Section 5.1 Review

The Binomial Setting

1) There are a fixed number n of observations

2) The n observations are independent

3) Each observation falls into just one of two categories, which, for convenience, we call “success” and “failure”.

4) The probability of a success, call it p, is the same for each observation

Binomial Distributions

The distribution of the count X of successes in the binomialsetting is called the binomial distribution with parametersn and p.

The parameter n is the number of observations

The parameter p is is the probability of a success on anyone observation.

The possible values of X are the whole numbers from 0to n.

We say that X is B(n,p)

Binomial Probability Tables(Table C, page T-7 to T-10)

Consider the previous transistor problem using the B(10,.1)distribution.

p

n K 0.10

10 0 .34871 .38742 .19373 .05744 .01125 .00156 .00017 .00008 .00009 .000010 .0000

Q: What is the probability that wewill pull an SRS from this distributionwith exactly one bad transistor?

A: P(X = 1) =0.3874

Binomial Mean and Standard Deviation

Q: If a count X is B(n,p), what are the mean X and the

standard deviation X ?

A: If a count X has the B(n,p) distribution, then :

X = np

X = np (1-p)

Sample ProportionsIn statistical sampling, we often want to estimate the proportion p of “successes” in a population.

Our estimator is the sample proportion of successes :

p =Count of successes in sample

Size of sample

X

n=

Mean and Standard Deviation of a Sample Proportion

Let be the sample proportion of successes in an SRSof size n drawn from a large population having populationproportion p of successes.

p

The mean and standard deviation of are the following :p

= pp

p

= p(1-p)

n

Normal Approximations for Counts and Proportions

• Draw an SRS of size n from a large population having population proportion p of successes.

• Let X be the count of successes in the sample, and = X / n be the sample proportion of successes.p

• When n is large, the sampling distributions of these statistics are approximately normal.

• X is approximately N( np, np(1-p) )

• is approximately Np ( p, ) p(1-p)

n

Note: Use this when np 10 and n(1-p) 10

End O’ Review

The Sampling Distribution of a Sample Mean

Recall :

1) Counts and proportions are discrete random variables that describe categorical data.

2) Measured data is usually described with continuous random variables.

The Sampling Distribution of a Sample MeanRecall :

3) Averages are less variable than individual observations

The Sampling Distribution of a Sample MeanRecall :

4) Averages are more normal than individual observations

1obs.

2obs.

10obs.

25obs.

Sample Mean

Q: What is a sample mean?

A: This is an estimate of the mean of the underlying population

Q: How do we find the sample mean?

A: 1) Select an SRS of size n from a population

2) Measure a variable X on each individual in the sample

3) Let each observation be labeled X1, X2, … , Xn

4) The sample mean is the average of the Xi’s

Note: If the sample is large, each Xi can be thought of as an independent random variable

Mean and Standard Deviation of a Sample Mean

Let be the mean of an SRS of size n from a populationhaving mean and standard deviation . The mean andstandard deviation of are:

x

x

= x

x = n

= = x

Mean and Standard Deviation of a Sample Mean

Example: The height of a randomly chosen Jedi varies according to the N(73, 2.8) distribution. If Yoda askedthe height of an SRS of 100 Jedi, what is the mean andheight of this sampling distribution?

73 inches

x = n

= 2.8

100

2.8

10= = 0.28

If a population has the N(, ) distribution, then the sample mean has the N( ) distribution.

The Sampling Distribution of x

Q: What is the shape of the distribution?

A: It depends on the population. So, if the population distribution is normal, then so is the distribution of the sample mean.

This leads to the following:

, / n

In the previous example, the sample mean of the heights of 100 Jedi has the N(73, 0.28) distribution.

The Sampling Distribution of x

• So, the sample mean from a normal distribution is normal.

• We can extend these ideas to the following :

Any linear combination of independent normal random variables is also normally distributed.

• In other words, if X and Y are independent normal random variables, then so is aX + bY.

• This means the following are normal:

2X + 7Y, 3X - 78Y, OX + 2Y, X - Y, etc.

Example: Ben and Anakin are playing in the local Jedi golftournament. Ben’s golf game X has the N(60, 4) distribution.

Anakin’s golf game Y has the N(76, 12) distribution.

Q: What is the probability that Anakin will score lower than Ben in the tourney? In other words, what is P(Y < X) ?

A:Notice that Y - X is a linear combination of two independent random normal variables, so Y - X is normal.

What is the mean of the variable Y - X ?

Y-X = Y - X = 76 - 60 = 16

Example: Ben and Anakin are playing in the local Jedi golftournament. Ben’s golf game X has the N(60, 4) distribution.

Anakin’s golf game Y has the N(76, 12) distribution.

Q: What is the probability that Anakin will score lower than Ben in the tourney? In other words, what is P(Y < X) ?

A:Notice that Y - X is a linear combination of two independent random normal variables, so Y - X is normal.

What is the standard deviation of the variable Y - X ?

2Y-X = 2

Y + 2X = 122 + 42 =160

Y-X = 160 = 12.65

So, Y - X has the N(16, 12.65) distribution.

Example: Ben and Anakin are playing in the local Jedi golftournament. Ben’s golf game X has the N(60, 4) distribution.

Anakin’s golf game Y has the N(76, 12) distribution.

Q: What is the probability that Anakin will score lower than Ben in the tourney? In other words, what is P( Y < X) ?

So, Y - X has the N(16, 12.65) distribution.

P(Y < X) = P( Y - X < 0)

= P ( )(Y - X) - 16

12.65< 0 - 16

12.65

= P ( Z < -1.26 )= .1038

So, Anakin will score lower about 10% of the time.

Central Limit Theorem• Draw an SRS of size n from any population with mean

and finite standard deviation .When n is large, the sampling distribution of the sample mean is :

approximately N (, ) / n

Notes:

• The distribution of a sum or average of many small random quantities is close to normal.

• This is true if the quantities are not independent or even if they have different distributions.

• How large does n have to be depends on the distribution.

The Sampling Distribution of x(Revisited)

How do we find the sampling distribution ?

1) Take repeated random samples of size n from a

population with mean

2) Find the sample mean for each sample

3) Collect all the sample means and display their distribution.

The Sampling Distribution of x(Revisited)

Homework28, 31, 32, 37, 45