Any one know how to do probabilities?

Researchers at a pharmaceutical company have found that the effective time duration of a safe dosage of a pain relief drug is normally distributed with mean 2 hours and standard deviation 0.3 hour. For a patient selected at random:


a) What is the probability that the drug will be effective for 2 hours or less?


b) What is the probability that the drug will be effective for 1 hour or less?


c) What is the probability that the drug will be effective for 3 hours or more?

Any one know how to do probabilities?
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


X ~ Normal( μx = 2 , σx² = 0.09 )


X ~ Normal( μx = 2 , σx = 0.3 )





Find P( X %26lt; 2 )


P( ( X - μ ) / σ %26lt; ( 2 - 2 ) / 0.3 )


= P( Z %26lt; 0 )


= 0.5





this is easy to find because for any normal random variable Y with mean μ, P(Y %26lt; μ) = P(Y %26gt; μ) = 0.5





Find P( X %26lt; 1 )


P( ( X - μ ) / σ %26lt; ( 1 - 2 ) / 0.3 )


= P( Z %26lt; -3.333333 )


= 0.0004290603





Find P( X %26gt; 3 )


P( ( X - μ ) / σ %26gt; ( 3 - 2 ) / 0.3 )


= P( Z %26gt; 3.333333 )


= P( Z %26lt; -3.333333 )


= 0.0004290603





P(X %26lt; 1) = P( X %26gt; 3) because you are looking at the area under the curve in the tails 3.33333... standard deviations above or below the mean.
Reply:Find the z-score then look up the probabilities on the chart in your book (standard normal probability tables)