Normal Distribution Question. Need help

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Homework Help Overview

The discussion revolves around a problem involving a normal distribution characterized by a mean of 80 and a standard deviation of 24. Participants are exploring how to determine the percentage of the distribution that lies outside the range of 32 to 116.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating probabilities using the normal distribution properties and standard normal tables. There are questions about specific calculations and interpretations of probability values.

Discussion Status

Some participants have provided guidance on using probability formulas and standard normal tables. There is an ongoing exploration of different methods and interpretations of the probability values involved, with no explicit consensus reached yet.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. There are references to specific probability values (0.0228 and 0.0668) that are significant to the calculations being discussed.

helix999
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[MENTOR note] Post moved from General Math forum hence no template.

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.
 
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helix999 said:
[MENTOR note] Post moved from General Math forum hence no template.

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.

Use ##P(a < X < b) = P(X < b) - P(X < a)##, and find both ##P(X < a)## and ##P(X < b)## from standard normal tables.
 
Ray Vickson said:
Use ##P(a < X < b) = P(X < b) - P(X < a)##, and find both ##P(X < a)## and ##P(X < b)## from standard normal tables.
The solution I have with me is:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?
 
Maybe start here: http://stattrek.com/m/probability-distributions/standard-normal.aspx
 
Try to break the problem into smaller pieces. What probability do you believe 0.5 - 0.0228 is specifying? (i.e. Probability that X is in what range?).
 
helix999 said:
The solution I have with me is:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?
Try both your method and the one suggested by Ray. They should give the same result. Drawing a picture might help as well.
It might also be interesting to note that (0.5 - 0.0228) + (0.5 - 0.0668) = 1 - 0.0228 - 0.0668. Do you know why those two numbers (0.0228 and 0.0668) are significant in this problem?
 

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