Lecture 4: The binomial distribution - Oxford Statistics · Lecture 4: The binomial distribution...

Lecture 4: The binomial distribution

4th of November 2015

Lecture 4: The binomial distribution 4th of November 2015 1 / 26

Combination and permutation (Recapitulatif)

Consider 7 students applying to a college for 3 places:

Abi Ben Claire Dave Emma Frank Gail

How many ways are there of choosing 3 students from 7 when

1 order is important (permutation)i.e. (Abi, Dave)6= (Dave, Abi)

2 in not important (combination)i.e. (Abi, Dave)= (Dave, Abi)


Permutations

There are 3 places to fill at the college and 7 students applied.

There will be 7× 6× 5 permutations:

7× 6× 5 =7× 6× 5× 4× 3× 2× 1

4× 3× 2× 1=

7!

4!= 7P3 .

In general, the number of permutations of r objects from n is

nPr =n!

(n− r)!.


Combinations

For every combination of 3 students, there will be 3! = 6 permutations.For example, the combination Dave, Claire and Abi gives rise to thepermutations

ACD ADC CAD CDA DAC DCA

If we are not interested in the order, the number of combinations of 3students among 7 is then

7C3 =7P3

3!.

In general, the number of combinations of r objects from n is

nCr =nPr

r!=

n!

(n− r)!r!


An example of the Binomial distribution

An unfair coin: P (Head) = 2/3 and P (Tail) = 1/3

Let X = No. of heads observed in 5 coin tosses

X can take on any of the values 0, 1, 2, 3, 4, 5

X is a discrete random variable

Some values of X will be more likely to occur than others. Each value of Xwill have a probability of occurring. What are these probabilities?


What is P (X = 1)?

One possible way of observing Head once is if we observe the pattern

HTTTT.

The probability of obtaining this pattern is

P(HTTTT) = 23 ×

13 ×

13 ×

13 ×

13


There are 32 possible patterns of Head and Tails we might observe.

HHHHH THHHH HTHHH HHTHH HHHTH HHHHTTTHHH THTHH THHTH THHHT HTTHH HTHTHHTHHT HHTTH HHTHT HHHTT TTTHH TTHTHTTHHT THTTH THTHT THHTT HTTTH HTTHT

HTHTT HHTTT HTTTT THTTT TTHTT TTTHT

TTTTH TTTTT

Five of the patterns contain just one Head.

The other 5 possible combinations all have the same probability so theprobability of obtaining one head in 5 coin tosses is

P(X = 1) = 5×(23 × (13)

4)≈ 0.0412


What about P (X = 2)?

This probability can be written as

P (X = 2) = No. of patterns × Probability of pattern

= 5C2 ×(23

)2×

(13

)3

= 10 × 4

243≈ 0.165

In general, the probability to observe x Head (and 5− x Tail) is

P (X = x) = 5Cx ×(23

)x×(13

)(5−x)


We can use this formula to tabulate the probabilities of each possible valueof X.

P(X = 0) = 5C0 ×(23

)0×(13

)5≈ 0.0041

P(X = 1) = 5C1 ×(23

)1×(13

)4≈ 0.0412

P(X = 2) = 5C2 ×(23

)2×(13

)3≈ 0.1646

P(X = 3) = 5C3 ×(23

)3×(13

)2≈ 0.3292

P(X = 4) = 5C4 ×(23

)4×(13

)1≈ 0.3292

P(X = 5) = 5C5 ×(23

)5×(13

)0≈ 0.1317


Distribution of probabilities across the possible values of X.

0 1 2 3 4 5

0.0

0.1

0.2

0.3

0.4

0.5

X

P(X)

This situation is a specific example of a Binomial distribution.


Key components of the binomial distribution

In general a Binomial distribution arises when we have the following 4conditions:

- n identical trials, e.g. 5 coin tosses

- 2 possible outcomes for each trial “success” and “failure”,e.g. Heads or Tails

- Trials are independent, e.g. each coin toss doesn’taffect the others

- P(“success”) = p is the same for each trial,e.g. P(Head) = 2/3 is the same for each trial


The binomial distribution

If we have the above 4 conditions then if we let

X = No. of “successes”

then the probability of observing x successes out of n trials is given by

P(X = x) = nCx px(1− p)(n−x) x = 0, 1, . . . , n

If the probabilities of X are distributed in this way, we write

X∼Bin(n, p)

n and p are called the parameters of the distribution. We say X follows abinomial distribution with parameters n and p.


Example 1

Suppose X ∼ Bin(10, 0.4), what is P(X = 7)?

Here we have: n = 10, p = 0.4, x = 7,

P(X = 7) = 10C7(0.4)7(1− 0.4)(10−7)

= (120)(0.4)7(0.6)3

≈ 0.0425


Example 2

Suppose Y ∼ Bin(8, 0.15), what is P(Y < 3)?

Here we have: n = 8, p = 0.15,

P(Y < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2)

= 8C0(0.15)0(0.85)8 + 8C1(0.15)

1(0.85)7 + 8C2(0.15)2(0.85)6

≈ 0.2725 + 0.3847 + 0.2376

≈ 0.8948

Note that 1− p = 0.85.


