MG 2513.JPG a Search Etteds 15.2 Probability Encoding 313 1) (weight ze 1501&l;
ID: 3227254 • Letter: M
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MG 2513.JPG a Search Etteds 15.2 Probability Encoding 313 1) (weight ze 1501&l; 0.7 2) (weight s 70 0.2 3) (weight 100l&)- 0,5 4) (weight s 220l& 0,84 5) (weight 130I&) 0.58 6) (weight sa 1801&) 0.79 weight 250l& 0.9 8) (weight s 100l&) 0.4 9) (weight 50I&) 0,05 FIGURE 15.3 John's Probability Encoding List I: We see that at 100 pounds, there was a little inconsistency. Note that John has num- bered your answers in the comect chronological order. Do you think your earlier or later answers are more or less reliable? John, show us what you were writing. (John holds up a list of nunbers wirh nota- rions as shown in Figure 15.3.) Mary, how did you feel about the process Min I found I really got into it after a while. I feel better about my later answers. I: John has, in fact, sketched a smooth cumulative probability distribution through your answers. (He holds up Figure 15.4.) answer of 100 pounds. Now He has drawn the curve much closer to your later Mary, is there anything you would like to change about your curve? Mary: No, that is the best I can do with my present information. Would it be oK if I lifted the desk a bit? I want to see if my impression of its weight is accurate. Not just yet, if you don't mind. John has the plot with the measure being assessed, weight the desk and with your name, the date, and the time. of time stamp important, do you think?Explanation / Answer
a) .5 fractile of Standard Normal Distribution= Median= 0 as P(Z<=0)=0.5
5 Fractiles for the Standard Normal Distribution
b) The mid range value is 0.
c) The chance of being below a value 1 is .8413 as P(Z<=1)=.8413
d)P(-.75<Z<=-.5)=P(Z<=-.5)-P(Z<=-.75)= 0.3085-0.2266=.0819
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