Transcript Fish Law table

Probability
Definition: randomness, chance, likelihood,
proportion, percentage, odds.
Not sure what will happen in a single event but, in
the long run, certain patterns emerge.
Probability is the mathematical ideal.
We use letters like X and Y to represent quantities.
These will be called random variables.
Probability Model
List the outcomes for a given event (experiment or
question) and associated probabilities.
S: sample space (contains all possible outcomes)
Event: single outcome or collection of outcomes
Example: pick a card out of a standard deck
S: sample space contains
Event: pick a
pick a
Basic Rules
1. Event A has probability P(A), which is
between 0 and 1 (inclusive).
2. Probability of entire sample space, P(S), is
.
3. Addition: If two events are disjoint (nothing
in common), then P(A or B) =
.
4. Complement: P(not A) =
Probability Model for standard deck of cards
52 cards, 4 suits (Diamonds, Hearts, Clubs, Spades)
Each suit as 13 cards: 2, 3, 4, 5, … , 9, 10, J, Q, K, A
P(picking any single card)=
A = event that a red 5 is picked
B = event that a club is picked
C = event that a face card (J, Q, K) is picked
P(A or B)=
P(not C) =
Discrete Model
If sample space is finite, the probability model
is called discrete.
List all outcomes and associated probabilities in a table.
Roll 2 six-sided dice and record the sum.
Sum
Prob.
2 3 4
1 1 1
36 18 12
P(sum  9) 
P(sum  6) 
5
1
9
6
5
36
7
1
6
8 9
5 1
36 9
10
1
12
11 12
1 1
18 36
Continuous Model
If sample space contains a range of values,
the probability model is called continuous.
Density curves record probability as the area under the
curve for a given range of outcomes.
So, total area under the curve will always equal 1.
Continuous Model – Example 1
The uniform distribution for any real number, X,
from 3 to 7 looks like:
1
P(X  6) 
4
1
3
5
4
7
Area 
P(X  3.2 or X  5.1) 
3
1
5 67
4
3
5 7
3.2 5.1
Continuous Model – Example 2
The symmetric triangular distribution for any real
number, X, from 0 to 8 looks like:
P(X  4) 
4
Area 
8
4
8
P(X  6) 
4
8
More Probability
Use Venn diagrams to visualize probability rules.
Sample space, S: rectangle
Events (A, B, C, …): circles inside
If events are disjoint, don’t overlap circles.
Keep track of # of outcomes in each region of the
rectangle.
Venn diagram - example
Example: pick a card out of a standard deck
S: sample space contains 52 outcomes (52 cards)
A = event that a red 5 is picked
B = event that a club is picked
C = event that a face card (J, Q, K) is picked
S
B
A
C
P(A or B)
S
A
B 10
2
28
P(B or C)
S
A
28
3
C
9
B 10
2
3
C
9
General Addition Rule
A = event that a red card is picked
B = event that a number card is picked
P(A or B)
S
A
B
General Addition: P(A or B) =
Conditional Probability Rule
Given a condition (you know something happened), how
does that change the chances of something else happening?
P(B|A)= probability of B given A
S
A
B
P(A and B)
P(B | A) 
P(A)
Venn Diagram of 70 students
C: owns a cat
D: owns a dog
S
C
30 10 20
10
P(C) 
P(C | D) 
D
General Multiplication Rule
P(A and B)
Rewrite P(B | A) 
to get:
P(A)
P(A and B)  P(A)  P(B | A)
Experiment: pick two cards out of a standard deck
P(1st  K and 2nd  A)
P(1st  K and 2nd  K )
Independent Events
Two (or more) events are independent if knowledge of
one event does not change the chances of the other.
Multiplication Rule for Independent Events:
P(rolling three 7' s in a row) 
For a cholesterol-lowering drug, there is a 5% chance that
a loss-of-sleep side effect will occur.
What are the chances that two people picked at random take the
drug and experience sleep loss?
