Download Chapter5.3to5.4

Binomial
Distributions
Chapter 5.3 – Probability Distributions and
Predictions
Mathematics of Data Management (Nelson)
MDM 4U
Authors: Gary Greer (with K. Myers)
Our Problem…






suppose students either like math or they don’t
suppose 5% of students like math
if you had 300 students, how likely would it be
that 20 of them liked math?
this can be modeled as a binomial distribution
in statistics it is important in looking at how
likely a situation is to have occurred randomly
if it is very unlikely to have occurred, it lends
support to the significance of a finding
Binomial Experiments

a binomial experiment is any experiment that
has the following properties:






there are n identical trials
there are two possible outcomes for each trial,
termed success and failure
the probability of success is p and the probability
of failure is 1-p
the probabilities remain constant from trial to trial
the trials are independent
repeated trials which are independent and
have 2 possible outcomes (success/failure)
are called Bernoulli Trials
Bernoulli?




Jakob Bernoulli (Basel,
December 27, 1654 August 16, 1705)
Swiss Mathematician
one of the great names in
probability theory
one of a family of great
minds in a variety of
subjects
Binomial Distributions



in a binomial experiment the number of
successes in n repeated Bernoulli Trials is a
discrete random variable (usually called X)
X is termed a binomial random variable and
its probability distribution is called a binomial
distribution
the following formula provides a method of
solving highly complex situations involving
probability
Binomial Probability Distribution


consider a binomial experiment in which there are
n Bernoulli trials, each with a probability of success
of p
the probability of k successes in the n trials is
given by:
n k
nk
P( X  k )    p 1  p 
k 
Example 1





Consider a game where a coin is flipped 5 times.
You win the game if you get exactly 3 heads. What
is the probability of winning?
we will let heads be a success
3
53
n=5
 5  1  
1
P  X  3     1  
p=½
2
 3  2  
k=3
3
2
5
1 1
1
 10      10  
2 2
2
10
5


32 16
Example 1 continued


suppose the game is changed so that you win if
you get at least 3 heads
what is the probability of winning now?
P X  3  P ( X  3)  P( X  4)  P ( X  5)
5  5  1 
    
16  4  2 
5
5
1
  
16 32 32
4
 1   5  1 
     
 2   5  2 
1

2
5
The Batting Example

the Expected Value of a binomial experiment
that consists of n Bernoulli trials with a
probability of success, p, on each trial is




E(X) = n(p)
Example: Consider a baseball player who
has a batting average of 0.292
this means that his probability of getting a hit
each time he is at bat is 0.292
let a hit be a success where p = 0.292
a. What is the probability of no hits in the
next 5 at bats?
p  0.292 , n  5, k  0
5
0
5
P X  0    0.292  0.708 
0
 110.178   0.178
so there is a 0.178 probabilit y that
there will be no hits in 5 times at bat
b. What is the probability of 2 hits in the
next 8 at bats?
p  0.292 , n  8, k  2
8
2
6
P X  2   0.292  0.708 
 2
 28 0.085 0.126   0.300
so there is a 0.300 probabilit y that
there will be 2 hits in 8 times at bat
c. What is the probability of at least 1 hit
in the next 10 at bats?
p  0.292 , n  10
P X  1  1  P X  1  1  P X  0
10 
0
10
 1   0.292  0.708 
0
 1  110.032   0.968
so there is a 0.968 probabilit y that
there will be at least 1 hit in 10 times at bat
d. What is the expected number of hits in
the next 10 at bats?




