Probability distributions - Examples Discrete Uniform A number is chosen at random from the natural...

Probability distributions - ExamplesDiscrete Uniform

A number is chosen at random from the natural numbers between 1 and 50.What is the probability that a number over 30 is chosen?What is the expected value of the number chosen and what is the variance of numbers chosen?

X is the number chosenX~DU (50)p(X>30) = 20/50 = 0.4E[X] = 51/2 =25.5Var(X) = 2499/12 = 208.25

Probability distributions - ExamplesBinomial

A marksman fires at a target 10 times, with probability of hitting the target 9/10.What is the probability that he hits the target exactly 8 times?What is the probability that he hits the target more than 6 times?What is the expected value of the number of hits and what is the variance of the number of hits?

X is the number of hits X~B(10, 0.9)p(X=8) = 0.194p(X>6) = 1 – p(X≤6) = 0.987E[X] = 10x0.9 = 9Var(X) = 10x0.9x0.1 = 0.9

Probability distributions - ExamplesGeometric

A marksman fires at a target with probability of hitting the target 9/10.What is the probability that he hits the target for the first time on his second shot?What is the probability that he takes at least 5 shots to hit the target for the first time?What is the expected value of the number of shots before he hits the target and what is the variance of the number of shots before hitting the target?

X is the number of shots before he hits the target X~Geo(0.9)p(X=2) = 0.09p(X≥5) = 1 – p(X≤4) = 0.0001E[X] = 10/9 = 1.11Var(X) = 0.123

Probability distributions - ExamplesPoisson

It is known that on average the number of people visiting a shop in an hour is 17. What is the probability that exactly 30 people visit the shop in a two hour period?What is the probability that less than 50 people visit the shop in the first three hours of opening?What is the expected number of people visiting the shop in an 8 hour day, and what is the variance of the number of people visiting?

X is the number of people visiting in 2 hours X~Po(34)p(X=30) = 0.0568Y is the number of people visiting in 3 hours Y~Po(51)p(Y<50) = p(Y≤49) = 0.426Z is the number of people visiting in 8 hours Z~Po(136)E[Z] = 136Var(Z) = 136

Probability distributions - ExamplesHypergeometric

There are 20 students studying HL maths, of which 14 are girls. If 8 students from the HL classes are chosen at random to visit the maths department of a particular university, what is the probability that exactly 3 girls are chosen?What is the expected number of girls chosen in the group and what is the variance of the number of girls?

X is the number of girls in the groupX~Hyp(8,14,20)p(X=3) = 364x6/125970 = 0.0173E[X] = 5.6Var(X) = 504/475

Probability distributions - ExamplesNegative Binomial

A marksman fires at a target with probability of hitting the target 9/10.What is the probability that he hits the target for the third time on his eighth shot?What is the expected value of the number of shots before he hits the target for the fifth time and what is the variance?

X is the number of shots before he hits the target for the third timeX~NB(3,0.9)p(X=8) = 0.000153Y is the number of shots before he hits the target for the fifth timeY~NB(5,0.9)E[Y] = 50/9Var(Y) = 50/81

Probability distributions - ExamplesContinuous Uniform

A number is chosen at random from the real numbers between 0 and 50.What is the probability that a number over 30 is chosen?What is the expected value of the number chosen and what is the variance of numbers chosen?

X is the number chosenX~U(0,50)p(X>30) = 20/50 = 0.4E[X] = 50/2 =25Var(X) = 2500/12 = 208

Probability distributions - ExamplesNormal

The lengths of worms are normally distributed with mean 7.2cm and standard deviation 2.3cmFind the probability that a worm chosen at random is greater than 8cm

X is the length of the wormX~N(7.2, 2.32)p(X>8) = 0.364

Another species of worm also have their lengths normally distributed and 40% of worms are longer than 8cm and 20% less than 5cm.Find the mean and standard deviation of the lengths of these worms

Y is the length of the worm Y~N(,)p(Y<8) = 0.6, p(Y<5) = 0.28 – = 0.2533475 – = -0.841621 = 7.31 and = 2.74

Remember Z = (Y – )/So (Y – )/ = (0.6)

Probability distributions - ExamplesExponential

It is known that on average the number of people visiting a shop in an hour is 18.