P(1st loses and 2nd loses )
P(three people lose sleep)
What are the chances that at least 1out of 3 loses sleep?
P(at least one ) 
The Normal Distribution
Use curves to describe overall pattern seen
in a histogram.
Curve will capture 100% of all observations.
Hence, there will be a total area of 1 below it.
Then the area under the curve for a given range
of values will represent the proportion
(percent, fraction) of observations that fall in
that range.
Curves and proportions
%
The proportion of scores
above 80 is roughly 26.8%.
v
20
40
60
80
100
%
The area under the density
curve for scores above 80
is roughly 0.261 =26.1%.
v
20
40
60
80
100
Mean and Medians
Location of the median on a density curve is where
area under is cut in half.
Location of the mean on a density curve is where the
length of the curve is cut in half.
On symmetric curves:
On skewed curves:
Normal curves are special kinds of density curves
• Symmetric, single peaked, bell-shaped
Use m(mu) and s(sigma) to talk about mean and std. dev.
m –
s –
m-s m m+s
68-95-99.7 Rule
• About 68% of data fall within
• About 95% of data fall within
• About 99.7% of data fall within
m - 3s m - 2s
m-s
m
m + s m + 2s m + 3s
Example 1
Grasshopper jumps can be described by a Normal
distribution with m = 12 inches and s = 2 inches.
About 68% of all jumps are
About where would you
between
inches.
find the top 2.5%?
68%
95%
99.7%
6
8
10
12
14
16
18
Example 1 – continued
What % falls below 14 inches?
6
8
10
12
14
16
18
6 8 10 12 14 16 18
6
8 10 12 14 16 18
What % of jumps are more than 14 inches?
What % of jumps are between 14 and 16 inches?
Finding values without 68-95-99.7
We use tables or calculators to find harder values, like where
is the top 10% or what percent falls below a given
observation.
N(m, s) means observations come from a Normal distribution
with a mean of m and a standard deviation of s.
Standardize observation x from N(m, s) by:
z
xm
s
The standardized value is called a
z-scores from example 1, N(12, 2)
x 14 
x 16 
x 10 
x 17 
z
xm
s
Two functions on the calculator
(found under 2nd VARS => DISTR)
• normalcdf( : will give area between two bounds for a
given m, s.
• invNorm(
: will give the observation that has a
particular area to its left for a given m, s.
• normalcdf(lower bound, upper bound, m, s)
• invNorm(area, m, s)
m-s
m
m+s
Using the calculator with grasshopper N(12, 2)
What % of jumps fall below 17 inches?
No lower bound, so:
6 8 10 12 14 16 18
normalcdf(lower bound, upper bound, m, s)
= normalcdf(
) = area below 17 =
What % of jumps fall above 11.5 inches?
First, find area
normalcdf(lower bound, upper bound, m, s)
= normalcdf(
Since total area is 1 and we have
we want
6 8 10 12 14 16 18
) = area
=
:
Using the table with grasshopper N(12, 2)
What % of jumps fall between 10 and 16.36 inches?
-
=
6 8 10 12 14 16 18
6 8 10 12 14 16 18
6 8 10 12 14 16 18
Area between = area below 16.36 – area below 10.
Calculator does this all at once with the normalcdf( function.
normalcdf(lower bound, upper bound, m, s)
= normalcdf(
) = area between =
Using the table with grasshopper N(12, 2)
What jumps fell in the top 10%?
10%
6 8 10 12 14 16 18
What observation has an area of .10 above it?
What observation has an area of .90 below it?
Use invNorm function to find that observation.
invNorm(area, m, s)
= invNorm(
) = value with .9 area below=
Using the table with grasshopper N(12, 2)
50%
Where do the middle 50% fall?
6 8 10 12 14 16 18
What observation has an area of
below it?
What observation has an area of
below it?