E(X) = n(p)
E(X) = (10)(0.292)
= 2.92 → 3
therefore the player can expect to get 3 hits
in the next 10 at bats
Exercises / Homework


Homework:
page 299 #1, 3, 7, 8, 9, 10, 11, 12
Normal Approximation of
the Binomial Distribution
Chapter 5.4 – Probability Distributions and
Predictions
Mathematics of Data Management (Nelson)
MDM 4U
Authors: Gary Greer (with K. Myers)
Recall…

the probability of k successes in n trials (where p is
the probability of success) is
n k
nk
P( X  k )    p 1  p 
k 

this formula can only be used if we have a binomial
distribution:


each trial is identical
the outcomes are either success or failure
This calculation is easy in simple cases…

find the probability of 30 heads in 50 trials
P(30 heads in 50 trials ) 
 50  30
 0.5 1  0.55030  0.042
 30 



so there is about a 4.2% chance
however, if we wanted to find out the probability of
tossing between 20 and 30 heads in 50 trial, we
would need to perform at least 10 of these
calculations
there is an easier way however
Graphing the Binomial Distribution



If the distribution is normal, we can solve
complex problems in the same way we did in
the last chapter
the question is: is the binomial distribution
a normal one?
it turns out that if the number of trials is
relatively large, the binomial distribution
approximates a normal curve
What does it look like?


when graphed
the distribution
of probabilities
of head looks
like this
what will the
mean be?
what will the
standard
deviation be?
Line Scatter Plot
0.12
0.10
probability

Binomial Distribution
0.08
0.06
0.04
0.02
0.00
0
5
10 15 20 25 30 35 40 45 50 55
heads
So how do we work with all this

it turns out that a binomial distribution can be
approximated by a normal distribution if:


n(p) > 5 and
n(1 – p) > 5
if this is the case, the distribution is approximated
by the normal distribution
N ( x ,  2 ) where x  np and   np(1  p)
But doesn’t a normal curve represent
continuous data and a binomial
distribution represent discrete data?



Yes!
so to use a normal approximation we have to
consider a range of values rather than
specific discrete values
for example the range of continuous values
between 4.5 and 5.5 can be represented by
the discrete value 5
Example 1



Tossing a coin 50 times, what is the probability that
you will get tails less than 20 times
let success be tails, so n = 50 and p = 0.5
now we can find the mean and the standard
deviation
x  50 (0.5)  25
  50 (0.5)(1  0.5)  12 .5  3.54
Example 1 continued






we will consider 0-19.5 (values below 20)
times, and use it to calculate a z-score
z = 19.5 – 25 = -1.55
3.54
therefore P(X < 19.5) = P(z < -1.55)
= 0.0606
there is a 6% chance of less than 20 tails in
50 attempts
In terms of the normal curve, it looks like
this

all the values less than
19.5 are found in the
shaded area
19.5
25.0
Example 2





Two dice are rolled and the sum recorded 40
times. What is the probability that a sum
greater than 6 occurs in at least half of the
trials?
let p be the probability of getting a sum
greater than 6
p = 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36
p = 7/12
now we can do some calculations
Example 2 continued
7
np  40   23 .3  5
 12 
P( x  20 )  ?
5
n(1  p)  40    16 .6  5
 12 
x  np  23. 3
  np(1 - p)  9.72  3.118
20 .5  23 .3
z
 0.91  0.8186
3.118

the probability of getting a sum greater than 6 on
at least half of the trials is 82%
Example 3



you have a drawer with one blue mitten, one
red mitten, one pink mitten and one green
mitten
if you closed your eyes and picked a mitten at
random 200 times (with replacement) what is
the probability of choosing the pink mitten
between 50 and 60 times?
so, success is considered to be drawing a
pink mitten, with n = 200 and p = 0.25
Example 3 Continued





check to see whether the normal
approximation can be used
np = 200(0.25) = 50
n(1 – p) = 200(0.75) = 150
since both of these are greater than 5 the
binomial distribution can be approximated by
the normal curve
now find the mean and standard deviation
Example 3 Continued
x  np  200 (0.25 )  50
  np1  p   200 (0.25)( 0.75 )  37 .5  6.124
49 .5  50
First Case z 
 0.081  0.4681
6.124
60 .5  50
Second Case z 
 1.715  0.9564
6.124

the probability of having between 50 and 60 pink
mittens drawn is 0.9564 – 0.4681 = 0.4883 or
about 49%
Exercises / Homework


Read the example on page 310
do Page 311 # 4-10