If there is no-one in the shop at present, find the probability that it is more than 10 minutes before someone comes into the shop.Find the expected amount of time before someone comes into the shop and the variance of that time.

X is the amount of time before someone comes into the shop X~Exp(0.3)p(X>10) = 0.0498E[X] = 10/3Var(X) = 100/9

10

10

30

0.498

x

x

e dx

e

e

10

xe dx 10

10

x

x

e dx

e

10

10

30

x

x

e dx

e

e

Expectation AlgebraThree coins are tossed and the number of heads is recorded

X is the number of headsX ~ B(3, 0.5)

Two dice are thrown and the number of sixes recordedY is the number of sixesY ~ B(2, 1/6)

x 0 1 2 3

P(X = x)

y 0 1 2

P(Y = y)

x 0 1 2 3

P(X = x) 1/8 3/8 3/8 1/8

y 0 1 2

P(Y = y) 25/36 10/36 1/36

x 0 1 2 3

P(X = x) 1/8 3/8 3/8 1/8

y 0 1 2

P(Y = y) 25/36 10/36 1/36

E[X] = 3/2 and Var(X) = 3/4

E[Y] = 1/3 and Var(Y) = 5/18

Now what if I want to consider the total number of heads and sixes in the two experiments

What will be the expectation and variance for the total?

Remember E[X] = npAnd Var(X) = npq

x 0 1 2 3

P(X = x) 1/8 3/8 3/8 1/8

y 0 1 2

P(Y = y) 25/36 10/36 1/36

P(Z = 0) = (1/8)(25/36) = 25/288

z = x + y

0 1 2 3 4 5

P(Z = z)

z = x + y

0 1 2 3 4 5

P(Z = z) 25/288

P(Z = 1) = (3/8)(25/36)+(1/8)(10/36)=85/288

z = x + y

0 1 2 3 4 5

P(Z = z) 25/288 85/288

P(Z = 3) = (1/8)(25/36)+(3/8)(10/36)+(3/8)(1/36)

=58/288

z = x + y

0 1 2 3 4 5

P(Z = z) 25/288 85/288 106/288

P(Z = 2) = (3/8)(25/36)+(3/8)(10/36)+(1/8)(1/36)

=106/288

z = x + y

0 1 2 3 4 5

P(Z = z) 25/288 85/288 106/288 58/288

P(Z = 4) = (1/8)(10/36)+(3/8)(1/36) = 13/288P(Z = 5) = (1/8)(1/36) = 1/288

z = x + y

0 1 2 3 4 5

P(Z = z) 25/288 85/288 106/288 58/288 13/288

z = x + y

0 1 2 3 4 5

P(Z = z) 25/288 85/288 106/288 58/288 13/288 1/288

So E[X + Y] = So E[X + Y] = 11/6and Var(X + Y) = 37/36

= E[X] + E[Y]= Var(X) + Var(Y)

In general:E[X + Y] = E[X] + E[Y]Var(X + Y) = Var(X) + Var(Y)

However more generallyE[aX ± bY] = aE[X] ± bE[Y]

Var(aX ± bY) = a2Var(X) + b2Var(Y)

Note only +

Example 1:The weights of apples are normally distributed with mean 380g and

standard deviation 45g and the weights of bananas are normally distributed with mean 240g and standard deviation 25g.

A bag contains three apples and 5 bananas. What is the probability that the bag weighs more than 2.5kg?

