Normal distribution and probability

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Homework Help Overview

The problem involves a normal distribution related to the time until the first failure of a printer brand, characterized by a mean of 1500 hours and a standard deviation of 20 hours. The question seeks to determine the percentage of printers that fail before 1000 hours of operation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the z-score for the given time and explore the implications of that score. There is a mention of using the cumulative distribution function (CDF) to find probabilities, and one participant questions the source of an integral related to the probability density function.

Discussion Status

The discussion includes attempts to calculate probabilities using z-scores and integrals. Some participants confirm that the output of their calculations aligns with expectations, while others express confusion about the integration process. There is no explicit consensus on the interpretation of results, but guidance on using software for calculations has been shared.

Contextual Notes

Participants note the limitations of standard z-tables and the need for software to handle extreme values in the normal distribution context. There is an acknowledgment of the assumptions underlying the normal distribution model being applied to this scenario.

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Homework Statement


The time until the first failure occurred in supplies ink to a particular printer brand, follows a normal distribution with μ=1500 and standard deviation(σ) 20 hours of operation. What percentage of these printers will be damaged before the end of 1000 hours of operation


Homework Equations



Z = (X - μ)/σ

The Attempt at a Solution



Ok first of all i found the z score, which is (1000-1500)/20 = -25

this gives me 0 percentage, actually i have the table in front of me and it gives values for minimum z = -3 or something

so, is there anything i can do to find the exact result?
 
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Well, the value is exactly:

[tex]\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{-25} e^{-\frac{x^2}{2}} dx[/tex]

Using software I get:

>>> from scipy.stats import norm
>>> norm.cdf(-25)
3.056696706382561e-138
 
Last edited:
where did you get this integral from?

it gives me ouput 0, which is the correct asnwer, I guess

edit:

ok i get it now, you integrated the probability density function

thanks
 
Oh, I forgot the [itex]\frac{1}{\sqrt{2\pi}}[/itex]. Edited now.
 

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