Unformatted text preview: > μ + 4)] = 2 P( X > μ + 4)
Follow the steps in part (a) to get: ( P( X > μ + 4) = P Z > 4 4 ) = 1 − P(Z < 1) = 1 − F(1) A look‐up in Appendix Table 1 gives: F(1) = 0.8413 The answer is: 2 P( X > μ + 4) = 2 (1 − 0.8413) = 0.3174 10 Chapter 7 (d) Suppose that a second (independent) random sample of 10 students was taken. Without doing the calculations, state whether the probabilities in part (a) would be higher, lower, or the same for the second sample. The standard error of X is: se( X ) = σ n An increase in the sample size from n=4 to n=10 gives a smaller standard error. This leads to more concentration about the population mean μ and so P( X > μ + 2 ) becomes lower. This is illustrated in the graph. PDF of X n = 10 n=4 11 μ μ+2 Chapter 7 The Central Limit Theorem The normal distribution is a convenient approximation in many applications. The Central Limit Theorem gives a justification for this. Let the random sample X 1 , X 2 , . . . , X n be a set of random variables that are independently and identically distributed with mean μ and variance σ . The random variables need no...
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- Spring '10