St1norm.1
  Prof Vinod Stat 1

 Normal dist. word problems.  Important hints in doing these right.
Step 1: 
  Begin by drawing one long vertical line and two horizontal lines. Name the upper horizontal line x axis
  And name the lower horizontal line as z axis.  
  Get the bearings on the graph for x.  The upper limit of x is found as:
  Upper limit for the bell curve =  mean of x + 4 times standard deviation of x
  Lower limit for the bell curve =  mean of x - 4 times the std. dev of x
  For example, if mean is 100 and standard deviation is 10 the limits are 100-40=60 to 100+40=140
 
  Normal distribution x~N(mean mu, variance sigma squared) upstairs
                  With range (mu MINUS 4* sigma)  to (mu PLUS 4* sigma)
  Standard Normal z~N(0,1) is downstairs with range -4 to +4 downstairs
  Note that in this notation ~ means “is distributed as”
  N means Normal, we give the two parameters mean and variance in parentheses
  separated by a comma.  This notation is used in the statistical literature.
  
  The bearings for z are: Lower limit at -4, center at 0 and upper limit at 4

  Draw two bell shaped curves. If you can’t draw well, it is ok. 
  Let us call the two bell-shaped curves as UPSTAIRS / Downstairs graphs.

Step 2: Translate the probability asked into area under normal curve 
      and shade the area on a graph  (usually upstairs graph showing x on horizontal axis)
     (If you can’t draw well, it is ok)


Step 3: Map the problem to the z distribution
  The bearings for z are: Lower limit at -4, center at 0 and upper limit at 4
  Shade the desired area Under the z curve

Step 4:  Use Tables to find the answer.  In using the Normal dist. table
  remember that:  It is symmetric, centered at zero
  Area is always positive, 
  (Area to the left of the middle)=( area to the right of the middle0 = 0.50
  Table only gives area between 0 and a positive number,  
  If you want tail area you must subtract from 0.5
  If you want two equal tails together, you can double one tail



  To get full credit in exam, you must show the graphs.........


  Normal Distribution word problems  AREAs under the Normal:
  Be sure to have a copy of a table having "area of a standard Normal
  distribution"  Inside cover page of your text.  Bring a xerox copy
  to class from now on every day, unless you plan to bring the text.
  You may not be admitted to class if you do not have the table!

  x is Normally distributed r. v. with mean mu=200, variance=100

   for example, x=number of dozens of eggs sold in a week.

  Find the probability that  xL < x < xU where xL is lower limit,
    xU is the upper limit.
  P( xL < x < xU )  This probability does  not change when we make a
  Z transformation of all terms of the inequality.  The probability remains
  Exactly the same.  This is what justifies the mapping from upstairs to downstairs.

  To solve these problems draw a picture of the Normal curve for x
    show where the mean lies for x and find the xL and xU on the figure.
  Compute the std.dev= sq. root of variance =10 here.

  Now z transform: zL= ( xL - mean ) / std.dev.
               
                   zU= ( xU - mean ) / std.dev.

  There are several different types or problems possible:
  1) Right tail, zL>0, zU is positive infinity
      Find the prob. of more than 220 dozen eggs.
           xL=220,  xU=+infinity, zL=[(220-200)/10] = 20/10 = 2
           zL=2,  zU=infinity means right tail from 2.
      Table gives the area from 0 to a positive number like 2 (=0.4772)
      Tail area to the right of 2 is 0.5 minus 0.4772  (Ans=0.0228)
  2)  Left of a positive zU number:  xL= negative infinity, xU= positive number.
     Find the prob of less than 220 dozen eggs.
     zL=neg infinity, zU=2,   Ans=0.5 + 0.4772 = 0.9772
     Why add the half here? left half is included.
 3)  Both zL=1 and zU=2 are positive.  
     Find the prob of selling between xL=210 and xU=220 doz. eggs.
     zL= (210-200)/10= 10/10 =1  Area from table bet. 0 and 1 is 0.3413
     zU= (220-200)/10 =2.   (tabel area 0 to 2 is 0.4772 as before)
     Ans subtract the smaller tabulated value 0.3413 from 0.4772. Ans=0.1359
  4) Both zL and zU are negative.
     Find the prob of selling bet. xL=180 and xU=190 doz eggs.
     zL=(180-200)/10= (-20/10)=-2  prob is 0.4772 from table (use symmetry)
    zU= (190-200)/10 = (-10/10)= -1, prob is 0.3413
     Ans as in (3) above 0.1359
 5) zL is negative  zU is positive.
    Find the prob of selling bet. xL=180 and xU=210 doz eggs.
    zL=-2, zU=1.  Add the tabulated areas. 0.4772+0.3413 =0.8185
 6)  zL is neg infinity and zU is a negative number.
    Find the prob. of less than xU=180 eggs sold.  zU=-2
    We want left tail area which equals right tail area (symmetry)
    Right tail is 0.0228  as in (1) above.  Ans=0.0228
 7)  Two tails areas.
    Find the prob of less than 180 or more than 210 doz eggs.
    Find P( x<180) + P ( x>210).
    (xL=neg infinity, xU= 180) plus prob(xL=210,xU= positive infinity)
    Here there are two ares to be found with two separate xL and xU.
    zU is -2, left tail is 0.5-0.4772=0.0228
    zL=1, right tail is 0.5-0.3413=0.1587.  The answer is 0.0228+0.1587=0.1815
 8)  Similar to (5), zL is negative and zU is positive
    But the absolute magnitudes (irrespective of sign) are the same.
    The complication arises because the mean and std.dev are 
    unknown numbrs!  No matter though.  We can still answer the question.
    Find the prob of selling within one standard deviation of the mean.
    zL=-1,  zU=1
    Ans= 0.3413 + 0.3413 = 0.6826
   Another question find the prob of selling within 1.96 standard deviations
   of the mean.  zL=-1.96 and zU=1.96,  Area from the table is 0.4750
   Ans= 0.4750+0.4750 = 0.95
  9)  Negation of the (8).  
    Find the prob of not selling within 1.96 std.deviations of the mean.
    Ans= 1- 0.95 = 0.05 = two tail areas. 
   Remember the key phrase  "Within so many standard deviations of the mean". 


