Statistics Help III?

The random variable X representing the number of cherries in a cherry puff has the following probability distribution:


x 4 5 6 7


P(X=x) .2 .4 .3 .1


(a) find the mean (mew) and the variance (omega ^2) of X


(b) find the mean (mew Xbar) and the variance (omega squared x) of the mean (xbar) for random samples of 36 cherry puffs.


(c) find the probability that the average number of cherries in 36 cherry puffs will be less than 5.5

Statistics Help III?
for any discrete random variable the expectation is:





E(X) = μ = ∑x * P(X = x)





the variance of a discrete random variable is:





Var(X) = σ² = ∑(x - μ)² * P(X = x) = {∑x² * P(X = x) }- μ²





the standard deviation is just the square root of the variance.





in this question we have:





E(X) = μ = 4 * 0.2 + 5 * 0.4 + 6 * 0.3 + 7 * 0.1 = 5.3


Var(X) = σ² = 0.81





== -- == -- == -- == -- == -- == -- == -- ==





E(Xbar)


= E(1/n ∑ Xi)


= 1/n * E(∑Xi)





expectation is a linear operator so we can take the sum out side of the argurement





= 1/n * ∑ E(Xi)





there are n terms in the sum and the E(Xi) is the same for all i





= 1/n * nE(Xi)


= E(Xi)





E(Xbar) = μ





--





Var(Xbar)


= Var(1/n * ∑Xi)


= 1/n^2 Var(∑Xi)





because the samples are independent we can take the sum out side of the Var argument. if the samples where not independent then this would not be possible.





= 1/n^2 * ∑Var(Xi)





there are n terms in the sum and the Var(Xi) is the same for all i





Var(Xbar) = 1/n * Var(X)





and Standard Deviation of Xbar is σ/sqrt(n)








In this question for a sample of size 36 from this population:





E(Xbar) = 36


Var(Xbar) = 0.81 / 36 = 0.0225





== -- == -- == -- == -- == -- == -- ==





(c)





The central limit theorem tells us that:





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)





so





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.








In this question we have


Xbar ~ Normal( μ = 5.3 , σ² = 0.81 / 36 )


Xbar ~ Normal( μ = 5.3 , σ² = 0.0225 )


Xbar ~ Normal( μ = 5.3 , σ = 0.9 / sqrt( 36 ) )


Xbar ~ Normal( μ = 5.3 , σ = 0.15 )





Find P( Xbar %26lt; 5.5 )


P( ( Xbar - μ ) / σ %26lt; ( 5.5 - 5.3 ) / 0.15 )


= P( Z %26lt; 1.333333 )


= 0.9087888
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