Given the discrete uniform population f(x) = 1 3 , x = 2, 4, 6, 0, elsewhere, find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4. assume the means are measured to the nearest tenth.

 First find the expectation and standard deviation of X 

Let X be any discrete random variable with probability mass function, pmf, f(x) = P(X = x) 

The expectation is: 

E(X) = μ = ∑x * P(X = x) 

The variance is: 

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

the standard deviation is the square root of the variance. 

E(X) = 4 
Var(X) = 8/3 
stddev(X) = 1.632993 

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 < 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. 

In this question we have 
Xbar ~ Normal( μ = 4 , σ² = 2.666667 / 54 ) 
Xbar ~ Normal( μ = 4 , σ² = 0.04938272 ) 
Xbar ~ Normal( μ = 4 , σ = 1.632993 / sqrt( 54 ) ) 
Xbar ~ Normal( μ = 4 , σ = 0.2222222 ) 

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. 

Find P( 4.1 < Xbar < 4.4 ) 
= P( ( 4.1 - 4 ) / 0.2222222 < ( Xbar - μ ) / σ < ( 4.4 - 4 ) / 0.2222222 ) 
= P( 0.45 < Z < 1.8 ) 
= P( Z < 1.8 ) - P( Z < 0.45 ) 
= 0.9640697 - 0.6736448 
= 0.2904249

2.) is the most sense