Z Tests across multiple Samples to produce Probabilities? So I've a problem I'm trying to solve. I'm more of a coder than a stats guy so this may be very basic!
Movie A rated 5.5 with 10 ratings(5,6,5,6,5,6,5,6,5,6), movie B rated 5.44 with 9 ratings (4,5,6,7,4,5,6,7,5), movie C rated 5.33 with 9 ratings (2,3,8,9,2,3,8,9,4)
What is the probability that C is rated higher than A or B when all three have 10 ratings. We are waiting on two ratings to be added - 1 for B and 1 for C. A is finished as it has ten already.
Normal distribution can be assumed as being true for the three movies.
I know the method of using a Z Score to work out the probability of an individual movie being above, below or in between a certain rating ranges. But when comparing two or more ratings how do you work out the probabilities as its a constantly moving target?
With my programming hat on the simplest method I can think of is a brute force method of looping through all combinations of Z Scores relative to each other but I imagine this won't scale well at all. My question can this be done and what is the cleanest way to achieve this?
 A: Note: This Answer was for a quite different original
version of the question.
First, here is an example comparing two movies:
For Movie X, 1000 subjects give an average
rating of 7.0 and, for Movie Y, 1100 subjects
give an average rating of 4.9.
Summaries of individual scores for the two movies
are given below:
summary(x)
    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  3.000   6.000   7.000   7.049   8.000  10.000 
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   4.000   5.000   4.977   6.000   9.000 

Boxplots of the scores (X bottom) show that about half
of the scores for X were 6, 7 or 8 and about half of the scores
for Y were 4, 5, or 6. The non-overlapping
'notches' in the sides of the boxes suggest that
there is statistically significant difference
in the popularity of the two movies.
boxplot(x, y, horizontal=T, notch=T, col="skyblue2")

For small samples, the Wilcoxon Rank Sum test has
difficulty giving definitive results, if there are
many individuals with the same scores ("ties"), as is often
true of movie ratings. However,
for large samples, results are reliable. The P-value
very near $0$ shows strong evidence of a statistically
significant difference. (Results from R.)
wilcox.test(x,y)

        Wilcoxon rank sum test 
        with continuity correction

data:  x and y
W = 902510, p-value < 2.2e-16
alternative hypothesis: 
 true location shift is not equal to 0

However, if 900 ratings for Movie W are much more diverse
than for Movie Y, it is more difficult to say which
movie is better, even though both have similar median ratings
(both around 5).
summary(w)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.000   3.000   5.000   4.991   7.000   9.000 

Boxplots (W on bottom) show overlapping notches. We
can say the scores are more diverse for Movie W, but it
is more complicated to try to claim one movie is more
popular than the other.
boxplot(w, y, horizontal=T, notch=T, col="skyblue2")


A Wilcoxon test is looking for a difference in 'location'
and finds no difference at the 5% level of significance.
wilcox.test(y,w)

        Wilcoxon rank sum test 
        with continuity correction

data:  y and w
W = 494870, p-value = 0.992
alternative hypothesis: 
 true location shift is not equal to 0

If the people who really like Movie W express
their opinions more loudly than the the the ones
who hate it, Movie W might turn out to make more
money than Movie Y.

Note: Here is R code for sampling the fictitious
scores for Movies W, X, and Y.
set.seed(2011)
x = rbinom(1000, 10, .7)
y = rbinom(1100, 10, .5)
w = sample(1:10, 900, rep=T)

