<p>Ch6 example to show how the Central Limit Theorem works.</p><p>Consider the population described by the probability distribution shown here. Assume the r.v. x is observed twice (n=2) and these observations are independent.</p><p> x 2 3 4 5 6   ____ and   ____ p(x) .1 .2 .2 .1 .4</p><p>A sampling distribution is shown below. Is it the correct sampling distribution for the sample mean, x ?</p><p> x 2 2.5 3 3.5 4 4.5 5 5.5 6 x = ___  x = ____ p( x ) .01 .04 .08 .1 .16 .2 .17 .08 .16</p><p>To answer the question, you must show that  x   and  x   / n .</p><p>Use your calculator to draw the histogram for the population, x. The graph looks skewed somewhat.</p><p>Use your calculator to draw the histogram for the sampling distribution for x . This distribution looks approximately normal. </p><p>Be sure you understand how the Central Limit Theorem is at work here. Even though the population above isn’t normally distributed, taking a small random sample of n=2 shows the sampling distribution taking on the normal distribution. This author says for n  30 , then the sampling distribution for x will be approximately normally distributed yet this example uses a sample of n=2. Look on p285 and see what sample sizes will bring about a sampling distribution for x that is about normal.</p>