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Assume that randomly selected subject given bone density test: Those test scores are normally distributed with mean of and standard deviation of Draw graph and find...

Question

Assume that randomly selected subject given bone density test: Those test scores are normally distributed with mean of and standard deviation of Draw graph and find the bone density test scores that can be used as cutoff values separating the owest 13% and highest 13%, indicating level that are too low or too high; respectively:Sketch the region: Choose the correct graph below:The bone density scores are (Use comma to separate answers as needed. Round to two decimal places as needed )

Assume that randomly selected subject given bone density test: Those test scores are normally distributed with mean of and standard deviation of Draw graph and find the bone density test scores that can be used as cutoff values separating the owest 13% and highest 13%, indicating level that are too low or too high; respectively: Sketch the region: Choose the correct graph below: The bone density scores are (Use comma to separate answers as needed. Round to two decimal places as needed )



Answers

Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of $1 .$ In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places. If bone density scores in the bottom $2 \%$ and the top $2 \%$ are used as cutoff points for levels that are too low or too high, find the two readings that are cutoff values.

Okay, Dokey. So here top 3% is 100% minus the 3% that's going to 30.97 00 lower. 3% is 30.300 So our task, then, is to go look in the chart for the thing closest 2.9700 And that's gonna get me, uh, the closest thing that that is 0.9699 So when I go and I look at that chart over and above, I get a Z score a 1.8 with the aid over there. So 1.88 The nice thing about this is whatever this is, it's going to be the same over here, just a different sign. So there's both of our scores.

Okay, so we're 17. We're gonna go find negative 1.23 in the table, A two. So you go to negative 1.2 the threes here and then wherever they meet, that's our answer. So that's going to be, Ah, 0.10 93 And it does say to draw a graph. So there's your normal curb and you're looking at 10%. So somewhere like this or so would be where the negative 1.23 would be. And since it's less than we shade that way

Number 32. Number three to we have to go find the score, probably for negative 4.27 And in our chart, we don't even see negative four point anything because it says anything less than negative. 3.99 or so is going to have the probability of 0.1 automatically. And then we do have a Z score, of course, for 2.34 So I go to 2.3, locate the four. They meet at 40.9904 So I'm gonna do 0.9904 minus 0.0 01 And that's gonna get me 10.9903 as a final answer on. Then I've got my curb to draw on. It's gonna be pretty much, you know, all of it. So with that being said, you're just gonna pretty much shade the whole thing, leaving maybe like a smidgen of some work. And that's gonna be the negative 4.2 seven. And this will be the 2.34

For a standard normal distribution, he has asked that find the Z score which got the one to be one or the 99. Mhm. The one from lies for that value of -2.33. And we know B 99 comes out to be two point quickly, because because it's symmetric since girl is symmetric, like the way it is. Yeah, this is derived from that table. True.


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