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# Chapter 8 - PowerPoint PPT Presentation

Chapter 8. Continuous Probability Distributions. Probability Density Functions…. Unlike a discrete random variable which we studied in Chapter 7, a continuous random variable is one that can assume an uncountable number of values.

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### Chapter 8

Continuous Probability Distributions

• Unlike a discrete random variable which we studied in Chapter 7, a continuous random variable is one that can assume an uncountable number of values.

•  We cannot list the possible values because there is an infinite number of them.

• Because there is an infinite number of values, the probability of each individual value is virtually 0.

• In a continuous setting (e.g. with time as a random variable), the probability the random variable of interest, say task length, takes exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.

• It is meaningful to talk about P(X ≤ 5).

• A function f(x) is called a probability density function (over the range a ≤ x ≤ b if it meets the following requirements:

• f(x) ≥ 0 for all x between a and b, and

• The total area under the curve between a and b is 1.0

f(x)

area=1

a

b

x

• The normal distribution is the most important of all probability distributions. The probability density function of a normal random variable is given by:

• It looks like this:

• Bell shaped,

• Symmetrical around the mean …

• Important things to note:

The normal distribution is fully defined by two parameters:

its standard deviation andmean

The normal distribution is bell shaped and

Normal distributions range from minus infinity to plus infinity

1

1

Standard Normal Distribution…

• A normal distribution whose mean is zero and standard deviation is one is called the standard normal distribution.

• As we shall see shortly, any normal distribution can be converted to a standard normal distribution with simple algebra. This makes calculations much easier.

• The normal distribution is described by two parameters:

• its mean and its standard deviation . Increasing the mean shifts the curve to the right…

• The normal distribution is described by two parameters:

• its mean and its standard deviation . Increasing the standard deviation “flattens” the curve…

• We can use the following function to convert any normal random variable to a standard normal random variable…

0

Some advice: always draw a picture!

• Example: The time required to build a computer is normally distributed with a mean of 50 minutes and a standard deviation of 10 minutes:

• What is the probability that a computer is assembled in a time between 45 and 60 minutes?

• Algebraically speaking, what is P(45 < X < 60) ?

0

• P(45 < X < 60) ?

…mean of 50 minutes and a

standard deviation of 10 minutes…

0

• OK, we’ve converted P(45 < X < 60) for a normal distribution with mean = 50 and standard deviation = 10

• to

• P(–.5 < Z < 1) [i.e. the standard normal distribution with mean = 0 and standard deviation = 1]

• so

• Where do we go from here?!

• P(–.5 < Z < 1) looks like this:

• The probability is the area

• under the curve…

• We will add up the

• two sections:

• P(–.5 < Z < 0) and

• P(0 < Z < 1)

• We can use Table 3 in Appendix B to look-up probabilities P(Z < z)

0

–.5 … 1

• Recap: The time required to build a computer is normally distributed with a mean of 50 minutes and a standard deviation of 10 minutes

• What is the probability that a computer is assembled in a time between 45 and 60 minutes?

P(45 < X < 60) = P(–.5 < Z < 1) = .5328

“Just over half the time, 53% or so, a computer will have an assembly time between 45 minutes and 1 hour”

Using the Normal Table (Table 3)…

• What is P(Z > 1.6) ?

P(Z < 1.6) = .9452

z

0

1.6

P(Z > 1.6) = 1.0 – P(Z < 1.6)

= 1.0 – .9452

= .0548

Using the Normal Table (Table 3)…

• What is P(0.9 < Z < 1.9) ?

P(Z < 0.9)

P(Z < 1.9)

z

0

0.9

1.9

P(0.9 < Z < 1.9) = P(Z < 1.9) – P(Z < 0.9)

=.9713 – .8159

= .1554

• What value of z corresponds to an area under the curve of 2.5%? That is, what is z.025 ?

Area = .50

Area = .025

Area = .50–.025 = .4750

If you do a “reverse look-up” on Table 3 for .9750,you will get the corresponding zA = 1.96

Since P(z > 1.96) = .025, we say: z.025 = 1.96

• Other Z values are

• Z.05 = 1.645

• Z.01 = 2.33

• Because z.025 = 1.96 and - z.025= -1.96, it follows that we can state

• P(-1.96 < Z < 1.96) = .95

• Recall that the empirical rule stated that approximately 95% would be within + 2 standard deviations. From now on we use 1.96 instead of 2.

• Similarly

• P(-1.645 < Z < 1.645) = .90

• Three other important continuous distributions which will be used extensively in later sections are introduced here:

• Student t Distribution, [will use in this class]

• Chi-Squared Distribution,

• F Distribution, [will use in this class]

• Exponential.

Student t Distribution…

• Much like the standard normal distribution, the Student t distribution is “mound” shaped and symmetrical about its mean of zero: Looks like the normal distribution after an elephant sat on it [flattened out/spread out more than a normal]

Student t Distribution…

• In much the same way that and define the normal distribution, , the degrees of freedom, defines the Student

• t Distribution:

• As the number of degrees of freedom increases, the t distribution approaches the standard normal distribution.

Figure 8.24

Determining Student t Values…

• The student t distribution is used extensively in statistical inference. Table 4 in Appendix B lists values of

• That is, values of a Student t random variable with degrees of freedom such that:

• The values for A are pre-determined

• “critical” values, typically in the

• 10%, 5%, 2.5%, 1% and 1/2% range.

Using the t table (Table 4) for values…

• For example, if we want the value of t with 10 degrees of freedom such that the tail area under the Student t curve is .05:

Area under the curve value (tA) : COLUMN

t.05,10

t.05,10=1.812

Degrees of Freedom : ROW

F Distribution…

• The F density function is given by:

• F > 0. Two parameters define this distribution, and like we’ve already seen these are again degrees of freedom.

• is the “numerator” degrees of freedom and

• is the “denominator” degrees of freedom.

• For example, what is the value of F for 5% of the area under the right hand “tail” of the curve, with a numerator degree of freedom of 3 and a denominator degree of freedom of 7?

• Solution: use the F look-up (Table 6)

There are different tables

for different values of A.

the correct table!!

F.05,3,7=4.35

F.05,3,7

Denominator Degrees of Freedom : ROW

Numerator Degrees of Freedom : COLUMN

• If the random variable Z has a standard normal distribution, calculate the following probabilities.

• P(Z > 1.7) =

• P(Z < 1.7) =

• P(Z > -1.7) =

• P(Z < -1.7) =

• P(-1.7 < Z < 1.7)

• If the random variable X has a normal distribution with

• mean 40 and std. dev. 5, calculate the following

• probabilities.

• P(X > 43) =

• P(X < 38) =

• P(X = 40) =

• P(X > 23) =

• The time (Y) it takes your professor to drive home each

• night is normally distributed with mean 15 minutes and

• standard deviation 2 minutes. Find the following

• probabilities. Draw a picture of the normal distribution and

• show (shade) the area that represents the probability you are

• calculating.

• P(Y > 25) =

• P( 11 < Y < 19) =

• P (Y < 18) =

• The manufacturing process used to make “heart pills” is

• known to have a standard deviation of 0.1 mg. of active ingredient.

• Doctors tell us that a patient who takes a pill with over 6 mg. of

• active ingredient may experience kidney problems. Since you want to

• protect against this (and most likely lawyers), you are asked to

• determine the “target” for the mean amount of active ingredient in each

• pill such that the probability of a pill containing over 6 mg. is 0.0035 (

• 0.35% ). You may assume that the amount of active ingredient in a pill

• is normally distributed.

• *Solve for the target value for the mean.

• *Draw a picture of the normal distribution you came up with and show the 3 sigma limits.