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Sample Mean Samples give information about random variables X - Random variable Sample size -n Expected value of the sample mean Expected value of the sample variance
Here x^ and s^2 are sample statistic, used to estimate the expected value
of sample mean & expected value of the sample variance
Justification (Extra) In the world of business applications, we usually must gather information about a random variable, X , by collecting a single random sample { x 1 , x 2 , ¼ , xn }. The sample size, n , may be too small to provide much information about the distribution of X****. Hence, we must learn what we can from the two sample statistics x and s^2. We know that the expected value of the sample mean is E ( X ) and that the expected value of the sample variance is V ( X ). Thus, we can estimate the two main parameters of X , E ( X ) @ x and V ( X ) @ s^2. As we have seen in our examples, interest is usually centered on E ( X ). This is the number that will influence our business decisions. For this reason, we need information on how accurately x approximates E ( X ). This, in turn, means that we must know something about the probability distribution of the sample mean, taken as a new random variable. E X @ x
V X @ s^2
As we saw in Variance , the Variance of the sample mean is = V ( X )/ n @ s^2 / n****. But V ( X ) @ s^2 n s V x V X n 2 ( ) ( ) / @ Taking the expected value of the sample mean to be approximately x , we have estimates for both the mean and the variance of the sample mean****. Unfortunately, we have also seen that the mean and variance of a random variable do not determine its probability distribution.
As the sample size, n, increases, the distributions of the
standardized sample means of any random variable
always approach the same fixed probability distribution
function.
We let Z be the continuous random variable whose probability is given by this universal distribution function. Z is called the standard normal random variable and fZ is called the standard normal probability density function****. The Central Limit Theorem If X is any random variable, then, as n increases without bound, the distribution of its standardized sample mean, , n x X X
approaches the distribution of the standard normal random variable, Z, whose p.d.f****. is
0. 5 z
f Z z e
- The graph of the standardized values appear to be approximately as follows Extra
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
of x. ( ii ) Compute all values for the p.m.f. of S^ 4. ( iii ) Compute the mean and standard deviation of S^ 4. Solution. ( i ) So, the mean is 0.5, variance is 0.0625, and standard deviation is 0. ( ii ) ( iii ) x f^ x (^ x ) x^^ ^ fx (^ x ) (^ x^ μx )^2 fx ( x ) 0.00 0.0625 0.0000 0. 0.25 0.2500 0.0625 0. 0.50 0.3750 0.1875 0. 0.75 0.2500 0.1875 0. 1.00 0.0625 0.0625 0. Sum 1.000^ μ^ x 0.5000 V^ (^ x ) 0. σ (^) x 0. Using
s ( x x )/ x
s (^) 4 f^ S 4 (^ s 4 ) s ( 0 . 5 )/. 25 (^2) 0. -1 0. 0 0. 1 0. 2 0. s 4 f^ S 4 (^) (^ s 4 ) s 4^ ^ fS 4 (^ s 4 ) ( ) 4 ( 4 ) 2 s (^) 4 μS 4 fS s -2 0.0625 -0.125 0. -1 0.2500 -0.250 0. 0 0.3750 0.000 0. 1 0.2500 0.125 0. 2 0.0625 0.250 0. Sum 1.000 μ^ S 4 0 V^ (^ S 4 ) 1. σ (^) S 4 1.
So, the mean is 0 and the standard deviation is 1. (This MUST be true for all standardized variables) Additional notes (Background Information)
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-3 -2 -1 0 1 2 3 - 1
Example-- Who’s performing better in sales? Bill = $10,000 sales in Region 1 or Janice = $5,000 sales in Region 2 Need to consider their relative standings in their regions-- maybe it’s harder to sell in Region 2 than 1 compare their z scores z scores especially useful when comparing “apples and oranges” e.g. job candidate’s scores on a written test with a 0-100 range and a performance test with a 1-10 range z scores not useful when you need to know the raw values Are sales quotas being met? How much profit was made last year? We will use z-scores mainly to use with the normal probability table and not as a statistic in itself On a recent exam, the scores were normally distributed with mean 50 and standard deviation 12. o What is the probability that a randomly selected student would get at least the average? o What is the probability that a randomly selected student would get between 50 and 75?
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o What score did only 16% of the students meet or exceed?
