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DISTRIBUTION OF THE SAMPLE MEAN X1 , X 2 , , X n sample from a distribution/population with mean μ and standard deviation σ. 1 n X Xi n i 1 We know: can take different values for different samples – sampling distribution. FACT1. The mean and standard deviation for the distribution of are given by: X and X X n The mean of X is the same as the population mean μ. So, unbiased for μ. The standard deviation of standard deviation. X is X is n times smaller than the population Averages have smaller variability than single observations! Law of Large Numbers Closer look at the standard deviation of : X X n As n=sample size increases, X 0; i.e. as n increases, spread of the sample mean decreases to zero. What random variable has spread zero? Constant! X comes arbitrarily close to μ 1.5 1.5 2.0 2.0 Conclusion- Law of Large Numbers: for large enough n. Distribution with mean=10, st.dv.= 2/100^0.5=0.2 1.0 1.0 Distribution with mean=10, st.dv.= 2/10^0.5=0.63 0.0 0.0 0.5 0.5 Distribution with mean=10, st.dv.=2 5 10 5 15 n=10, n=100 10 15 DISTRIBUTION OF THE SAMPLE MEAN – NORMAL DATA X1 , X 2 , , Xn sample from a Normal distribution, N(μ, σ ). From FACT 1, We know: FACT 2: If X 1 , X 2 , distribution. X and X n , X n are from N(μ, σ ), then X has a N(μ, σ/ √n ) NOTE: Since X is normally distributed, with μX = μ and standardize it: X X n( X ) Z . / n n , then we may SAMPLING DISTRIBUTION OF THE SAMPLE MEAN EXAMPLE: Students in an university have a weight distribution that is known to be N(150, 20). Let X1, X2, …, X16 represent the weights of 16 randomly selected students from this university. If X is the average weight for this sample, find P( X > 160). Solution: Since the sample came from a normal distribution, by Fact 2, the sample mean has a normal distribution as well. X ~N(μ, σ/ √n )=N(150, 20/ √16)=N(150, 5). Thus, P( X 160) P( X 150 160 150 ) P( Z 2) 1 P( Z 2) 1 0.9772 0.0228. 5 5 EXAMPLE, CONTD. An elevator at this university has a capacity of 1500 pounds. What is the probability that 9 students who enter the elevator will have a safe ride, i.e. their total weight is less than 1,500 lb? Solution: Again, by Fact 2, the sample mean has a normal distribution: X ~N(μ, σ/ √n )=N(150, 20/ √9)=N(150, 6.67). Also, P( Total weight < 1500)=P( X <1500/9)=P( X <166.67). So, X 150 166.67 150 P( X 166.67) P( ) P(Z 2.5) 0.9938. 6.67 6.67 DISTRIBUTION OF THE SAMPLE MEAN, CONTD. EXAMPLE. Suppose X is the score on a test and X~N(500, 100). Let X1, X2, …X16 be a sample of scores for 16 individuals and X their average score. Find P( 550 < X ≤ 600). Solution: Since the data come from a normal distribution, by Fact 2, X has a normal distribution with mean X 500 and X / n 100 / 16 25. Thus, P(550 < X ≤ 600) = P( 550 500 X 500 600 500 ) 25 25 25 = P(2 < Z ≤ 4) = P(Z ≤ 4) - P(Z ≤ 2 ) = = 1 – 0.9772 = 0.0228. The CENTRAL LIMIT THEOREM (CLT) What if the data does not come from the normal distribution? FACT 3. (CLT): If X1, X2, …Xn are any set of observations with mean μ and standard deviation σ, their sample mean X , has approximately normal N(μ, σ/√n) distribution, if n is sufficiently large. How large is sufficiently large? Depends on the distribution the data comes from. Definitely n should be at least 20 before we use this approximation. Difference between Fact 2 and Fact 3: Fact 2 holds only for samples from Normal distribution and gives exact distribution of X . Fact 3 holds for samples from any distribution, but gives an approximate distribution for X . The Central Limit Theorem contd. Example. Suppose X1, X2, …, X25 are lifetimes of electronic components, with μ=700 hours and σ=10 hours. Find P( X ≤ 702), where X is the sample mean of the lifetimes of 25 components. Solution. Usually lifetime data is skewed to the right, so not normal (Why?) Since n=25 (reasonably large), we will use CLT and the normal approximation of the distribution of the sample mean: X So, has approx. a N(μ, σ/√n) = N(700, 10/√25) = N(700, 2) distr. X 700 702 700 P( X 702) P( ) P(Z 1) 0.8413. 2 2

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