07+Sampling+distribution

# 3 3 the variance of the random variable x equals v

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Unformatted text preview: ((0 − 4)2 + (2 − 4)2 + (10 − 4)2 ) = . 3 3 the variance of the random variable x equals V ar(x) = (x − µ)2 P (x) = (0 − 4)2 · 1 2 1 2 + (1 − 4)2 · + (2 − 4)2 · + (5 − 4)2 · + 9 9 9 9 2 28 σ 2 21 +(6 − 4) · + (10 − 4) · = =. 9 9 3 2 It can be shown that the value V ar(x) is the smallest variance among all unbiased estimators of µ, i.e. x is the minimum variance unbiased estimator for µ. 2 3. If a random sample is selected from a population with a normal distribution, then the sampling distribution of x is normal. Central Limit Theorem. If a random sample of size n is selected from any population, then for suﬃciently large n the sampling distribution of x is approximately normal (the larger n, the better will be the normal approximation). We typically use the Central Limit Theorem when n > 30. If the population distribution is not extremely skewed, then even smaller n may be "large enough". 3 A. Zhensykbaev Sampling distributions Example. 5,246 savings accounts have a mean balance µ = \$1, 000 and σ = \$240. Auditors took at random a sample of 64 accounts and said they will certify the bank’s report if the sample mean balance is within \$60 of the reported \$1,000. What is the probability that the auditors will not certify the report? Solution. Regardless of the true probability distribution of balances due to the central limit theorem we may deduce that the sample mean of 64 accounts will be normally distributed with µ = \$1, 000 and \$240 σ σs = √ = √ = \$30. n 64 Thus, \$60 = 2σs . Reduce to standard normal random variable (z -score) z= x−µ σs means x1 = 1, 000 − 60 and x2 = 1, 000 + 60, z -scores are z1 = 1, 060 − 1, 000 = 2, 30 z2 = 940 − 1, 000 = −2. 30 Now we may use the Table IV (McClave and Benson) of normal probabilities: P1 = P2 = 0.5 − 0.4772 = 0.0228. On the pic.1...
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