SEC. 2:  
SAMPLING DIST OF MEANS PROBLEMS 
  Here the random variable is xbar (or mean)~ N(mu,  sigma square/n )
  Thus sampling distribution has the additional wrinkle that the 
  standard deviation of the mean is sigma divided by square root of n
  Otherwise these are very similar to word problems for Normal. 

 What is a sampling distribution?  It is a probability distribution
 of a statistic like mean, variance, standard deviation, range, median, etc.
 Recall that things we calculate from a sample are called statistic's
 A sampling distribution is defined over all possible values of the statistic
 computed from all possible samples.  If population size is N and sample size
 is n there are NCn or N choose n ways.  E.g. N=5, n=2, NCn=5C2= 5!/(2!*3!)=10
 These all possible ways of selecting a sample of size 2 from a population
Of size 5
 Thus the sample space S over which the sampling distribution is defined.
 The sampling distribution is somewhat hard to get exactly.  

Fortunately
great statisticians have proved that if x is Normal (mean mu, variance sigma^2)
then sample mean xbar is also Normal with same mean and variance 
  sigma^2/n
hence standard deviation of xbar is sigma/(square root of n)

 EXAMPLE
 Average score is unknown ( mu is unknown)
 standard dev is 130
 a sample of 460 is obtained  (n=460)
 sd(xbar)= sd/square-root(n)
          =130/ sqrt(460)
          =6.06

  Find the prob that the sample mean will “differ” from the
  unknown population mean (mu) by less than 12
  Here we use absolute value |  |to recognize that it
  Can “differ” by being smaller or by being larger.

  P( | xbar MINUS mu| < 12 ) = ?
   (mu+12) - mu/ 6.06  = 1.98
   Prob 0 to 1.98 is  0.4761
   prob -1.98 to zero is also 0.4761
  Ans= 0.9522


SEC. 3
  Approximating the Binomial dist by the Normal
  Think of the integers 0, 1,2, ..,  n  of the Normal as pillars centered at
  the integers.  First translate the prob. problem into shaded area
   First find p the probability of one success in one trial. Let q=1-p  
  Before starting, we need to know if Normal distribution can approximate
  The binomial.  This involves two tests.
  TEST1) np> or =5   AND TEST2) nq> or =5 
  both tests must be satisfied
  if Normal is to be a good approximation to the Binomial problem.

CORRECTION FOR CONTINUITY is needed since Binomial is discrete and Normal
is a continuous distribution.  What is this correction?  
You need to split the difference between neighboring integers
  allocating 0.5 to the left and 0.5 to the right pillar.
  for example if the desired pillars are 1 to 3, 1 begins at 0.5 and 
   3 ends at 3.5
Now use steps 2,3,4 above.  

Remember that the MEAN OF BINOMIAL is np
and VARIANCE IS npq=np(1-p)
don’t forget the square root of npq
These are needed in finding the right variable x ~approx~ N(np, npq)

CENTRAL LIMIT THEOREM
What is central limit theorem?  This is a powerful result by Polya
in 1920's showing that even if x is not Normal, if n>30
the process of averaging (is so helpful) that it
yields Normality of xbar.


NORMAL DENSITY IN Microsoft EXCEL software INSTEAD OF from THE TABLES
FUNCTION NAME IS NORMSDIST
IT HAS 4 INPUTS
The first input is x =the value of x for which we want to compute the cumulative density
Second input is the mean
The third input is the standard deviation
The fourth input is a value  True if you want cumulative probability or False if you want the height of probability curve

NORMSINV function computes the inverse of NORMSDIST
It has 3 inputs
First input is cumulative probability,
Second input is the mean
The third input is the standard deviation