section 5.1
DESCRIPTION
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 The probabilities must have a sum of 1. - PowerPoint PPT PresentationTRANSCRIPT
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 – The probabilities must have a sum of 1.
x 4 6 8 10 12
P(x) 1/4 1/4 0 1/8 3/8
x 1 2 3 4 5
P(x) 0.4 0.3 0.1 0.3 1.4
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:
• Expected Value = = 6.35
x 4 5 6 7 8
P(x) 0.2 0.2 0.1 0.05 0.45
Variance and Standard Deviation• Variance: where mean is the expected value.• Standard Deviation: square root of the
variancex 2 3 4 5 6
P(x) 0.5 0.05 0.05 0.1 0.3
x P(x)
2 0.5 -1.65 2.7225 1.361253 0.05 -0.65 0.4225 0.0211254 0.05 .35 0.1225 0.0061255 0.1 1.35 1.8225 0.182256 0.3 2.35 5.5225 1.65675
Sum = 3.2275 = Variance
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: • is the value of the profit or what you receive• 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.
Solution1. 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 and .4. Fill in formula.
5. So for each round that is played the is an expected gain of $2.08.
6. 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.
Creating Probability Distribution w/ Tree Diagram
• The number of tails in 4 tosses of a coin.
x 0 1 2 3 4
P(x)
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 to solve for x.• Some problems you will have two solutions.
Word Problems