Outline

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Outline

• Two kinds of random variables
• Discrete random variables
• Continuous random variables
• Symmetric distributions
• Normal distributions
• The standard normal distribution
• Using the standard normal distribution
• The normal approximation to the binomial

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1. Two kinds of random variables

1. Discrete (DRV)

• Outcomes have countable values
• Possible values can be listed
• E.g., of people in this room
• Possible values can be listed might be 51 or
  52 or 53

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1. Two kinds of random variables

1. Discrete RV
2. Continuous RV

• Not countable
• Consists of points in an interval
• E.g., time till coffee break

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1. Two kinds of random variables

• The form of the probability distribution for a
  CRV is a smooth curve.

• Such a distribution may also be called a
• Frequency Distribution
• Probability Density Function

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1. Two kinds of random variables

• In the graph of a CRV, the X axis is whatever
  you are measuring
• E.g., exam scores, mood scores, of widgets
  produced per hour.

• The Y axis measures the frequency of scores.

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X
The Y-axis measures frequency. It is usually not
shown.
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2. Symmetric Distributions

• In a symmetric CRV, 50 of the area under the
  curve is in each half of the distribution.
• P(x ?) P(x ?) .5

• Note Because points on a CRV are infinitely
  thin, we can only measure the probability of
  intervals of X values
• We cant measure or compute the probability of
  individual X values.

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A symmetric distribution which is not
mound-shaped. The two sections (above and below
the mean) each contain 50 of the observations.
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µ
A mound-shaped symmetric distribution (the normal
distribution)
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3. Normal Distributions

• A very important set of CRVs has mound-shaped
  and symmetric probability distributions. These
  are called normal distributions

• Many naturally-occurring variables are normally
  distributed.

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3. Normal Distributions

• Are perfectly symmetrical around their mean, ?.

• Have standard deviation, ?, which measures the
  spread of a distribution
• ? is an index of variability around the mean.

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3. Normal distributions

• There are an infinite number of normal
  distributions

• They are distinguished on the basis of their
  mean (µ) and standard deviation (s)

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3. Standard Normal Distribution

• The standard normal distribution is a special
  one produced by converting raw scores into Z
  scores

• Thus, µ 0 and s 1

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3. Standard Normal Distribution

• The area under the curve between ? and some
  value X ? has been calculated for the standard
  normal distribution and is given in Table IV of
  the text.
• E.g., for Z 1.62, area .4474

• Because distribution is symmetric, table values
  can also be used for scores below the mean (Z
  scores below 0).

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.4474
X
Z 1.62
Z 0
?
Area gives the probability of finding a score
between the mean and X when you make an
observation
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?
X
Z -1.62
For Z lt 0, same values can be used as for Z gt 0
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5. Using the Standard Normal Distribution

• Suppose the average height for Canadian women is
  µ 160 cm, with ? 15 cm.
• What is the probability that the next Canadian
  woman we meet is more than 175 cm tall?

• Note two things
• 1. this is a question about a single case
• 2. it specifies an interval.

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Using the Standard Normal Distribution
We need this area
Table gives this area
160
175
cm
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µ
Remember that area above the mean, ?, is half
(.5) of the distribution.
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Using the Standard Normal Distribution
Call this shaded area P. We can get P from Table
IV
160
175
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Using the Standard Normal Distribution

• Z X - ? 175-160
• ? 15
• 1.00
• Now, look up Z 1.00 in the table.
• Corresponding area is .3413.

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• Value of Z that marks one end of the interval
  you want to find probability that a randomly
  selected case has a score that falls in this
  interval
• Other end of the interval is at µ ( 0)

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Using the Standard Normal Distribution
This area is .3413
So this area must be .5 .3413 .1587
160
175
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Using the Standard Normal Distribution
This area is .3413
So this area must be .5 .3413 .1587
Z 0
Z 1.0
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Using the Standard Normal Distribution

• What is the probability that the next Canadian
  woman we meet is more than 175 cm tall?
• Answer .1587

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Binomial Random Variable Method 3

• When n is large and p is not too close to 0 or 1,
  we can use the normal approximation to the
  binomial probability distribution to work out
  binomial probabilities.
• How can you tell if the normal approximation is
  appropriate in a given case?
• Use the approximation if np 5 and nq 5

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BRV Method 3
Normal curve
Histogram
X
The histogram shows the probabilities of
different values of the BRV X. Because area gives
probability, we can use the area under a section
of the normal curve to approximate the area of
some part of the histogram
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BRV Method 3

• In order to use the normal approximation, we have
  to be able to compute the mean and standard
  deviation for the BRV (in order to work out Z).
• µ np ( of observations times P(Success))
• s vnpq
• Theres just one other issue to deal with

