Understanding Z Distribution for Lightbulb Lifetime

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Discussion Overview

The discussion revolves around understanding the application of the Z Distribution in determining the expected fraction of lightbulbs with lifetimes within a specific range, based on a given average and standard deviation. Participants are exploring the calculations involved in finding the area under the curve of the Z Distribution and the reasoning behind the addition of areas from the Z table.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant describes a problem involving the average lifetime and standard deviation of lightbulbs and expresses confusion about the method used to find the area between two Z-scores.
  • Another participant clarifies that the area is obtained from -0.5 to +0.6, which justifies the need to add the two areas from the Z table.
  • A participant questions the equality of areas on either side of the y-axis, indicating a misunderstanding of how areas are calculated in relation to the Z Distribution.
  • Further clarification is provided regarding the properties of the standard normal distribution, specifically the relationship between positive and negative Z-scores.
  • One participant acknowledges a correction regarding the properties of the Z Distribution, confirming the intended meaning of the earlier statement.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the properties of the Z Distribution, but there remains confusion about the specific calculations and the rationale for adding versus subtracting areas. The discussion reflects differing understandings of the concepts involved.

Contextual Notes

Some participants express uncertainty about the relationship between areas under the curve and the properties of the standard normal distribution, indicating potential gaps in understanding the underlying principles.

hyddro
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So, I have been studying for my chemistry quiz and I came across a problem that I can't really get, or rather, part of the solution.

The problem itself is easy, it's just the way my teacher solved what confuses me. Here it is and thanks in advance.

There was an histogram for lightbulbs where <x> (average lifetime) = 845.2 hr and s (standar deviation) = 94.2 hr.

b) What fraction of the bulbs is expected to have a lifetime between 798.1 and 901.7?

So here is how i would approach the problem. Given the Z Distribution table.

Z1 = (x - <x>)/s = (798.1 - 845.2)/94.2 = -0.5 or just 0.5; since the graph is the same on either side of the y axis, the area from 0 to 0.5 and from 0 to -0.5 will be the same.
Z2 = (901.7 - 845.2)/94.2 = 0.6

So i go to the table, and I find that the area from 0 to 0.5 is .1915 and the area from 0 to 0.6 is .2258.

Now here is where i get confused. In order to find the area in between (and eventually the number of bulbs in between 798.1 and 901.7) my professor ADDED the two areas. That blew my mind, I thought you were supposed to substract the area from 0 to 0.5 from the area from 0 to 0.6 to get the area in between... but no, she added them and the book does the same thing but they don't explain why, please any help? I would like to know why the added them Tahnks!
 
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You are getting the area from -0.5 to +0.6. That's why you need to add.

You would subtract if you were getting the area from +0.5 to +0.6.
 
hi, but aren't the areas the same though? I mean, positive cause they are, you know, areas? lol sorry but i don't get it.
 
The areas are the same when you look at them around the origin (i.e. the y-axis).

For a standard normal you have the property that P(Z < a) = P(Z > a) where a is a positive real number.
 
chiro said:
The areas are the same when you look at them around the origin (i.e. the y-axis).

For a standard normal you have the property that P(Z < a) = P(Z > a) where a is a positive real number.

I think you meant P(Z < -a) = P(Z > a).
For his problem it is P(-a < Z < 0) = P(0 < Z < a) for a > 0.
 
Yeah mathman, sorry that's what I meant!
 

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