Download Chapter 6: Probability

Chapter 6: Probability
Probability
• Probability is a method for measuring and
quantifying the likelihood of obtaining a specific
sample from a specific population.
• We define probability as a fraction or a
proportion.
• The probability of any specific outcome is
determined by a ratio comparing the frequency
of occurrence for that outcome relative to the
total number of possible outcomes.
Rolling a die
• Event A: even number, greater than 4, or less
than 3, etc.
Tossing a coin three times
• Event A: one head occurs
a deck of cards
• 52 cards: clubs and spades, diamonds and
hearts (13 cards in each suit: A, 2, 3, 4, 5, 6, 7,
8, 9, 10, J, Q, K)
Probability (cont'd.)
• Whenever the scores in a population are
variable, it is impossible to predict with perfect
accuracy exactly which score(s) will be obtained
when you take a sample from the population.
– In this situation, researchers rely on probability to
determine the relative likelihood for specific
samples.
– Thus, although you may not be able to predict
exactly which value(s) will be obtained for a
sample, it is possible to determine which
outcomes have high probability and which have
low probability.
Probability and Sampling
• To assure that the definition of probability is
accurate, the use of random sampling is
necessary.
– Random sampling requires that each member of
a population has an equal chance of being
selected.
– Independent random sampling includes the
conditions of random sampling and further
requires that the probability of being selected
remains constant for each selection
Random Sampling in this textbook
(sampling with replacement)
• The definition of random sampling in this
textbook is similar to “sampling with
replacement”, it requires:
(1) each individual in the population has an equal
chance of being selected.
(2) the probability of being selected stays constant
from one selection to the next. (~ with
replacement)
Random Sampling without Replacement
and.....
• a random sampling
• the probability of being selected changes from
one selection to the next...
• There are many types of random sampling: e.g.
systematic random sampling, stratified random
sampling, or cluster sampling....
Random Sampling
• for each subject, the probability of being
selected = 1/12
With Replacement (independent)
• 4 black and 5 red balls
dependent
independent ↓
Without Replacement (dependent)
• 2 blue and 3 red marbles in the bag
Example 6.1 (p. 168)
• N=10, X=1,1,2,3,3,4,4,4,5,6
• random sample, n=1
• P(X>4) = ?
• P(X <5) = ?
ex. 1-3 (p. 169)
• 1. 19 females, 8 males, 4/19 males and 3/8
females have no siblings
• 2. 10 red, 30 blue marbles in a jar
(random sampling = with replacement!!)
• 3. N=10, n=1, Fig 6.2(p. 168)
Probability (cont'd.)
• When a population of scores is represented by a
frequency distribution, probabilities can be
defined by proportions of the distribution.
• Probability values are expressed by a fraction or
proportion.
• In graphs, probability can be defined as a
proportion of area under the curve.
Probability and the Normal Distribution
Fig 6.4 (p. 171) check z table (col. D)
• For z=1.00, the table’s
value is 0.3413; times 2
is 0.6826.
• For z=2.00, the table’s
value is 0.4772; times 2
is 0.9544.
• For z=3.00, the table’s
value is 0.4987; times 2
is 0.9974.
Probability and the Normal Distribution
• If a vertical line is drawn through a normal
distribution, several things occur.
– The line divides the distribution into two sections.
The larger section is called the body and the
smaller section is called the tail.
– The exact location of the line can be specified by
a z-score.
Normal Distribution
• body v.s. tail
μ = 500
example 6.2 (p. 171-172)
• SAT scores, μ=500, σ=100
• P(X>700) = ?
1st : X=700  z=(700-500)/100=2
2nd : P(z>2) = 0.5-0.4772 = 2.28%
Probability and the Normal
Distribution (cont'd.)
• The unit normal table lists several different
proportions corresponding to each z-score
location. (p. 699 - 702)
– Column A of the table lists z-score values.
– For each z-score location, columns B and C list the
proportions in the body and tail, respectively.
– Finally, column D lists the proportion between the
mean and the z-score location.
• Because probability is equivalent to proportion,
the table values can also be used to determine
probabilities.
Probability and the Normal Distribution
(cont'd.)
• To find the probability corresponding to a
particular score (X value), you first transform the
score into a z-score, then look up the z-score in
the table and read across the row to find the
appropriate proportion/probability.
