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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.
- 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…

- 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 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 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

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…

- The normal distribution is described by two parameters:
- its mean and its standard deviation . Increasing the mean shifts the curve to the right…

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…

- 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 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 Probabilities…

- P(45 < X < 60) ?

…mean of 50 minutes and a

standard deviation of 10 minutes…

0

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 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 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)…

- 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

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 Z…

- Other Z values are
- Z.05 = 1.645
- Z.01 = 2.33

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…

- 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.

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

- 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

- 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: 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

- 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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