Solution:W = A1 + A2 + A3 + B1 + B2 + B3 + B4 + B5

and Ak~N(380,452), Bk~N(240,252)

E[W] = E[A1 + A2 + A3 + B1 + B2 + B3 + B4 + B5]

= E[A1] + E[A2] + E[A3] + E[B1] + E[B2] + E[B3] + E[B4] + E[B5]

=3 x 380 + 5 x 240 = 2340Var(W) = Var(A1 + A2 + A3 + B1 + B2 + B3 + B4 + B5)

= Var(A1) + Var(A2) + Var(A3) + Var(B1) + Var(B2) + Var(B3) + Var(B4) + Var(B5)

=3 x 452 + 5 x 252 = 9200So W~N(2340, 9200)P(W>2500) = 0.0476



Apples cost 3 pesos per gram and bananas cost 5 pesos per gramIf I buy an apple and a banana at random, what is the probability I

spend more than 2500 pesos?Solution:C = 3A + 5B

and A~N(380,452), B~N(240,252)E[C] = E[3A + 5B]= 3E[A] + 5E[B]=3 x 380 + 5 x 240 = 2340Var(C) = Var(3A + 5B)= 9Var(A) + 25Var(B)

=9 x 452 + 25 x 252 = 33850So W~N(2340, 33850)P(W>2500) = 0.192

Compare this to example 1



If I buy an apple and a banana at random, what is the probability that the apple is more than twice as heavy as the banana?

Solution:D = A - 2B

and A~N(380,452), B~N(240,252)E[D] = E[A - 2B]= E[A] - 2E[B]=380 – 2 x 240 = -100Var(D) = Var(A - 2B)= Var(A) + 4Var(B)

=452 + 4 x 252 = 4525So D~N(-100, 4525)P(D>0) = 0.0686

The distribution of sample means

When making inferences about a POPULATION we usually have to do so from the data provided by a SAMPLE.

For example if I want to estimate the number of children per family in Switzerland, I may take a sample of 50 families to make my estimate.

Suppose the mean of my sample is 2.6 children per familyI can ESTIMATE the mean of the population (all families in Switzerland) to be 2.6 children per family

If someone else did the same thing, they would very likely find a different estimate for the mean number of children per family

If many people did the same thing, we would have many different sample means.

The CENTRAL LIMIT THEOREM states thatthe distribution of sample means of size n is approximately

Normal

with mean and standard deviation

where and are the mean and standard deviation of the population

n

The distribution of sample means

So

The Central Limit Theorem is exact if the sample is from a Normally distributed population, and if it not a Normally distributed population, the Central Limit Theorem still applies as an approximation – the larger the value of n, the better the approximation.

We can see that as n increases, so the standard deviation decreases.

2

~ ,X Nn

Using the Central Limit theorem

Example 1On a certain beach, turtles go each year to lay their eggs. The number of eggs in a nest is known to be normally distributed with

mean 114.78 and standard deviation 22.4(a)Calculate the probability that the mean number of eggs in a sample

of 10 nests is greater than 120(b)If a sample of 5 nests is taken, calculate the value of a, such that the

probability of the mean of the sample is within a of the mean is 90%

Solution(a)

(b)

222.4~ 114.78,

10X N

120 0.231p X

222.4~ 114.78,

5X N

114.78 114.78 0.9p a X a

114.78 0.95p X a

0.9522.4/ 5

ap Z

1.64

22.4/ 5a

16.6a

The Proportion of success in a large sample

An egg manufacturer claims that eggs delivered to a supermarket have no more than 5% that are broken. On a certain day, of 1000 eggs delivered to the supermarket, 80 are found to be broken. What is the probability of this happening? Briefly comment on the manufacturer’s claim

Method 1 – exactX are the number of broken eggs in the deliveryX ~ B(1000, 0.05)p(X ≥ 80) = 3.49 × 10-5

It seems very unlikely that the manufacturer’s claim is correct

The Proportion of success in a large sampleMethod 2 – Using the central limit theoremX are the number of broken eggs in the deliveryX ~ B(1000, 0.05) so E[X] = 50 and Var(X) = 47.5

Now if

then

and

So

if n is large (by the Central Limit Theorem)

Xp

n

1 1XE p E E X np p

n n n

2 2

1 1X pqVar p Var Var X npq

n n n n

~ ,pq

p N pn


So in this example

So

It still seems very unlikely that the manufacturer’s claim is correct.

