Discrete probability distributions

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SUMMARY

The discussion focuses on discrete probability distributions, specifically regarding the performance of batteries advertised to last an average of 100 minutes with a standard deviation of 5 minutes. The user successfully calculated the interval containing at least 90% of the performance periods as (85, 115) minutes. The user speculates on the likelihood of batteries failing before 80 minutes, questioning the statistical implications of this interval. The need for understanding normal distribution tables is highlighted for further analysis.

PREREQUISITES
  • Understanding of normal distribution and its properties
  • Familiarity with standard deviation and mean calculations
  • Knowledge of confidence intervals in statistics
  • Ability to interpret statistical tables for normal distributions
NEXT STEPS
  • Study the concept of confidence intervals in statistics
  • Learn how to use Z-scores to determine probabilities in normal distributions
  • Explore statistical tables for normal distributions
  • Investigate the implications of standard deviation on performance metrics
USEFUL FOR

Students studying statistics, data analysts, and anyone interested in understanding battery performance metrics and their statistical analysis.

Mdhiggenz
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Homework Statement



A certain manufacture advertises batteries that will run under a 75 amp discharge test for an average of 100 minutes, with standard deviation of 5 minutes.

a. find an interval that must contain at least 90% of the performance periods fr batteries of this type.

b. would you expect many batteries to die out in less than 80min why or why not?

I solved a. to be (85,115)

however I'm not quite sure what to do about B. I would guess and say that know due to the interval which is from 85 to 115 which would mean that the battery life generally last in that range however, I'm just speculating.


Thanks

Higgenz


Homework Equations





The Attempt at a Solution

 
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How many standard deviations would 80 min be from the average? Do you have any tables which tell you the percentage of a normal distribution that would be that far or further below the mean?
 

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