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ChE253K Spring09 HW07 Solutions.1

# ChE253K Spring09 HW07 Solutions.1 - ChE 253K Applied...

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ChE 253K -- Applied Statistics -- Homework 07 1. Text Problem 5.22 1 Point. (a) 1 std deviation (b) 2 std deviations 1 Point. (c) 3 std deviations (d) 4 std deviations 2. Text Problem 5.19 2 Points. (a) <1.50 (b) <-1.20 2 Points. (c) >2.16 (d) >-1.75 (e) between –1.20 and –1.75 2 Points. (f) between –1.20 and +1.50 (g) How would you solve Part (e) if ...? 3. ... please find the range for proportions of the distribution. 2 Points. (a) Top 10% of the distribution (b) Bottom 10% of the distribution 2 Points. (c) The 50% of the distribution between the 1st and 3rd Quartiles. Total: 12 Points

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Standard Normal Distribution 1. Text Problem 5.22 Parts (a) thru (c) from the chart. (a) P( μ ± 1 σ ) = 2(34%) = 68% (b) P( μ ± 2 σ ) = 2(34% + 13.5%) = 95% (c) P( μ ± 3 σ ) = 2(34% + 13.5% + 2.35%) = 99.7% Part (d) from Appx Table 3 (d) P( μ ± 4 σ ) = F(+4) – F(–4) = = 0.99997 - 0.00003 = 0.99994 Note: The problem statement says the variable has a "normal distribution," but it does not say it is a "standard normal distribution." Why can probabilities be found from the standard normal distribution? Because distances are given in units of std deviations away from the mean, i.e. they are transformed into z-scores, and thereby the normal
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ChE253K Spring09 HW07 Solutions.1 - ChE 253K Applied...

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