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) females, 8 males, 4/19 males and 3/8 females have no siblings 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 ; times 2 is For z=2.00, the table’s value is ; times 2 is For z=3.00, the table’s value is ; times 2 is

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 ) SAT scores, μ=500, σ=100 P(X>700) = ? 1 st : X=700  z=( )/100=2 2 nd : P(z>2) = = 2.28%

Probability and the Normal Distribution (cont'd.) The unit normal table lists several different proportions corresponding to each z-score location. (p ) –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.)

example 6.3A (p. 175) P(z > 1) = ? Fig. 6.8 (a) check z table (p. 700) col. C: P(z < -1) = check col. D: P(z < -1) = 0.5 – P(0<z<1) = 0.5 – =

example 6.3B (p. 175) P( z < 1.5) = ? Fig. 6.8 (b) check col. D in z table P(z < 1.5) = P(0<z<1.5) = = check z table (p. 701) col. B = P(z < 1.5)= or check col. C = P(z 1.5)= P(z < 1.5) = 1 – =

example 6.3C (p ) 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 – = col. C = P(z < -0.5) =

example 6.4A (p. 176) P( z > z 0 ) = 0.1, z 0 = ? 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) = check col. C of p.700 (probability = )  z 0 = 1.28

example 6.4B (p. 176) P( |z| < z 0 ) = 0.6, z 0 = ? Fig 6.9 (b) check col. D of p.700 (probability = ) P(-? < z < ?) = 0.6  P(0 < z < ?) = 0.3  find P(0<z<0.84) =  z 0 =  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) = check col. C of p.700 (probability = )  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) =  z0 = check col. B of p.699 (probability = )  z0 = 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) =  z0 =  1.04 check col. D of p.700 (probability = )  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 = ( )/15 = 1.33  P(z < 1.33) = ?  check D  P(z < 1.33) = 0.5+P(0<z<1.33) = = The probability of randomly selecting a person with IQ less than 120 is The proportion of the individuals in the population with IQ less than 120 is

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

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( – ) – P(z > 0.7) check col. B, C  = – = or check col. D P( – 0.3 < z < 0.7) = P( 0< z < 0.3) + P( 0< z < 0.7) = =

example 6.7 (p. 180) (interstate highway) driving speed X miles/hour μ = 58, σ=10 (approximately normal) P(65 < X < 75) = ?  P( ) – P(z > 1.7) check col. C  = – = or check col. D P( 0.7 < z < 1.7) = P( 0< z < 1.7) – P( 0< z < 0.7) = =

example 6.8 (p. 182) time commuting to work: minutes/day μ = 24.3, σ=10 (assumed normal) P( X > X 0 ) = 0.1, X 0 =?  P(z > z 0 ) = 0.1, z 0 = ? 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) =  z 0 =1.28  X 0 = 1.28* = % 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) = , P(0 < z < 1.65) =  z0 =1.645  X0 = 1.645* =  z0 =  X0 = * = % of people spend between 7.85 and 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 > ?) = P(0<z<?) = 0.6  P(0<z<?) = 0.1 P(0 < z < 0.25) = = P( < z < 0) X 0 = (-0.25)* = 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) = X 0 = 1.28* = 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) =  X 0 = 0.67* = 567, X 0 = – 0.67* = 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 ) 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 xn=2n=3n=4n=5n= P(each) 1/4 1/81/16 1/32 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 ) 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 =  P(z > 1.17) = 0.5 – P(0< z <1.17) = 0.5 – = 0.121

example 6.11 (p ) 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 – =

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 * = 0.135

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.