• To find the score (X value) corresponding to a
particular proportion, you first look up the
proportion in the table, read across the row to
find the corresponding z-score, and then
transform the z-score into an X value.
Probability and the Normal Distribution
(cont'd.)
• The normal distribution is symmetrical; therefore,
the proportions will be the same for the positive
and negative values of a specific z-score.
• Proportions are always positive, even if the
corresponding z-score is negative.
• A negative z-score means that the tail of the
distribution is on the left side and the body is on
the right, and vice versa for a positive z-score.
Probability and the Normal Distribution
(cont'd.)
+ 0.25
example 6.3A (p. 175)
• P(z > 1) = ?
Fig. 6.8 (a)
check z table (p. 700)
col. C: P(z < -1) = 0.1587
check col. D: P(z < -1) = 0.5 – P(0<z<1)
= 0.5 – 0.3413 = 0.1587
example 6.3B (p. 175)
• P( z < 1.5) = ?
Fig. 6.8 (b)
check col. D in z table
P(z < 1.5) = 0.5 + P(0<z<1.5) = 0.5+0.4332 =
0.9332
check z table (p. 701) col. B = P(z < 1.5)=0.9332
or check col. C = P(z < - 1.5) = P(z > 1.5)=0.0668
P(z < 1.5) = 1 – 0.0668 = 0.9332
example 6.3C (p. 175-176)
• P(z < -0.5) = ?
Fig. 6.8 (c)
check z table (p. 700)
check col. D in z table
P(z < -0.5) = 0.5 – P(0<z<0.5) = 0.5 – 0.1915 =
0.3085
col. C = P(z < -0.5) = 0.3085
example 6.4A (p. 176)
• P( z > z0) = 0.1, z0 = ?
Fig 6.9 (a)
check col. D in z table
P(z > ?) = 0.5 – P(0<z<?) = 0.1
P(0<z<?) = 0.4  find P(0<z<1.28) = 0.3997
check col. C of p.700 (probability = 0.1003)
 z0 = 1.28
example 6.4B (p. 176)
• P( |z| < z0) = 0.6, z0 = ?
Fig 6.9 (b)
check col. D of p.700 (probability = 0.2995)
P(-? < z < ?) = 0.6P(0 < z < ?) = 0.3
 find P(0<z<0.84) = 0.2995
 z0 =  0.84
ex. 2(p. 177)
a. P( z > z0) = 0.2, z0 = ?
check col. D in z table  P(z > ?) = 0.5 – P(0<z<?) = 0.2
P(0<z<?) = 0.3  find P(0<z<0.84) = 0.2995
check col. C of p.700 (probability = 0.2005)
 z0 = 0.84
b. P( z > z0) = 0.6, z0 = ?
check col. D in z table  P(z > ?) = 0.5 – P(0<z<?) = 0.6
P(0<z<?) = 0.1  find P(0<z<0.25) = 0.0987  z0 = -0.25
check col. B of p.699 (probability = 0.5987)
 z0 = -0.25
c. P( |z| < z0) = 0.7, z0 = ?
check col. D in z table  2*P(0< z < ?) = 0.7 P(0<z<?) = 0.35
 find P(0<z<1.04) = 0.3508  z0 =  1.04
check col. D of p.700 (probability = 0.3508)
 z0 =  1.04
6.3 From X to z (p.178)
• If X is normally distributed, then you can use z
score and z distribution to find the probability.
• If X is not from a normal distribution, you should
not just assume it’s normal and use z table to
find the probability for X.
example 6.5 (p. 178)
• IQ scores ~ Normal, μ = 100, σ=15
• P(X < 120) = ?
• z = (120-100)/15 = 1.33
P(z < 1.33) = ?
 check D  P(z < 1.33) = 0.5+P(0<z<1.33) =
0.5+0.4082 = 0.9082
• The probability of randomly selecting a person
with IQ less than 120 is 0.9082.
• The proportion of the individuals in the
population with IQ less than 120 is 0.9082.
Percentiles and Percentile Ranks
• The percentile rank for a specific X value is the
percentage (%) of individuals with scores at or
below that value.
• When a score is referred to by its rank, the score
X is called a percentile. The percentile rank for
a score in a normal distribution is simply the
proportion to the left of the score.
