Statistics Homework?

This question has to do with central limit theorem.





Suppose a random number generator generates real numbers with uniform frequency between 1 and 5. If we randomly sample 75 random numbers, determine the following.





a. X=


b. Xbar ~ "blank" X~ "blank"





c. draw and label both distributions from b.


(you don't really have to answer this if there is no way to draw it online)





d. the probability that the sample mean is no more than 2.5





e. find the probability that any given random number is no more than 2.5





f. find the 45th percentile for the average random number generated.





can you please show some work. THANK YOU.

Statistics Homework?
Let X1, X2, ... , Xn be a simple random sample from a population with mean μ and variance σ².





Let Xbar be the sample mean = 1/n * ∑Xi


Let Sn be the sum of sample observations: Sn = ∑Xi





then, if n is sufficiently large:





Xbar has the normal distribution with mean μ and variance σ² / n


Xbar ~ Normal(μ , σ² / n)





Sn has the normal distribution with mean nμ and variance nσ²


Sn ~ Normal(nμ , nσ²)





The great thing is that it does not matter what the under lying distribution is, the central limit theorem holds. It was proven by Markov using continuing fractions.





if the sample comes from a uniform distribution the sufficient sample size is as small as 12


if the sample comes from an exponential distribution the sufficient sample size could be several hundred to several thousand.





if the data comes from a normal distribution to start with then any sample size is sufficient.


for n %26lt; 30, if the sample is from a normal distribution we use the Student t statistic to estimate the distribution. We do this because the Student t takes into account the uncertainty in the estimate for the standard deviation.





if we now the population standard deviation then we can use the z statistic from the beginning.


the value of 30 was empirically defined because at around that sample size, the quantiles of the student t are very close the quantiles of the standard normal.














P(X = 1) = 0.20


P(X = 2) = 0.20


P(X = 3) = 0.20


P(X = 4) = 0.20


P(X = 5) = 0.20





E(X) = 3


Var(X) = 2








Xbar ~ Normal(μ = 3 , σ² / n = 2/75)








e) P(X %26lt; 2.5 ) = P(X = 1) + P(X = 2) = 0.40








f) 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...





We know from the standard normal that


P(Z %26lt; z) = 0.45


= P(Z %26gt; z) = 0.55


when z = -0.1256613





x = μ + z * σ


x = 3 + -0.1256613 * 0.1632993


x = 2.979480





P(X %26lt; 2.979480 ) = 0.45


P(X %26gt; 2.979480 ) = 0.55