# Tutor.com Probabilities Session

##### Jan. 14, 2013

Session Transcript - Math - Statistics, 1/12/2013 12:34AM - Tutor.comSession Date: 1/12/2013 12:34AM
Length: 25.8 minute(s)
Subject: Math - Statistics

System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

AL (Customer)
[00:00:00] Find the area under the normal distribution curve to the right of 2.45 and to the left of â€“ 2.08
[00:00:10] how are you tonight?

Linda S (Tutor)
[00:00:16] Great, thanks!
[00:00:25] OK, so how can I help you with this?

AL (Customer)
[00:00:42] I don't know how to solve this problem
[00:00:51] I am stuck on it

Linda S (Tutor)
[00:01:03] OK, is this just one problem, or is it 2?
[00:01:17] Separate problems?

AL (Customer)
[00:01:35] I think they are separate problem

Linda S (Tutor)
[00:02:14] OK. Well, basically, you just need a standard normal distribution table to look up the given z-scores and find the areas. Do you have a table handy?

AL (Customer)
[00:02:35] no I don't

Linda S (Tutor)
[00:03:26] Hmm, okay, usually they have one included in any Stats textbook. Let me try to find one online.

AL (Customer)
[00:03:47] ok thank you

[00:04:43] URL > http://wps.aw.com/aw_triola_stats_series/
[00:05:17] Well, if you want to go to that site, then click on the white book at the left, then click on the green button that says 'Formula & Table Card'.
[00:05:49] The table we want to use is on pp. 5-6 of that formula card.

AL (Customer)
[00:07:11] I see it

Linda S (Tutor)
[00:07:22] OK, perfect.

AL (Customer)
[00:07:22] thank you for that website

Linda S (Tutor)
[00:08:05] Sure. So, the table gives the areas to the LEFT of a given z-score. For a z-score of -2.08 then, just find that z-score in the table and you can read off the area to the left.
[00:11:04] I'm working on this...

AL (Customer)
[00:10:28] me too I am trying to figure it out too

Linda S (Tutor)
[00:10:45] Well, do you know how to read the table?

AL (Customer)
[00:11:29] not really

Linda S (Tutor)
[00:11:47] OK. Hmm, let me copy part of it over to our window.

AL (Customer)
[00:12:00] thank you

Linda S (Tutor)
[00:13:23] OK, so we are looking for the area to the LEFT of z = -2.08. So first, we need to find z = -2.08 in the table, and we use a combination of the left-most column, and the very top row.
[00:14:15] So, where that row (-2.0) and that column (.08) intersect, that means z = -2.08

AL (Customer)
[00:14:22] I see so the point is 0.188

Linda S (Tutor)
[00:14:39] Exactly!
[00:14:53] And, that corresponds to the area to the LEFT of z = -2.08, and that is what the problem asked for.

AL (Customer)
[00:15:07] I see

Linda S (Tutor)
[00:15:18] OK, let me start another window for the other part of the problem.

AL (Customer)
[00:15:27] ok thank you

Linda S (Tutor)
[00:15:48] In this case, we want the area to the RIGHT of z = 2.45. We use the second page of the table, which has positive z-scores.

AL (Customer)
[00:16:19] i see
[00:16:56] the answer would be .9929

Linda S (Tutor)
[00:17:16] But be careful, because the table always gives the area to the LEFT! So, 0.9929 is the area to the LEFT, but what you want is the area to the RIGHT.

AL (Customer)
[00:17:30] ok

Linda S (Tutor)
[00:18:21] So in this case, how do you find the area to the right? What is the total area under the curve?

AL (Customer)
[00:19:32] I am a little confused
[00:19:49] sorry

Linda S (Tutor)
[00:19:54] OK. Well, by definition, the area under the curve is equal to 1.
[00:20:08] Therefore, you use the rule of complements to find the area to the right of z = 2.45.
[00:20:29] Since we know that the area to the left is 0.9929, then the area to the right must equal 1 minus 0.9929.

AL (Customer)
[00:21:07] i see so it is the same answer

Linda S (Tutor)
[00:21:24] Well, it is 1 - 0.9929.
[00:21:46] Did you calculate that?

AL (Customer)
[00:22:06] no i am going to do it now

Linda S (Tutor)
[00:22:14] Sure, go ahead.

AL (Customer)
[00:22:43] I got 0.0071

Linda S (Tutor)
[00:22:48] Yes, exactly!
[00:23:07] OK, does that all make sense?

AL (Customer)
[00:23:34] a little bit now

Linda S (Tutor)
[00:23:49] OK, sounds good.
[00:24:09] Well, I'd better get going, so have a great night!

AL (Customer)
[00:24:39] so that is the answer for the problem

Linda S (Tutor)
[00:24:47] Yes, exactly.

AL (Customer)
[00:24:48] thank you for the help

Linda S (Tutor)
[00:24:51] You bet!
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AL (Customer)
[00:25:04] so made it look easy

Linda S (Tutor)
[00:25:13] ha ha , bye

AL (Customer)
[00:25:20] thank you and you have a good night