will have a life of 103 to 112 hours.

<Solutions>:

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<Example 5.6>: (Advance)

z is a standard normal random variable. Find z0 such that

(a) P(z > z0) = 0.05.

(b) P(z > z0) = 0.8413.

(c) P(z z0) = 0.3264.

(d) P(z z0 = 0.9495.

<Solutions>:

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Sec 5.4: The Exponential Distribution

The exponential distribution (also knows as the negative exponential distribution) is a good probability model to study the amount of time or distance between the occurrences of two random events. For example, we can use the exponential distribution model to study the random variables such as

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(1) The time between two arrival cars in front of a ticket booth;

(2) The waiting time between two customers in a restaurant;

(3) The length of time between two fatal car accidents in a busy intersection;

(4) The distance between two disease trees in a national park.

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Suppose X be an exponential random variable with parameter The probabilitydensity function of Xis

Both the mean and the standard deviation of x are . We can use Table V to obtain the probabilities such as P(x > a) and P(x < a).

f x e when xx( ) ./ 10

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<Example 5.7>: (Basic)

Suppose x is an exponential random variable with variance equal to 4.

(a) Find mean of x.

(b) Compute P(x > 2).

(c) Compute P(4 > x > 2).

(d) Find P(x 3).

<Solutions>:

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<Example 5. 8>: (Intermediate)

An auto manufacturer wants to determine the warranty coverage on the power train of its new compact car. A warranty is desired such that no more than 9.5% of the cars will have to use it. Assume that the time between power train failures is exponentialiy distributioned. How many miles should the warranty coverage last when the power train fails at a mean rate of one every 500,000 miles driven.

<Solutions>: