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Stat 201 Handout -- updated version
(June 4, 2003)
More on Coin Tossing and the Normal distribution
Review: Examples from class
1. Toss coin 5 times. Random variable of interest is X=# heads.
Computed and graphically displayed probability distribution for X.
Demonstrated idea of P(X=x) being equal to the area associated with x under the probability
distribution histogram.
Introduced idea of discrete r.v.'s and continuous r.v.'s.
2. Randomly guess answers on exam. R.v. of interest is X=exam mark out of 100. Know X has normal
distribution with mean 21.7 and SD 8.49.
Discussed P(X=50)=0. (Similarly for P(X=x) for any single value of x.)
Computed P(pass)=P(X >= 50):
Need to standardize X (change scale to Z-scale):
--------------|---------|------------------|////////////->
x
21.7
30.19
50
--------------|---------|------------------|////////////->
z
0
1
50-21.7
--------8.49
P(X>=50)=P( (X-21.7)/8.49 >= (50-21.7)/8.49 ) = P(Z
>=3.33)
Use normal table on pp. 538-539 to find areas under the normal curve:
areas listed are left tails -- correspond to P(Z<=z) for given values of z
Read off P(Z<=3.33)=P(Z<3.33)=0.9996. So P(X>=50)=P(Z>=3.33)=1-P
(Z<3.33)=1-0.9996=0.0004.
Binomial Distribution
For Example 1 above, we have a special name for the distribution of X. We say "X has a binomial distribution
with n=5 and p=0.5." We write X~Bin(n=5,p=0.5).
Definition
A binomial distribution results from a random operation involving n independent success-failure
trials with a common probability of success p. The total number of successes has a Bin(n,p)
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distribution.
More examples of binomial r.v.'s
a. For any given beef-cow, there is a 0.2 probability for it to have BSE (mad cow disease). Your herd has 38
animals. X is the total number of cows in your herd that have BSE. So X~Bin(n=38,p=0.2).
b. From Example a: take Y to be the total number of healthy animals in your herd. So Y~Bin(n=38,p=0.8).
c. I toss a coin 5 times. You toss a coin 10 times. Assume you and I toss the same coin in the exact same
way so that our tosses are entirely independent. Let X=total H for me, Y=total H for you. Then X+Y~Bin
(n=15,p=0.5). (So X+Y is just like the total # heads if you were to toss this coin 15 times.)
d. I toss a penny 5 times. You toss a two-nie 10 times. I throw the coin in the air and let it land. You spin the
coin then smack it flat. Although my trials and your trials have a common probability of successs (i.e.
0.5), X+Y is unlikely to have a Bin(15,0.5) distribution --- my 5 tosses are dependent (through my way of
tossing), and your 10 tosses are dependent (through your way of tossing).
FACT
If X~Bin(n,p), then E(X)=np (n times p) and V(X)=np(1-p).
Using this fact, it'll save you some trouble of using the formulas E(X)=sum of {x P(X=x)} and V(X)=sum of
{(x-E(X))
2
P(X=x)}.
Linear Combinations of R.V's
Adding and/or multiplying constants to r.v.'s gives a linear combination of r.v.'s.
From Example c: Suppose I'm told to double the amount of tosses, and you're told to add 5 heads to your total
count (out of 10 tosses). Then the total number of heads is W=2X+Y+5. This is a linear combination of X and Y.
FACTS
For any two r.v.'s X and Y:
E(aX+b)=a E(X)+b where a,b are any non-random numbers
e.g. E(2X)=2 E(X)
E(X+Y)=E(X)+E(Y)
SD(aX+b)=a SD(X) where a,b are any non-random numbers
e.g. SD(Y+5)=SD(Y)
V(X+Y)=V(X-Y)=V(X)+V(Y) if X,Y are independent
From Example c: What is the total number of heads (W) you expect to see, and give or take how many?
E(W)=E(2X+Y+5)=2 E(X)+E(Y)+5 = 2 * (5*0.5) + (10*0.5) + 5 = 15
If we assume your tosses and mine are independent, then V(W)=V(2X+Y+5)= 4 V(X)+V(Y) = 4 *
(5*0.5*0.5) + (10*0.5*0.5) = 7.5. So SD(W)=2.74.
So we expect to see 15 heads, give or take 2.74 (i.e. between 2 and 3).
Q: What if the tosses are dependent due to the different ways we toss the coin?
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A: We can't just add the variances, but you won't need to know what to do in Stat 201.
FACT
Linear combinations of normal random variables always have normal distributions!
From Example 2: Suppose you guess all your answers on both your midterm and final exams. Both marks have
a normal distribution. For the midterm, the distribution has mean 21.7 and SD 8.49 (as above). For the final, the
mean and SD are 20.2 and 6.87. Your course grade is the total of those exams marks (altogether out of 200).
What is the probability that you pass the course?
Let X=midterm mark, Y=final exam mark.
P(pass course)=P(X+Y>=100). We know X+Y has normal distribution (because both X and Y are
normal).
Now we need to standardize the quantities so we can look up the normal table on pp. 538-9.
E(X+Y)=E(X)+E(Y)=21.7+20.2=41.9
V(X+Y)=??
Are X and Y independent? -- Here, we can assume so, as both times you are randomly guessing the
answers (i.e. your final exam performance is unaffected by what mark you got previously).
So V(X+Y)=V(X)+V(Y)= 8.492+6.872. So SD(X+Y)=10.92.
--------------|---------|------------------|////////////-> x+y
41.9
52.82
100
--------------|---------|------------------|////////////-> z
0
1
100-41.9
---------10.92
P(X+Y>=100)=P( (X+Y-41.9)/10.92 >= (100-41.9)/10.92 ) = P(Z >=5.32).
This probability is practically 0 (infinitesimal area for the normal right tail beyond 5 SD's from the
mean)!
Percentile calculations for the Normal Distribution
For a normal r.v. W with mean 41.9 and SD 10.92, we write W~N(mean=41.9,SD=10.92).
My Stat 201 class has about 200 students. If all of you randomly guess your answers on the two exams, then I
only expect 0.0004 of you, or 0.04% of you, or fewer than one to pass the course! Suppose I want to scale the
marks so that the top 5% of you (i.e. top 10 students in the course) will get a pass. But I must now report to the
Department what will be a passing mark for my course.
Q: What mark should I assign as the passing mark?
A: This is a percentile calculation.
Let w* be the passing mark I'll assign.
--------------|---------|-----------|////////////-> w
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41.9
52.82
w*
--------------|---------|-----------|////////////-> z
0
1
w*-41.9
--------10.92
I want the shaded part of the axis (beyond the passing mark w*) to correspond to 5% of the normal
distribution.
From the normal table on pp. 538-9, a left tail area of 95% corresponds to a z score of between 1.64
and 1.65. So let's say I'll use 1.645 as my z score. That is, P(Z<=1.645)=0.95. So P(Z>=1.645)
=0.05. So 1.645 is the standardized 95th percentile of the marks distribution. Of course, the z-scale
is no good for me. I need to covert it back to the w-scale:
0.05=P(W>=w*)=P( (W-41.9)/10.92 >= (w*-41.9)/10.92) )=P(Z>=1.645)
So I want
w*-41.9
--------- = 1.645
10.92
So w*=59.86.
Thus, anyone whose final grade is at least 59.86 (out of 200) will pass the course.
This is, of course, hypothetical!!!!
Normal Approximations for the Binomial Distribution
Recall in Example 1 I mentioned the possibility of using the 68-95 rule to find probabilities for the coin-tossing
experiment. In general, normal approximations work well for binomial distributions whose mean (=np) and
variance (=np(1-p)) are at least 10.
From Example c, W=2X+Y+5, E(W)=15, SD(W)=2.74. Here, the normal approximations should work
reasonably well.
Q: What is the probability that the total number of heads exceeds 20?
A: Two ways: based on W-5~Bin(n=20, p=0.5) or W approximately ~ N(mean=15, SD=2.74).
The first way is clumsy -- we need to (1) constantly remind ourselves about the extra 5 heads
tagged on to the final count, and (2) find and sum all probabilities from W=20, W=21, ..., W=25.
The second is much easier, once you've mastered computing normal probabilities:
P(W>20) ~= P( Z> (20-15)/2.74 ) = ...
http://www.stat.sfu.ca/~sgchiu/Grace/S201/Handouts/h3.html
3/26/2011