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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

Chapter 8

Continuous Probability Distributions

probability density functions
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.
  •  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).
probability density function
Probability Density Function…
  • 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
The Normal Distribution…
  • 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 …
the normal distribution1
The Normal Distribution…
  • Important things to note:

The normal distribution is fully defined by two parameters:

its standard deviation andmean

The normal distribution is bell shaped and

symmetrical about the mean

Normal distributions range from minus infinity to plus infinity

standard normal distribution
0

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.
normal distribution
Normal Distribution…
  • The normal distribution is described by two parameters:
  • its mean and its standard deviation . Increasing the mean shifts the curve to the right…
normal distribution1
Normal Distribution…
  • The normal distribution is described by two parameters:
  • its mean and its standard deviation . Increasing the standard deviation “flattens” the curve…
calculating normal probabilities
Calculating Normal Probabilities…
  • 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!

calculating normal probabilities1
Calculating Normal Probabilities…
  • 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

calculating normal probabilities2
Calculating Normal Probabilities…
  • P(45 < X < 60) ?

…mean of 50 minutes and a

standard deviation of 10 minutes…

0

calculating normal probabilities3
Calculating Normal Probabilities…
  • 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?!
calculating normal probabilities4
Calculating Normal Probabilities…
  • 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

calculating normal probabilities5
Calculating Normal Probabilities…
  • 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
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 31
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

finding values of z
Finding Values of Z…
  • 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

finding values of z1
Finding Values of Z…
  • Other Z values are
  • Z.05 = 1.645
  • Z.01 = 2.33
using the values of z
Using the values of Z
  • 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
other continuous distributions
Other Continuous Distributions…
  • 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
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 distribution1
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
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
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
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.
determining values of f
Determining Values of F…
  • 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.

Make sure you start with

the correct table!!

F.05,3,7=4.35

F.05,3,7

Denominator Degrees of Freedom : ROW

Numerator Degrees of Freedom : COLUMN

problem normal distribution
Problem: Normal Distribution
  • 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)
problem normal
Problem: Normal
  • 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) =
problem normal1
Problem: Normal
  • 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) =
problem normal targeting mean
Problem: Normal – Targeting Mean
  • 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.
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