• in ex. 6.5: the percentile rank for IQ = 120 is
0.9082
example 6.6 (p. 179)
• (interstate highway) driving speed X miles/hour
• μ = 58, σ=10 (it is approximately normal)
• P(55 < X < 65) = ?
P( – 0.3 < z < 0.7) = P(z > -0.3) – P(z > 0.7)
check col. B, C
=0.6179 – 0.2420 = 0.3759
or check col. D
P( – 0.3 < z < 0.7) = P( 0< z < 0.3) + P( 0< z < 0.7)
= 0.1179+0.2580 = 0.3759
example 6.7 (p. 180)
• (interstate highway) driving speed X miles/hour
• μ = 58, σ=10 (approximately normal)
• P(65 < X < 75) = ?
P( 0.7 < z < 1.7) = P(z > 0.7) – P(z > 1.7)
check col. C
=0.2420 – 0.0446 = 0.1974
or check col. D
P( 0.7 < z < 1.7) = P( 0< z < 1.7) – P( 0< z < 0.7)
= 0.4554+0.2580 = 0.1974
example 6.8 (p. 182)
• time commuting to work: minutes/day
• μ = 24.3, σ=10 (assumed normal)
• P( X > X0) = 0.1, X0 =?
P(z > z0) = 0.1, z0 = ?
check col. D in z table
P(z > ?) = 0.5 – P(0<z<?) = 0.1
P(0<z<?) = 0.4  find P(0<z<1.28) = 0.3997
z0 =1.28  X0 = 1.28*10 + 24.3 = 37.1
10% of people spend more than 37.1 minutes of
commuting time each day
example 6.9 (p. 183)
• time commuting to work: minutes/day
• μ = 24.3, σ=10 (assumed normal)
• P( |X| < X0) = P( |z| < z0) = 0.9, X0 = ?
• check col. D for 45% P(0 < z < ?) = 0.45
• P(0 < z < 1.64) = 0.4495, P(0 < z < 1.65) = 0.4505
 z0 =1.645  X0 = 1.645*10 + 24.3 = 40.75
 z0 =-1.645  X0 = -1.645*10 + 24.3 = 7.85
90% of people spend between 7.85 and 40.75 minutes of
commuting time each day
ex. 2 (p. 184)
•
•
•
•
•
•
•
•
•
•
•
•
•
SAT(math): μ = 500, σ=100 (assumed normal)
a. P( X > X0) = 0.6, minimum score: X0 =?
P(z > ?) = 0.5 + P(0<z<?) = 0.6  P(0<z<?) = 0.1
P(0 < z < 0.25) = 0.0987 = P( -0.25 < z < 0)
X0 = (-0.25)*100+500 = 475
b. P( X > X0) = 0.1, minimum score: X0 =?
P(z > ?) = 0.5 – P(0<z<?) = 0.1  P(0<z<?) = 0.4
P(0 < z < 1.28) = 0.3997
X0 = 1.28*100+500 = 628
c. P( |X| < X0) = P( |z| < z0) = 0.5, X0 = ?
P( – ? < z < ?) = 2*P( 0< z < ?) = 0.5  P( 0< z < ?) = 0.25
P(0 < z < 0.67) = 0.2486X0 = 0.67*100+500 = 567,
X0 = – 0.67*100+500 = 433
ex. 3 (p. 184)
• positively skewed, μ = 40, σ=10
• find P(X > 45) = ? lack the info about the degree
of skewness  impossible to find P(X > 45)
The Relative Positions of the Mean,
Median and the Mode
Probability and the Binomial
Distribution
• Binomial distributions are formed by a series of
observations (for example, 100 coin tosses) for
which there are exactly two possible outcomes
(heads and tails)
• The two outcomes are identified as A and B,
with probabilities of p(A) = p and p(B) = q.
• p + q = 1.00
• The distribution shows the probability for each
value of X, where X is the number of
occurrences of A in a series of n observations.
Probability and the Binomial Distribution (cont.)
• The outcome of each trial is classified into one of
two mutually exclusive categories—a success
or a failure
• The random variable, x, is the number of
successes in a fixed number of trials (n).
