# Z Distribution for lightbulls

## Main Question or Discussion Point

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 dont explain why, please any help? I would like to know why the added them Tahnks!

## Answers and Replies

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mathman
Science Advisor
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 dont get it.

chiro
Science Advisor
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.

mathman
Science Advisor
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.

chiro
Science Advisor
Yeah mathman, sorry that's what I meant!