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Question Help 5.3.29 Find the z-scores for which 28% of the distributions area lies between -z and z Click to view page 1 of
X Standard Normal Table (Page 1) 0.04 3.4 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 -3.3 0.0003 0
-X i Standard Normal Table (Page 1) 1.9 U.0233 U.U239 0.0244 U.0250 0.0256 0.0262 0.0268 0.0274 0.0281 U.0287 -1.8 0.0294 0.0
- X i Standard Normal Table (Page 2) 0.04 0.06 0.07 0.00 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0
X Standard Normal Table (Page 2) 1.5 U.9332 U.9345 0.9357 0.9370 0.938Z 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463
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Answer #1

Answer)

First we need to find the value of alpha

Alpha = 100-28 = 72 (0.72)

Now we know that, standard normal z table is symmetrical in nature

So, 0.72/2 would lie on both sides

0.36

From z table, p(z<-0.36) = 0.3595 ~ 0.36

And due to symmetrical nature

P(z>0.36) = 0.36

So, 28% lies in between -0.36 and 0.36

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