Understanding Z Distribution for Lightbulb Lifetime

In summary, the problem is easy but the way to solve it confuses the student. The area in between 798.1 and 901.7 is .1915 and the area from 0 to 0.6 is .2258. So in order to find the number of bulbs in between those ranges, my professor added the two areas.
  • #1
hyddro
74
2
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
Yeah mathman, sorry that's what I meant!
 

1. What is a Z Distribution for lightbulbs?

A Z Distribution for lightbulbs is a statistical probability distribution that is used to model the likelihood of different outcomes for lightbulbs. It is based on the concept of a normal distribution and is often used in quality control and manufacturing processes.

2. How is a Z Distribution for lightbulbs calculated?

A Z Distribution for lightbulbs is calculated using the mean and standard deviation of a sample of lightbulbs. The formula for calculating the Z Distribution is (x - μ)/σ, where x is the value of a single lightbulb, μ is the mean of the sample, and σ is the standard deviation of the sample.

3. What is the purpose of using a Z Distribution for lightbulbs?

The purpose of using a Z Distribution for lightbulbs is to determine the probability of a lightbulb having a certain characteristic or attribute. This can help manufacturers identify potential issues with their production processes and improve the overall quality of their products.

4. Can a Z Distribution for lightbulbs be used for any type of lightbulb?

Yes, a Z Distribution for lightbulbs can be used for any type of lightbulb as long as the data follows a normal distribution. However, it may be more useful for lightbulbs with specific characteristics or attributes that need to be closely monitored for quality control purposes.

5. Are there any limitations to using a Z Distribution for lightbulbs?

One limitation of using a Z Distribution for lightbulbs is that it assumes a normal distribution of data. If the data does not follow a normal distribution, the results may not be accurate. Additionally, the Z Distribution may not be suitable for identifying outliers or extreme values in the data.

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