statistics question 2...best answer = 10 points!!?

Z Formula:

z=(x-xbar)/s

Standard error:

E=s/sqrt(n)

The formula for the mean uses the standard error rather than sample standard deviation.

z=(6.2-10)/(0.6/100^(1/2))

z=-63.333, look in your normal distribution table for a high negative z and use that as your approximation. Note that the question is asking for the values greater which means the values to the right. I don't have a normal distribution chart in front of me atm, but I can say that the percentage should be close to 100%=99.999% .

The deviation is 6.2 - 10 = -3.8. The standard deviation of the population is .6 so the sampling distribution standard deviation is .6/sqrt(100) = .6/10 = .06. That means we have a z value of:

-3.8 /.06 = -380/6 = -63.3333

This is z value represents such an incredibily small probability that we cannot easily calculate it.

For all practical purposes, there is no chance that you would see a sample average of 6.2 if the population has a normal distribution and the sample has a mean of 10 and a standard deviation of .6. You can say that all of your samples will have an average greater than 6.2.

For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)

You can translate into standard normal units by:

Z = ( X - μ ) / σ

Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.

If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed

with mean μ and standard deviation σ /√(n)

An applet for finding the values

http://www-stat.stanford.edu/~naras/jsm/FindProbab...

calculator

http://stattrek.com/Tables/normal.aspx

how to read the tables

http://rlbroderson.tripod.com/statistics/norm_prob...

In this question we have

Xbar ~ Normal( μ = 10 , σ² = 0.36 / 100 )

Xbar ~ Normal( μ = 10 , σ² = 0.0036 )

Xbar ~ Normal( μ = 10 , σ = 0.6 / sqrt( 100 ) )

Xbar ~ Normal( μ = 10 , σ = 0.06 )

Find P( Xbar > 6.2 )

P( ( Xbar - μ ) / σ > ( 6.2 - 10 ) / 0.06 )

= P( Z > -63.33333 )

= P( Z < 63.33333 )

= 1

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