~Recall: a random variable, discrete & continuous (see topics of Interest)

~Probability Distributions: Ties probability  with statistics, chance occurrences  of events are at the heart of statistics.

~Note: These are also called probability spaces. a probability is assigned or computed for each value of x, then the results are displayed in one of 3 ways, by a chart or table or by a formula for P(x). This formula is a function that will give probabilities for each x value.

~Note: Charts & tables are more popular in basic statistics.

~Probability Histograms: For all practical purposes, these are the same as the relative frequency  histograms covered in lesson #2

~What is the requirement for a probability distribution?   Simply that all probabilities make sense and  their sum is exactly 1.

~Ex:  Does the following define a probability distribution?   
          x |  0     1     2      3
     P(x) | 1/8  3/8  3/8  1/8                    Ans:  yes

~Ex: Does the following function define a probability distribution?            
        P(x) = x/4, for x=0,1,2,3               Ans:  No

~Finding the Mean, variance, & standard deviation of a probability distribution.

~Mean: Just multiply x with P(x) for each x, then add the result,
              i.e., m = ∑[xP(x)]

~Variance: There are 2 equivalent forms:  s²= ∑[(x-m)²P(x)]  or         
s² = ∑[x²P(x)] - m² (be careful here, ∑ is only with the first term)

~Standard deviation: Simply, the positive square root of the variance:        
  So, s = sqr(either expression for s² above).

~Note: The TI-83 gives m as x bar and  s as s_x.  Use the same procedure as in the past, enter x’s in L1 and P(x) in L2.

~Ex: 3 light bulbs are selected  from a box containing 12, of which 3 are defective. Let x=# of defective bulbs in our sample of 3.

Obviously, x=0,1,2, or 3. Find the mean & standard deviation of the # of defective bulbs we expect to get.

We need a probability distribution for the # of defects,
i.e., P(0), P(1), P(2), & P(3) when our sample contains 3 bulbs.

P(0)= # of ways  we can have 3 non-defective bulbs divided by the # of ways we can choose any 3 bulbs.

P(0)=9C3/12C3= 84/220,  P(1)=(3)9C2/220 = 108/220,
P(2) =(9)3C2/220= 27/220, P(3)= 3C3/220= 1/220.
Now, enter the x values in L1 & the corresponding probabilities in L2.  

This will give you   m = .75  and   s = .68.

~Round-off rule:  One should only round-off at the end of a calculation, since, if you don’t, your answers will be far off in many cases.

~At the end, we usually round-off our answers to one more place than the original data for x. Hence, m= .8 and s = .7 would be fine here.

~Unusual results:  Recall, more than 2 standard deviations on either side of the mean is a guideline. This is know as the range rule of thumb.

~Note:  In the example we just covered, x=3 (3 defective bulbs when sampling 3) is unusual to get.

~Using probabilities to test for unusual results: Be careful here, low probabilities do not necessarily mean results are unusual.

~Ex: a coin is flipped 1000 times. The # of heads observed was 501, however, P(exactly 501 heads in 1000 ) is approx. .03.(will do later)

~Note: Better to say  at least 501 heads. That probability is roughly .49 (will do later)

~Note: So, keep this in mind, when using a probability to determine whether or not an event is unusual. Generally speaking, when
probabilities come out .05 or less for an event, we usually conclude that the event is unusual (taking into account the situation  described)

~Expected value:  Also called Mathematical expectation, or Theoretical mean or Theoretical return.    (Symbolized by E)

~Let x₁,x₂,..., x_k  be outcomes with probabilities of P(x₁), P(x₂), ..., P(x_k), then E=x₁P(x₁)+x₂P(x₂)+...+x_kP(x_k)= ∑xP(x)=m

~Note:  So, the expected value is just the mean of the probability distribution.

~Ex:  You roll a die. If a 1 appears, you are paid $5, if a 2 appears, you must pay $7, if a 3 appears, you  collect $3, if a 4 appears, you
collect $10, if a 5 appears, you must pay $5, if a 6 appears, you collect $2. What is the expected value? Would you  pay $2 to play?

Ans:  $1.33, NO

~Binomial Probability distributions: There is a 2 outcome category (yes, no),(girl, boy),(success, failure), & so on.

~Note: Must meet all of the following: 1) n=# of trials is fixed, 2) each trial is independent of the others, 3) probabilities stay the same for each

~Binomial Probability formula:  Let r=# of successes, p=probability of a success, q= probability of a failure. (note that q=1-p)

Then, P( r ) = nCr (p)^r( q )^(n-r).

~Ex: If the probability of a girl baby born is .6, find the probability of a family of 6 having 4 girls.

~Let having a girl be a success, then  P(4)= 6C4 (.6)⁴(.4)²= .31104.  
TI-83 will get it for us: 2nd, VARS, binompdr(, insert n,p,r, Enter.
                                                                                                                                          
~Note: These types of problems are called Bernoulli trials.  
(technically, a Bernoulli trial consists a single success or failure, indicated by a 0 & 1, along with their probabilities)
       Also, nCr is denoted as  (      ) , where n & r are  
      written in a column (where r is under n), in many textbooks.