Use invNorm function to find those observations.
invNorm(area, m, s)
= invNorm(
) = value with
area below =
invNorm(area, m, s)
= invNorm(
) = value with
area below =
1,2,3 standard deviations away
accurate to two decimal places
68-95-99.7 Rule
Sampling Distributions
Know the entire population:
(parameter)
Know only a sample (SRS):
(statistic)
Law of Large Numbers
- As you increase the sample size, sample mean gets closer
to population mean
Population = 3, 3, 8, 15, 20, 21, 22, 31, 39
Sample of size 1= 8
Sample of size 2= 8, 22
Sample of size 3= 8, 22, 31
Sample of size 4= 8, 22, 31, 3
Sample of size 5= 8, 22, 31, 3, 20
Population of 7 people and their weights (in pounds)
m  156
122, 140, 150, 155, 160, 170, 195
Samples of size 1: {122}, {140}, {150}, {155}, {160}, {170}, {195}
Mark off the sample mean for each sample with an “x”
x
120 130
x
140
x x x
150
160
x
170
x
180
190
200
Samples of size 2: {122, 140}, {122, 150}, {122, 155}, {122, 160}, {122, 170},
{122, 195}, {140, 150}, …, (170, 195}. There are 21 possible samples.
Mark off the sample mean for each sample with an “x”
x
x x x x x x x x x x xxx x x x x x x x
120 130
140
150
160
170
180
190
200
Population of 7 people (continued)
140, 122, 160, 195, 150, 155, 170
m  156
Samples of size 1:
x
x
120 130
140
x x x
150
160
x
170
x
180
Samples of size 2:
x
x x x x x x x x x x xxx x x x x x x
130
140
150
160
170
190
200
x
180
190
Samples of size 6: 7 possible sample of this size. {122, 140, 160, 150, 155, 170}, …
x  149.5
x x xxxx x
140
150
160
170
180
Sampling distribution of x
Sampling from a large population with mean m and
standard deviation s:
samples of size n will have their sample means distributed
with a mean m and standard deviation s over root n.
If population is N(m, s), then
s 

x is N  m ,
.
n

If population is not Normal but n is large, then
 s 
x is approximat ely N  m ,
.
n

mx  m
s
sx 
n
Ex. 1 - Weight of eggs is N(65, 3)
Your egg carton holds 9 eggs, so consider each carton as a
random sample of 9 eggs. Let X be the weight of a single
egg in grams and X be average weight of your carton.
P( X  67)
What is the sampling distribution for your carton’s average weight?
mx 
P( X  67)
sx 
Weight of eggs is N(65, 3) – continued
Mean weight of carton is N(65,1)
Convert 67 to a z-score
for the carton:
Convert 67 to a z-score
for a single egg:
56
59
62
65
67 68
71
74
Ex. 2 - Length of trout is N(17.5, 2.5)
Your local waters contain a multitude of trout. Let X be the
length of a single fish in inches and X be average length
of your daily catch of five fish.
P(16  X  19)
What is the sampling distribution for your daily catch?
mx 
P(16  X  19)
sx 
Trout length is N(17.5, 2.5) – continued
Mean length of daily catch is N(17.5,1.118)
Convert 16 to a z-score
for the daily catch:
Convert 16 to a z-score
for a single fish:
10
12.5
15
17.5
20
22.5
25
Trout length is N(17.5, 2.5) – continued
Mean length of daily catch is N(17.5,1.118)
10 12.5 15
17.5 20
22.5 25
10 12.5
15
17.5
20
22.5 25
Ex 3 - Length of trout is N(10, 2)
Your fishing pond has another type of trout. Let X be the
length of a single fish in inches taken at random and X be
average length of a sample of 16 fish.
P( X  10.72)
What is the sampling distribution for a sample of 16 fish?
mx 
P( X  10.72)
sx 
Trout length is N(10, 2) – continued
Mean length of 16 fish is N(10,0.5)
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6
8
10
12
14
16
4
6
8
10
12
14
16