Please explain this Statistics question :-) thanks!?

Harley-Davidson motorcycles make up 20% of all the motorcycles registered in the United States. You plan to interview an SRS of 527 motorcycle owners.


What is the approximate distribution of the proportion of the motorcycle owners in your sample who own Harleys?





b)How likely is your sample to contain 22% or more who own Harleys? Do a normal probability calculation to answer this questions.





(c) How likely is your sample to contain atleast 17% who own Harleys? Do a normal probability calculation to answer this question.

Please explain this Statistics question :-) thanks!?
the proportion is approximately normal with mean 0.20 and standard deviation sqrt( 0.20 * (1 - 0.20 ) / 527) = 0.01742427





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





Moving from the standard normal back to the original distribuiton using:


X = μ + Z * 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


X ~ Normal( μx = 0.2 , σx² = 0.0003036052 )


X ~ Normal( μx = 0.2 , σx = 0.01742427 )








Find P( X %26gt; 0.22 )


P( ( X - μ ) / σ %26gt; ( 0.22 - 0.2 ) / 0.01742427 )


= P( Z %26gt; 1.147824 )


= P( Z %26lt; -1.147824 )


= 0.1255206





Find P( X %26gt; 0.17 )


P( ( X - μ ) / σ %26gt; ( 0.17 - 0.2 ) / 0.01742427 )


= P( Z %26gt; -1.721736 )


= P( Z %26lt; 1.721736 )


= 0.9574414
Reply:I use activ stats for these problems. you enter in the population parameter and drag the flag and the density tool tells you the %. thank god for that package.