Sampling Distribution of theSampling Distribution of the

Sample Mean Sample Mean X

Example

Let X denote the lifetime of aa battery

Suppose the distribution of battery battery lifetimes has

– mean lifetime, = 400 = 400 hours– standard deviation of lifetimes, 4040

hours

A Sample of Size nn is Taken• Calculate the mean of these nn batteries

• Then another sample of size n batteries is taken– The mean of this second sample of nn batteries is

calculated

• This is done again, and again, and again, and ….

.σX

μX

deviation, standard a and , mean a on,distributi a has X

).X (big X variable random the of nobservatio an is these of Each

values. ) x (little x many now are There

X ofDeviation Standard andMean The

:has which ,X of ondistributi sampling the

called is s'x possible all of collection The

nX

X

σσ

μμ

small gets large gets n As:NoteX

σ

Distribution of X

on.distributi normal a having

to is Xcloser the n, size sample thelarger the

unknown) is ondistributi its(or normalnot is XIf

:if normal is batteries) n of life (average X•X (Life of aa battery) is normal, and•σ is known

Central Limit Theorem

Even if X does not have a normal distribution,Even if X does not have a normal distribution,

will be approximately normal if n is large.will be approximately normal if n is large.X

n = 30 is usually large enough to use this approximation.

Example When Distribution of X is Normal

X = the life of aa battery

• Assume battery life is:– Distributed normal – Mean battery life = 400 hours– Standard deviation of battery life = 40 hours

• We choose a battery at random– The battery lasts 350 hours

• This is an observation of X

Observation of X

XX

= 40= 40

μμ = 400 = 400x = 350x = 350

The Random Variable

• Suppose random samples of size n = 4 batteries are selected independently and their sample means calculated

• These are observations of the random variable

X

X

AN OBSERVATION OF

• Suppose 4 batteries are selected and their lives are: 420, 450, 380, 390– Their average = (420 + 450 + 380 + 390)/4 = 410

• This is an observation of a random variable for the Sample Mean

X

X

DISTRIBUTION OF WHEN X IS NORMAL

X

normal is X

:normal is Xwhen True

n

σσ

:true Always

X

X

__XX

Sampling Distribution and anObserved Value

400 μX

204

40 X

410 x

Observation of XWhen X is Not Normal

= 400

= 40

x = 350

f(x)

XX

AN OBSERVATION OF

• Suppose 4 batteries are selected and their lives are: (410, 450, 360, 360)– Their average = (410 + 450 + 360 + 360)/4 = 395

• This is an observation of a random variable for the Sample Mean

• But the sample size is small so we do not know the distribution of -- we can’t plot it

X

X

X

Using A Larger Sample Size

• Suppose 100 batteries are selected and their lives are: (420, 450, 380, 350,…., 415)– Their average = (420 + … + 415)/100 = 408

• This is an observation of the random variable (where n = 100)

• Because this is a large sample, the distribution of is approximately normal

X

X

__XX

Sampling Distribution and anObserved Value

400 μX

4100

40 X

408 x

EXAMPLES

• Answer the following questions assuming:– Battery life is distributed normal– Battery life distribution is not normal (or

unknown)

• What is the probability:– a random battery will last longer than 408 hours?– the average of life of 16 batteries be longer than

408 hours?– the average life of 100 batteries will be longer than

408 hours?

P(X > 408)X normal

= 40

400 X400 X0 Z0 Z.20.20

.5793.5793

408408

1 - .5793 = 1 - .5793 = .4207.4207

P(X >408)X Not Normal

= 400

= 40

x = 408 X

?

_ P(X >408) X Normal, n = 16

10

16

40 σX

__XX400400

knownσ normal, is X

because normal is X

4084080 Z0 Z.80.80

.7881.7881

1 - .7881 1 - .7881 = .2119= .2119

_ P(X >408) X Not Normal, n = 16

• Can’t do• X is Not Normal and n is small

– So we do not know the distribution of the Sample Mean

X

4100

40 σX

__XX400400

_ P(X >408) X Normal, n = 100

knownσ normal, is X

because normal is X

0 Z0 Z2.002.00408408

.9772.9772

1 - .9772 = .0228 .0228

4100

40 σX

__XX400400

_ P(X >408) X Not Normal, n = 100

0 Z0 Z2.002.00408408

.9772.9772

1 - .9772 = .0228 .0228

Theorem)Limit (Central large isn

because normalely approximat is X

Using Excel• As long as it can be assumed that the distribution of

the sample mean is normal, NORMDIST and NORMINV can be used to give probabilities except:– Instead of using , put in /n

• Let Excel do the arithmetic

– Example: Find the probability the average of 100 batteries exceeds 408 hours

nσ/

Review

• Given a distribution X, with and known -- for samples of size n:

• If X is normal and is known

• If X is not normal

n

σ σ μ μ

XX

ondistributi normal a has X

???? is X ofon distributi theOtherwise

large isn if normalely approximat is X