Note that the answer is a little inaccurate due to the approximation of the CLT

~ 0.05,0.0000475p N

60.08 6.72 10 p p


This can also be found on the GDC using the 1-proportion Z testOn the STAT menu find 1-PropZTestSet

and Calculate

The Proportion of success in a large sampleAnother ExampleA drug company claims that a new drug cures 75 % of patients suffering

from a certain disease. However, a medical committee believes that less than 75 % are cured. To test the drug company’s claim, a trial is carried out in which 100 patients suffering from the disease are given the new drug. It is found that 68 of these patients are cured. Does the medical committee have the evidence it requires to refute the drug company’s claim?

SolutionWe assume 75% of patients are cured by the drug

Method 1Number of patients cured is X

X ~ B(100, 0.75)p(X ≤ 68) = 0.0693



SolutionAssuming 75% of patients are cured by the drug

Method 2 ~ 0.75,0.001875p N

0.68 0.0530p p



SolutionAssuming 75% of patients are cured by the drug

Method 3

From GDC using 1-PropZTestp = 0.0530

Hypothesis TestingIn the previous example, what could you conclude?Was the drug company’s claim valid?Would the medical committee have sufficient evidence

for a lawsuit?

We need to develop a more formal approach to testing the hypothesis

First we write the hypothesis in a formal way.We say that the NULL HYPOTHESIS is that the

company’s claim is correct – i.e. that 75% of patients are cured

Also the ALTERNATIVE HYPOTHESIS is that LESS than 75% are cured

Note that we are not interested in this case if MORE than 75% are cured

This is called a 1-TAIL TEST

Hypothesis TestingWe write: H0: 75% of patients cured by the drug

H1: Less than 75% of patients cured by the drug

OR H0: p=0.75

H1: p<0.75

Hypothesis TestingH0: p=0.75

H1: p<0.75

Now we can find the probability of the result found as before (by any of the methods)

Next we decide on a SIGNIFICANCE LEVELIf we need to be very sure (e.g. to take out a lawsuit)

then we need a low significance level. A higher level if it is not so important

Note that you will usually be given the significance level

Usually it is 10%, 5%, or 1%

~ 0.75,0.001875p N

0.68 0.0530p p

Hypothesis TestingH0: p=0.75

H1: p<0.75

So if we test at the 10% level (probably not sure enough for a lawsuit)Then as 5.3% < 10%, we would reject H0 in favour of

H1. The company’s claim does not seem reasonable

On the other hand if we test at the 1% levelThen as 5.3% > 1%, there is insufficient evidence to

reject H0 and we accept the company’s claim that 75% of patients are cured by the drug

0.68 0.0530p p

Hypothesis TestingIn the previous example the standard deviation was known, as the distribution of number of patients etc. was Binomial. In other cases, the standard deviation may be known (by other sources) or estimated from the sample.

These cases give rise to different types of hypothesis testing

Hypothesis TestingFor example

A sample of 10 crates of items is taken from a factory which states that the mean number of defective items per crate is 12. The sample mean of the ten crates is found to be 13. Knowing that the standard deviation of the number of defective items is 1.6, test the hypothesis that the mean is more than 12:

(a) at the 10% level of significance

(b) at the 1% level of significance

Hypothesis TestingH0: = 12

H1: > 12

by the central limit theorem

(a) so testing at the 10% level of significance we see that 2.5%<10% so we reject H0 in favour of H1, the mean number of defective items is greater than 12

(b) testing at the 1% level of significance we see that 2.5%>1% so there is insufficient evidence to reject H0. We accept that the manufacturer’s claim that the mean number of defective items is 12

21.6~ 12,

10X N

13 0.0241p X

Hypothesis TestingOr!

H0: = 12

H1: > 12



13 0.0241p X

Hypothesis TestingAnother example

A sample of 5 crates of items is taken from a factory which states that the mean number of defective items per crate is 12. The sample is 12, 13, 11, 14, 13. Knowing that the standard deviation of the number of defective items is 1.6, test the hypothesis at the 5% level of significance

SolutionH0: = 12 H1: > 12

Sample mean is 12.6 (from GDC)

20%>5% so there is insufficient evidence to reject H0. We accept that the mean number of defective items is 12

21.6~ 12,

5X N

12.6 0.201p X

Or from GDC using Z-Test

Hypothesis TestingNow if the standard deviation of the population is NOT known, and has to be estimated from the sample, then the distribution is altered. Rather than being Normal, the distribution follows that of a Student-t distribution.

