Week 6 Mon Oct 4 - WEEK 6, Monday, Oct 4 Today: Cover...

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1 WEEK 6, Monday, Oct 4 Today: Cover Sections 4-5, 4-7 . Finding percentiles of Normal Distributions (Chap 4, Sec 4-5). TI-83 . Use invNorm(.90, μ , σ ) and get 90 th percentile of the Normal distribution N( μ , σ ). Excel . Use = norminv(.90, μ , σ ) and get 90 th percentile of the Normal distribution N( μ , σ ). Percentiles using z-Table: Calculating the position x where P ( X < x ) = 0.90 is a two-step process . First step is to use the z-Table to find the position z where P ( Z < z ) = 0.90, that is, to find the 90 th percentile of the Standard Normal distribution N(0, 1). It happens to be the value z = 1.28. Second step is to calculate the position x by the formula . 28 . 1 z x (The textbook calls this equation the inverse transformation.) The position x is the 90 th percentile of the Normal distribution N( , ). Example 1. Finding a 90 th Percentile of N(165, 10) . What is the 90 th percentile of the distribution of weights of healthy, male, St. Bernard, age 6, dogs assuming that distribution has mean 165 lbs, standard deviation 10 lbs and is a Normal Distribution? Draw the density function and then the arrow that cuts the area to 90% below and 10% above. It gives the 90 th percentile of the weight distribution. Make a guess for the x -position of the arrow.
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Mean = 165 lbs, StDev = 10 lbs 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 135 140 145 150 155 160 165 170 175 180 185 190 195 x (lbs) f(y) TI-83 . Use invNorm(.9, 165, 10) and get 177.8155 lbs . Excel . Use = norminv(.9, 165, 10) and you get 177.8155 lbs . z-Table . From the table the position has z-score 1.28. The inverse of the standardization function z x is z x . Hence, in this example the raw score x corresponding to z = 1.28 is x = 165 lbs + (1.28) 10 lbs = 177.8 lbs Using z-Tables, 120 + (.15)(44) = 126.6.
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Week 6 Mon Oct 4 - WEEK 6, Monday, Oct 4 Today: Cover...

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