Statistics Question II?

The height of 1000 students are approximately normally distributed with a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. If 200 random samples of size 35 are drawn from this population and the means recorded to the nearest tenth of a centimeter, determine


(a) the mean and standard deviation of the sampling distribution of x (X bar)


(b) the number of sample means that fall between 172.5 and 175.8 centimeters inclusive


(c) the number of sample means falling below 172 centimeters

Statistics Question II?
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/...





calculator


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





how to read the tables


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





In this question we have


Xbar ~ Normal( μ = 174.5 , σ² = 47.61 / 35 )


Xbar ~ Normal( μ = 174.5 , σ² = 1.360286 )


Xbar ~ Normal( μ = 174.5 , σ = 6.9 / sqrt( 35 ) )


Xbar ~ Normal( μ = 174.5 , σ = 1.166313 )








Find P( 172.5 %26lt; Xbar %26lt; 175.8 )


= P( ( 172.5 - 174.5 ) / 1.166313 %26lt; ( Xbar - μ ) / σ %26lt; ( 175.8 - 174.5 ) / 1.166313 )


= P( -1.714806 %26lt; Z %26lt; 1.114624 )


= P( Z %26lt; 1.114624 ) - P( Z %26lt; -1.714806 )


= 0.8674942 - 0.04319042


= 0.8243037





we expect 0.8243037 * 200 = 164.8607 of the 200 sample means to have a value within this interval.





Find P( Xbar %26lt; 172 )


P( ( Xbar - μ ) / σ %26lt; ( 172 - 174.5 ) / 1.166313 )


= P( Z %26lt; -2.143507 )


= 0.01603620





we expect to see 0.01603620 * 200 = 3.20724 means below 172 cm
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