The standardised Student-t distribution has parameter , known as the degrees of freedom

The degrees of freedom is generally n - 1

Hypothesis TestingFor example

A sample of 10 crates of items is taken from a factory which states that the mean number of defective items per crate is 12. The sample mean of the ten crates is found to be 13 and the estimate of the standard deviation of the number of defective items is 1.6, test the hypothesis that the mean is more than 12:

(a) at the 10% level of significance

(b) at the 1% level of significance

Hypothesis TestingH0: = 12

H1: > 12



~ 9

10

XT t

S

1013 0.0398

1.6p X p T

Note the degrees of freedom is 9and S is the estimate of the population standard deviation

13 12 10

1.6 1.6

10

Or use T-Test on the GDC

Hypothesis TestingError Types

Falsely rejecting H0, is known as a Type I error

Falsely accepting H0, is known as a Type II error

So in the example used previously:

..testing at the 10% level of significance we see that 2.5%<10% so we reject H0 in favour of H1, the mean number of defective items is greater than 12.

There is a 2.5% risk of committing a type I error

Confidence intervals

A confidence interval is an interval in which we are certain to a given probability of the mean lying in that interval.For example if a 95% confidence interval is given, then there would be a 95% probability that the mean is in that interval.

Example 1

On a given busy day, 1000 eggs are delivered to a supermarket and 80 are broken. Find the 95% confidence interval for the mean percentage of broken eggs.

Confidence intervalsExample 1On a given busy day, 1000 eggs are delivered to a supermarket and 80 are broken. Find the 95% confidence interval for the mean percentage of broken eggs.Method 1As before

We can estimate p as 0.08 so Now if Then

so and the confidence interval is [6.32%, 9.68 %]

~ ,pq

p N pn

~ 0.08,0.0000736p N 0.08 0.08 0.95 p a p a

0.08 0.975 p p a

0.0168a

Confidence intervalsExample 1On a given busy day, 1000 eggs are delivered to a supermarket and 7% are broken. Find the 95% confidence interval for the mean percentage of broken eggs.Method 2From formula book

For a 95% confidence interval we take z=InvNorm(0.975)=1.96 So applying the formula we find the confidence interval to be [6.32%, 9.68%]Method 3From GDC using 1-PropZInt, the interval is [6.32%, 9.68%]

Confidence intervalsExample 2

To test a drug company’s claim, a trial is carried out in which 100 patients suffering from a disease are given a new drug. It is found that 68 of these patients are cured. Find a 90% confidence interval for the proportion of patients cured by the drug.

Method 1

a = 0.0767

So the confidence interval is [0.603, 0.757]

Or either of the other two methods!

~ 0.68,0.002176p N

0.68 0.68 0.90 0.68 0.95p a p a p p a

Example 3A sample of 10 crates of items is taken from a factory and the mean of the ten crates is found to be 13. Knowing that the standard deviation of the number of defective items is 1.6, find a 95% confidence interval for the mean number of defective items in a crate.

Solution

So the confidence interval is [12.01, 13.99]•Or use the formula in the formula book•Or use ZInterval on the GDC

21.6~ 13,

10X N

13 0.975p X a

0.99167a =10

0.9751.6

ap Z

Confidence intervals

Again if the standard deviation of the population is NOT known, and has to be estimated from the sample, then the distribution is altered to that of a Student-t distribution.

Example 4A sample of 10 crates of items is taken from a factory and the sample mean of the ten crates is found to be 13 and the estimate of the standard deviation of the number of defective items is 1.6. Find a 90% confidence interval for the mean number of defective items.

SolutionAs before

so a = 0.927 (1.833 × 1.6 / 10)(from tables or solver)

So the confidence interval is [12.07, 13.93]• Or use the formula in the formula book• Or use TInterval on the GDC

( )13

~ 91.610

XT t

-=

100.95

1.6a

p Tæ ö÷ç ÷< =ç ÷ç ÷çè ø

( )13 0.95p X a< + =

The 2– Goodness of fit testA dice is thrown 60 times and the following frequencies are obtainedScore 1 2 3 4 5 6Frequency 12 13 10 11 11 3

Is there evidence to suggest that the dice is biased?How can we calculate the probability of results as extreme as these in order to test a hypothesis?

