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Binomial Distributions
Chapter 5.3 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U
Our Problem… Students either like math or they don’t Suppose 5% of students like math 300 students, how likely is it that 20 like math? Can be modeled as a binomial distribution How likely it is that a situation has occurred
randomly? More significant if it is unlikely to have occurred
randomly
Binomial Experiments a binomial experiment is any experiment that
has the following properties: n identical trials two possible outcomes for each trial, termed
success and failure the probability of success is p the probability of failure is 1-p the probabilities remain constant the trials are independent
Bernoulli Trials - repeated independent trials with 2 possible outcomes (success/failure)
Bernoulli?
Jakob Bernoulli (Basel, December 27, 1654 - August 16, 1705)
Swiss Mathematician one of the great names in
probability theory
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:
knk ppkn
kXP
1)(
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 n = 5 p = ½ k = 3
165
3210
2110
21
2110
211
21
35
3
523
353
XP
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?
21
321
325
165
21
21
55
21
21
45
165
)5()4()3(3504
XPXPXPXP
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 lifetime 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?
batat times5in hits no be willtherey thatprobabilit 0.178 a is thereso
178.0178.011
708.0292.005
0
0,5,292.0
50
XP
knp
b. What is the probability of 2 hits in the next 8 at bats?
batat times8in hits 2 be willtherey thatprobabilit 0.300 a is thereso
300.0126.0085.028
708.0292.028
2
2,8,292.0
62
XP
knp
c. What is the probability of at least 1 hit in the next 10 at bats?
batat times10in hit 1least at be willtherey thatprobabilit 0.968 a is thereso
968.0032.0111
708.0292.00
101
0111110,292.0
100
XPXPXPnp
d. What is the expected number of hits in the next 10 at bats? E(X) = np E(X) = (10)(0.292) = 2.92 → 3 therefore the player can expect to get 3 hits
in the next 10 at bats
MSIP/ Homework
p. 299 #1, 3, 7, 8, 9, 10, 11, 12
Normal Approximation of the Binomial Distribution
Chapter 5.4 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U
Recall… the probability of k successes in n trials (where p is
the probability of success) is
this formula can only be used if we have a binomial distribution: each trial is identical the outcomes are either success or failure
knk ppkn
kXP
1)(
This calculation is easy in simple cases… Find the probability of 30 heads in 50 trials
So there is about a 4.2% chance However, if we wanted to find the probability of
tossing between 20 and 30 heads in 50 trials, we would need to perform 11 of these calculations
But…there is an easier way
042.05.015.03050
) trials50in heads 30(
305030
P
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 heads looks like this
what will the mean be?
what will the standard deviation be?
prob
abili
ty
0.00
0.02
0.04
0.06
0.08
0.10
0.12
heads0 5 10 15 20 25 30 35 40 45 50 55
Binomial Distribution Line Scatter Plot
So how do we work with all this it turns out that a binomial distribution can be
approximated by a normal distribution if: n x p > 5 and n x (1 – p) > 5
if this is the case, the distribution is approximated by the normal distribution
)1( and where),( 2 pnpnpxxN
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
The interval for a value is from 0.5 below to 0.5 above, i.e., the interval for 10 goes from 9.5 to 10.5
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 n x p = 50(0.5) = 25 > 5 n x (1 - p) = 50(1 - 0.5) = 25 > 5 now we can find the mean and the standard deviation
54.35.12)5.01)(5.0(50
25)5.0(50
x
Example 1 continued we will consider 0-19.5 times (values below
20 - the interval from 19.5-20.5), 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 over 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
the probability of getting a sum greater than 6 on at least half of the trials is 100 – 18 = 82%
7 540 23.3 5 (1 ) 40 16.6 512 12
( 20) ? x np 23.3
np(1-p) 9.72 3.118
20.5 23.3 0.91 0.1814 18.14%3.118
np n p
P x
z or
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 (inclusive)?
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
the probability of having between 50 and 60 pink mittens (inclusive) drawn is 0.9564 – 0.4681 = 0.4883 or about 49%
9564.0715.1124.6
505.60 Case
4681.0081.0124.6
505.49 Case
124.65.37)75.0)(25.0(2001
50)25.0(200
zSecond
zFirst
pnp
npx
MSIP/ Homework
Read the example on page 310 do p. 311 # 4-10