Example 3

Suppose W ∼ Bin(50, 0.12), what is P(W > 2)?

Here we have: n = 50, p = 0.12,

P(W > 2) = P(W = 3) + P(W = 4) + . . .+ P(W = 50)

= 1− P(W ≤ 2)

= 1−(P(W = 0) + P(W = 1) + P(W = 2)

)= 1−

(50C0(0.12)

0(0.88)50 + 50C1(0.12)1(0.88)49

+50C2(0.12)2(0.88)48

)≈ 1−

(0.00168 + 0.01142 + 0.03817

)≈ 0.94874

Note that 1− p = 0.88.


Different values of n and p lead to different distributions with differentshapes:

0 2 4 6 8 10

0.0

0.1

0.2

0.3

0.4

0.5

n=10 p=0.5

X

P(X)

0 2 4 6 8 10

0.0

0.1

0.2

0.3

0.4

0.5

n=10 p=0.1

X

P(X)

0 2 4 6 8 10

0.0

0.1

0.2

0.3

0.4

0.5

n=10 p=0.7

XP(X)


Expected mean and expected standarddeviation

We have seen in the first lecture that the sample mean andstandard deviation can be used to summarize the shape of a dataset.

In the case of a probability distribution we have no data as such so wemust use the probabilities to calculate the expected mean andstandard deviation.


Example: X ∼ Bin(5, 2/3)

Consider the example of the Binomial distribution we saw above

x 0 1 2 3 4 5

P(X = x) 0.004 0.041 0.165 0.329 0.329 0.132

The expected mean value of the distribution, denoted µ can be calculatedas

µ = 0× (0.004) + 1× (0.041) + 2× (0.165) + 3× (0.329)

+4× (0.329) + 5× (0.132)

= 3.333


Expected mean and expected standarddeviation

In general, there is a formula for the mean of a Binomial distribution.There is also a formula for the standard deviation, σ.

If X ∼ Bin(n, p) then

µ = np

σ =√npq where q = 1− p

In the example above, X ∼ Bin(5, 2/3) and so the mean and standarddeviation are given by

µ = np = 5× (2/3) = 3.333

andσ =√npq = 5× (2/3)× (1/3) = 1.111


Testing a hypotheses using the Binomialdistribution – An example

Consider the following simple situation:

You have a six-sided die, and you have the impression that it’s somehowbeen weighted so that the number 1 comes up more frequently than itshould.

How would you decide whether this impression is correct?


You could do a careful experiment, where you roll the die 60 times, andcount how often the 1 comes up.

Suppose you do the experiment, and the 1 comes up 30 times (and othernumbers come up 30 times all together).

If the die is unbiased, you expect the 1 to come up one time in six, i.e. 10times. Therefore 30 times seems high. But is it too high?

There are two possible hypotheses:

1 The die is biased.

2 Just by chance we got more 1’s than expected.

How do we decide between these possibilities?


Perform an hypothesis test.

Hypothesis: The die is fair. All 6 outcomes have the same probability.

Experiment: We roll the die 60 times.

Sample: We obtain 60 outcomes and the 1 comes out 30 times.

Assuming our hypothesis is true the experiment we carried out satisfies theconditions of the Binomial distribution

n identical trials, i.e. 60 die rolls.

2 possible outcomes for each trial: “1” and “not 1”.

Trials are independent.

P(“success”) = 1/6 is the same for each trial


We define X = No. of 1’s that come up.

We observed X = 30.

We can calculate the probability of observing X=30 if our hypothesis istrue, i.e. if X∼Bin(60,1/6):

P (X = 30) = 60C30

(1

6

)30(5

6

)60−30

≈ 2.25× 10−9.

Conclusion:Under the hypothesis that the die is fair, the probability that the numberof 1’s come up 30 times in this experiment is very low. Therefore we mayconclude that the die has been biased.


Hypothesis testing

Now we summarise the general approach:

posit a hypothesis

design and carry out an experiment to collect a sample of data

test to see if the sample is consistent with the hypothesis

Testing the hypothesis:Assuming our hypothesis is true what is the probability that we would haveobserved such a sample or a sample more extreme, i.e. is our sample quiteunlikely to have occurred under the assumptions of our hypothesis?


Example: Drug efficiency

Until recently an average of 60 out 100 patients have survived aparticular severe infection.

When a new drug was administered to 15 patients with the infection,12 of them survived.

Does this provide evidence that the new drug is effective?


Hypothesis: The drug is not effective,i.e. the probability of surviving is still p = 0.6.

Experiment: We test the drug on 15 patients with the infection.

Sample: 12 patients survived.

Let X denote the number of patients who survived.

Under our hypothesis, X∼ Bin(15,0.6)

We compute the probability that we would have observed such a sampleassuming our hypothesis is true:

P (X = 12) = 15C12 (0.6)12 (0.4)15−12 ≈ 0.063.

There is more than 6% chance of observing such a number of survivingpatients if the drug in not effective. Therefore it may be just by chancethat we observe such a number of patients who survived.


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