Download Section 5.1

Section 5.1
Discrete Probability
Probability Distributions
• A probability distribution is a table that
consists of outcomes and their probabilities.
• To be a probability distribution it must have
the following properties:
– Each probability must be 0 ≤ 𝑥 ≤ 1
– The probabilities must have a sum of 1.
x
1
P(x) 0.4
2
3
4
5
x
4
6
8
0.3
0.1
0.3
1.4
P(x) 1/4 1/4 0
10
12
1/8 3/8
Discrete vs. Continuous
• Discrete – can be counted, whole numbers
• Continuous – cannot be counted, fractions,
decimals
Expected Value
• Expected value is the same as a weighted
mean.
• Formula: (𝑥 ∙ 𝑃 𝑥 )
x
4
5
6
7
8
P(x)
0.2
0.2
0.1
0.05
0.45
• Expected Value = 4 0.2 + 5 0.2 + 6 0.1 +
7 0.05 + 8(0.45)= 6.35
Variance and Standard Deviation
• Variance: [(𝑥 − 𝜇)2 ∙ 𝑃(𝑥)] where mean is
the expected value.
• Standard Deviation: square root of the
x
2
3
4
5
6
variance
P(x)
0.5
0.05 0.05
0.1
x
P(x)
𝒙 − 𝟑. 𝟔𝟓
(𝒙 − 𝟑. 𝟔𝟓)𝟐
(𝒙 − 𝟑. 𝟔𝟓)𝟐 ∙ 𝑷(𝒙)
2
0.5
-1.65
2.7225
1.36125
3
0.05
-0.65
0.4225
0.021125
4
0.05
.35
0.1225
0.006125
5
0.1
1.35
1.8225
0.18225
6
0.3
2.35
5.5225
1.65675
0.3
Sum = 3.2275 = Variance
𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 =
3.2275 = 1.80
Profit and Loss w/ Probability
• To determine the profit or loss using
probability you will use the expected value for
each event.
• Formula: Profit minus loss: [𝑥𝑝 ∙ 𝑃 𝑥𝑝 ] −
[𝑥𝐿 ∙ 𝑃 𝑥𝐿 ]
• 𝑥𝑝 is the value of the profit or what you
receive
• 𝑥𝐿 is the value of the loss or what you pay.
Example
• If you draw a card with a value of 2 or less
from a standard deck of cards, I will pay you
$303. If not, you pay me $23. (Aces are the
highest card in the deck)
• Find the expected value of the proposition.
Solution
1. Find the probability of drawing a card with a
value of 2 or less.
2. Find the value of drawing a card greater than 2.
3. Determine 𝑥𝑝 and 𝑥𝐿 .
4. Fill in formula.
5. 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑙𝑜𝑠𝑠 𝑜𝑟 𝑔𝑎𝑖𝑛 = 303 4 52 −
23 48 52 = 23.31 − 21.23 = 2.08
6. So for each round that is played the is an
expected gain of $2.08.
7. If there is a loss, the value would be negative.
Example (part 2)
• If you played the same game 948 times, how
much would you expect to win or lose?
Solution (part 2)
• Take the profit or loss from one round and
multiply by the number of times played.
• $2.08 948 = $1971.84
Creating Probability Distribution w/
Tree Diagram
• The number of tails in 4 tosses of a coin.
x
0
1
2
3
4
P(x)
1
16
4
1
=
16 4
6
3
=
16 8
4
1
=
16 4
1
16
Pascal’s Triangle
2 tosses
3 tosses
4 tosses
Example
1
16
4
16
6
16
4
16
1
16
Make sure you simplify all fractions.
To get the total you can add the numbers in the row or take 2 to the power
of the number of times you are choosing or flipping.
Finding Probability
a.
b.
c.
d.
𝑃
𝑃
𝑃
𝑃
≥7
≤7
>7
<7
A = .5
B = .9
C = .2
D = .5
Section 6-1
Introduction to Normal Curve
Normal Curve
Example
Section 6-2
Finding area under the Normal Curve
Area Under a Normal Curve
• Using z-scores (standard scores) we can find
the area under the curve or the probability
that a score falls below, above, or between
two values.
• The area under the curve is 1.
• The mean (or z=0) is the halfway point, or has
an area of .5000.
• Values are listed to four decimal places.
To How the Area under the Curve
• If asked for the area to the left, find the value
in the chart.
• If asked for the area to the right, find the value
and subtract from 1. Alternate Method: Find
the opposite z-score and use that value.
• If asked for the area between two z-scores,
find the values and subtract.
• If asked for the area to the right and to the left
of two numbers, find the values and add.
1 - z-score
Alternate Method
Examples
• Find the area:
– To the left of z=2.45
– To the right of z=2.45
– Between z=-1.5 and z=1.65
– To the left of z=1.55 and to the right of z=2.65
– To the left of z=-2.13 and to the right of z=2.13
Solutions
•
•
•
•
•
.9929
.0071
.9960-.0668=.9292
.0606+.0013=.0619
.0166+.0166=.0332
Problems with greater than and less
than
• Some problems will have greater than or less
than symbols.
• P(z<1.5) is the same as to the left of z=1.5
• P(z>-2.3) is the same as to the right of z=-2.3
• P(-1.24<z<1.05) is the same as between z=1.24 and z=1.05
• P(z<1.02 and z>.02) is the same as to the left
of z=1.02 and to the right of z=.02
Section 6-3
Finding area after finding the z-score
How to solve
• Find the z-score with the given information
• Determine if the value is to the left, right,
between, or to the left and right.
• Look up values in the chart and use directions
from 6-2.
Examples
Solutions
•
•
•
•
P(0<z<1.5) = .4332
P(z<0) = .5000
P(z>2) = .0228
P(-.75<z<0.5) = .4649
Section 6-4
Finding Z and X
Finding Z
• If the value is to the left:
– Find the probability in the chart and the z-score
that corresponds with it.
• If the value is to the right:
– Subtract the value from one, find the probability
and the z-score that corresponds with it.
OR
– Find the value and the corresponding z-score and
change the sign.
Finding Z
• If the value is between:
– Divide the area by 2, then add .5, then find the
corresponding z-score.
OR
– Subtract the area from 1, divide by two, then find
the corresponding z-score.
• If the value is to the right and left:
– Divide the area by 2, then find the corresponding
z-score.
Examples
• Find the z-score that corresponds with:
– Area of .1292 to the left
– Area of .3594 to the right
– Area of .7154 between
– Area of .8180 to the left and the right
Solutions
•
•
•
•
-1.13
.36
-1.07 and 1.07
-.23 and .23
Word Problems
• Determine if the problem is looking for less
than, greater than, between, or less than and
greater than.
• Find the z-score(s).
• Use the formula 𝑧 =
𝑥−𝜇
𝜎
to solve for x.
• Some problems you will have two solutions.
Word Problems