• The probability of success and failure stay
the same for each trial. P(success) = P(A) = p
• The trials are independent, meaning that the
outcome of one trial does not affect the outcome
of any other trial. (with replacement)
example 6.10 (p. 185-186)
•
•
•
•
toss a fair coin twice
X: # of heads
HH, HT, TH, TT
each has probability = ¼
• P(X=0) = P(X=2) = ¼
• P(X=1) = ½
• P(X≥1) = P(X=1) + P(X=2) = ¾
binomial distribution is a discrete probability
distribution.
Probability after n Coin tosses p.186
x : number of heads after tossing a fair coin n times
x
0
1
2
3
4
5
6
P(each)
n=2
1
2
1
n=3
1
3
3
1
n=4
1
4
6
4
1
n=5
1
5
10
10
5
1
1/4
1/8
1/16
1/32
n=6
1
6
15
20
15
6
1
1/64
Binomial – Shapes or Skewness for Varying 
and n = 10，  = 0.5 (symmetric)
The shape of a binomial distribution changes as n and 
change. Here we have n = 10
Binomial – Shapes or Skewness for Constant  and
Varying n：  = 0.1，n larger  more symmetric
Mean and Variance of a
Binomial Distribution
Knowing the number of trials, n, and the probability of
a success, , for a binomial distribution, we can
compute the mean and variance of the distribution.
Note: here P(success) = P(A) = p =  for the population
and P(failure) = P(B) = q = 1 – p = 1 - 
Probability and the Binomial
Distribution (cont'd.) p. 187
• When pn and qn are both greater than 10, the
binomial distribution is closely approximated by
a normal distribution with a mean of μ = pn
and a standard deviation of σ = npq.
• q=1-p
• In this situation, a z-score can be computed for
each value of X and the unit normal table can be
used to determine probabilities for specific
outcomes. Binomial  Normal distribution
example 6.11 (p. 188-189)
• p = ¼ and q = ¾ , n = 48
μ = np = 48*0.25 = 12,
σ = npq = 48*0.25*0.75 = 9 = 3
x: # of correct answer (guess/without ESP)
P(x>15) = ?
X: normal approximation: P(X >15.5) = ?
z = (15.5 – 12)/3 = 1.167
P(z > 1.17)
= 0.5 – P(0< z <1.17)
= 0.5 – 0.3790 = 0.121
example 6.11 (p. 188-189)
• p = ¼ and q = ¾ , n = 48, μ = 12, σ = 3
x: # of correct answer (without ESP)
P(x < 15) = ?  P(X < 14.5) = P(z < 0.83)
P(x ≥ 15) = ?  P(X > 14.5) = P(z > 0.83)
P(x ≤ 15) = ?  P(X < 15.5) = P(z < 1.17)
Ex. 2 (p. 189)
9 possible outcomes: RR, RP, RS, PR, PP, PS,
SR, SP, SS
• n = 72 trials, x: # of ties, p = 1/3, P(x > 28)=?
 μ = np = 24, σ = npq = 4
P( X > 28.5) = P(z > 1.125) ≈ P(z > 1.13)
= 0.5 – P(0 < z < 1.13) = 0.5 – 0.3708 = 0.1292
Ex. 3 (p. 189)
Toss a fair coin 36 times, n = 36
• x: # of heads, p = 1/2, P(x = 18)=?
 μ = np = 18, σ = npq = 3
P( x = 18) ≈
P (17.5 < X < 18.5) = P(-0.17< z <0.17)
= 2*P(0 < z < 0.17) = 2 * 0.0675 = 0.135
Probability and Inferential Statistics
• Probability is important because it establishes a
link between samples and populations.
• For any known population, it is possible to
determine the probability of obtaining any
specific sample.
• In later chapters, we will use this link as the
foundation for inferential statistics.
Probability and Inferential Statistics
(cont'd.)
• The general goal of inferential statistics is to use
the information from a sample to reach a general
conclusion (inference) about an unknown
population.
• Typically a researcher begins with a sample.
Probability and Inferential Statistics
(cont'd.)
• If the sample has a high probability of being
obtained from a specific population, then the
researcher can conclude that the sample is likely
to have come from that population.
• If the sample has a very low probability of being
obtained from a specific population, then it is
reasonable for the researcher to conclude that
the specific population is probably not the
source for the sample.