We use the fact that the quantity approximately follows a 2 probability distribution

2Observed Frequency Expected frequency

Expected Frequency

Note that if any of the expected frequencies are less than 5, then the chi-squared distribution may not be a reasonable approximation

The 2– Goodness of fit testSo following standard hypothesis testing proceduresH0: the dice is fair – i.e. X~UD(6) where X is the score

H1: the dice is biased – i.e. the distribution is not uniform

So

taking = 5

So testing at the 5%

significance level, 27%>5% so there is

insuficientevidence to reject H0. We

acceptthat the dice is not biased.

Observed frequency

Expected Frequency

1 12

2 13

3 10

4 11

5 11

6 3

Observed frequency

Expected Frequency

1 12 10

2 13 10

3 10 10

4 11 10

5 11 10

6 3 10

22 o ecalc

e

f f

f

2 6.4 0.269calcp

22 6.4o ecalc

e

f f

f

Also this can be done on a TI84

More on Degrees of freedom

= no of classes – no of restrictions – no of estimated parameters

So often = n – 1 as there is a restriction on the total frequency (as in the previous example)

Another example

Data is collected as below:

Test at the 5% significance level to see if the data is from a Poisson distribution..

H0: X~Po() H1: X is not Poisson

Estimate from the data to be 1.66 = 6 – 1 – 1 = 4Expected frequencies are 9.51, 15.78, 13.10, 7.25, 3.01, 1.36

x 0 1 2 3 4 5

frequency

10 16 12 7 3 2

Note that last expected frequency is that of ≥5

Now we notice that the last two expected frequencies are less than 5 and so will have to be grouped together

Regrouping the table

So we change = 4 – 1 – 1 = 2

So 94%>5% so there is insufficient evidence to reject H0. We accept that the data is distributed by a Poisson distribution.

x 0 1 2 ≥ 3

Observed frequency

10 16 12 12

Expected frequency

9.51 15.78 13.10 11.61

2 0.134calc

( )2 0.134 0.935calcp c > =

The 2– Independence test100 Tennis players are studies to see if their style is dependent on their nationalityThe results were as follows

Is there evidence to suggest that tennis style is dependent on nationality?Again we use the fact that the quantity

approximately follows a 2 probability distribution

2Observed Frequency Expected frequency

Expected Frequency

Eastern Europe

Western Europe

Americas

Serve/Volley 8 15 5

Baseline 11 38 23

The 2– Independence testThe expected frequencies can be calculated

Now, as before following procedure for significance testingH0: Style is independent of nationality

H1: Style depends on nationality = (row – 1)(col – 1) = 2

Testing at the 10% significance level, 19%>10% so there is insufficient evidence to reject H0. We accept that tennis style is independent of nationality.

Eastern Europe

Western Europe

Americas Total

Serve/Volley 8 15 5 28

Baseline 11 38 23 72

19 53 28 100

Eastern Europe

Western Europe

Americas Total

Serve/Volley 28*19/100=5.32 28*53/100=14.84

28*28/100=7.84 28

Baseline 72*19/100=13.68

72*53/100=38.16

72*28/100=20.16

72

19 53 28

2 3.306calc 2 3.306 0.191calcp

Or we can use 2-Test on the GDC

The 2– Independence testPut in the observed frequencies into matrix A

H0: Style is independent of nationality

H1: Style depends on nationality(with df=2)

(and expected frequencies are in matrix B if required)Testing at the 10% significance level, 19%>10% so there is insufficient evidence to reject H0. We accept that tennis style is independent of nationality.

2 3.306calc 2 3.306 0.191calcp

Probability distributions - Examples Discrete Uniform A number is chosen at random from the natural...

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Transcript of Probability distributions - Examples Discrete Uniform A number is chosen at random from the natural...