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BRV Method 3
Notice how the normal curve misses one top corner
of the rectangle for 7 and overstates the other
top corner these two errors cancel each other.
1 2 3 4 5 6 7 8 9 10
X
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BRV Method 3
1 2 3 4 5 6 7 8 9 10
X
Notice how the rectangle for 7 runs from 6.5 to
7.5. The probability of X values up to and
including 7 is given by the area to the left of
7.5.
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BRV Example 1 from last week

• Air Canada keeps telling us that arrival and
  departure times at Pearson International are
  improving. Right now, the statistics show that
  60 of the Air Canada planes coming into Pearson
  do arrive on time. (This actually is an
  improvement over 10 years ago when only 42 of
  the Air Canada planes arrived on time at
  Pearson.) The problem is that when a plane
  arrives on time, it often has to circle the
  airport because theres still a plane in its gate
  (a plane which didnt leave on time). Statistics
  also show that 50 of the planes that arrive on
  time have to circle at least once, while only 35
  of the planes that arrive late have to circle at
  least once.

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BRV Example 1 from last week

• c) Of the next 80 Air Canada planes that arrive
  at Pearson, whats the probability that between
  40 and 45 (inclusive) have to circle at least
  once?

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BRV Example 1

• First, we check to see whether we can use the
  normal approximation. To do this, we need to know
  the probability that a plane has to circle at
  least once
• P(C) P(C n Late) P(C n Not Late)
• P(L) P(C L) P(L) P(CL)
• .35 (.40) .6 (.5)
• .44

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BRV Example 1

• Now we can do the check
• n 80. p .44 and q (1 p) .56
• np 80 (.44) 35.2 gt 5
• nq 80 (.56) 44.8 gt 5
• Thus, its OK to use the normal approximation.

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BRV Example 1

• µ np 80 (.44) 35.2
• s vnpq v80(.44)(.56) 4.44
• Correction for continuity to get area for 40 and
  up, we use X 39.5. To get area for 45 and
  below, we use X 45.5

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35.2
40
45
40 and up starts at 39.5
45 and below starts at 45.5
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BRV Example 1

• Z39.5 39.5 35.2 .97
• 4.44
• Z45.5 45.5 35.2 2.32
• 4.44
• P(.97 Z 2.32) .4894 - .3340 .1554.
• That is the probability that between 40 and 45
  (inclusive) of the next 80 planes have to circle
  at least once.

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From here down .5 .4894 .9894
35.2
40
45
From here down .5 .3340 .8340
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Using the normal approximation, we estimate the
combined area of all these rectangles to be .9894
.8340 .1554
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CRV Example 1

• Wind speed in Windy City is normally-distributed
  and the middle 40 of wind speeds is bounded by
  23.9 and 29.3.
• a. Wind speed would be expected to be lower than
  what value only 5 of the time?

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What value of Z is associated with p .20? From
Table, Z 0.53
Since distribution is symmetric, µ midpoint
between 23.9 and 29.3. This is 26.6.
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CRV Example 1a

• Since Z X - µ then, s X - µ
• s 29.3 26.6 5.094
• 0.53
• Z for p .45 is 1.645 (from Table)
• Thus, required X 26.6 1.645 (5.094) 18.22

s
Z
44
.05
.45
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CRV Example 1

• Wind speed in Windy City is normally-distributed
  and the middle 40 of wind speeds is bounded by
  23.9 and 29.3.
• b. UV radiation in Windy City is normally
  distributed with a mean of 10.4 and a variance of
  6.25. Wind speed and UV radiation are independent
  of each other. A "bad day" in Windy City is any
  day on which either wind speed exceeds 35.0 or UV
  radiation exceeds 15. What is the probability of
  a bad day in Windy City?

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CRV Example 1b
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CRV Example 1b

• Z 15 10.4 1.84
• 2.5
• P(Z lt 1.84) .4671 (from Table)
• P(X gt 15) P(Z gt 1.84) .5 .4671 .0329

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CRV Example 1b
5.094
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CRV Example 1b

• Z 35 26.6 1.65
• 5.094
• P(Z lt 1.65) .4505 (from Table)
• P(X gt 35) P(Z gt 1.65) .5 .4505 .0495

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CRV Example 1b

• P(Bad Day) P(UV gt 15) or (Wind gt 35)
• .0329 .0495 (.0329)(.0495)
• .0808
• (From Additive Rule of Probability)

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Review

• Area under curve gives probability of finding X
  in a given interval.
• Area under the curve for Standard Normal
  Distribution is given in Table IV.
• For area under the curve for other
  normally-distributed variables first compute
• Z X - ?
• ?
• Then look up Z in Table IV.