Unformatted text preview: etry
= 1 − F(0.5)
A look‐up in Appendix Table 1 gives: F(0.5) = 0.6915 The answer is: P( X > μ + 2 ) = 1 − 0.6915 = 0.3085 7 Chapter 7 (b) What is the probability that the sample mean is more than 3 hours below the population mean ? That is, find: P( X < μ − 3) Note: P( X < μ − 3) < P( X < μ − 2) = P( X > μ + 2) This is illustrated in the graph below. PDF of X μ−3 μ−2 μ μ+2 Therefore, the answer must be smaller than the probability calculated for part (a). 8 Chapter 7 The solution method follows similar steps to part (a): ⎛ X − μ (μ − 3) − μ ⎞
P( X < μ − 3) = P⎜
⎟
⎜ se( X ) < se( X ) ⎟
⎠
⎝
= P Z < − 34 ( ) = 1 − P(Z < 0.75) = 1 − F(0.75)
A look‐up in Appendix Table 1 gives: F(0.75) = 0.7734 The answer is: P( X < μ − 3) = 1 − 0.7734 = 0.2266 9 Chapter 7 (c) What is the probability that the sample mean differs from the population mean by more than 4 hours ? That is, find: lower tail upper tail ↓ ↓ 1 − P(μ − 4 < X < μ + 4) = P( X < μ − 4) + P( X...
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 Spring '10
 WHISTLER

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