Solve the problem that involves Normal distribution

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Homework Help Overview

The discussion revolves around a problem involving the normal distribution, specifically focusing on calculating probabilities and Z-scores related to a given dataset. Participants are analyzing part (c) of the problem, which requires understanding the properties of the normal distribution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate probabilities associated with specific Z-scores and are discussing the meaning of probability in the context of continuous distributions. There are questions about the validity of certain probability statements and the correct interpretation of Z-scores.

Discussion Status

Several participants have offered insights into the calculations and interpretations of the probabilities involved. There is an ongoing exploration of different methods to approach the problem, with some participants questioning the assumptions made in earlier posts. The discussion is active, with multiple interpretations being considered.

Contextual Notes

Participants are aware of the requirement for answers to be expressed to five significant figures, and there are indications of corrections being made to earlier calculations. The conversation reflects a collaborative effort to clarify misunderstandings regarding the use of probabilities in the context of the normal distribution.

chwala
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Homework Statement
See attached
Relevant Equations
stats
1654172872696.png
1654172908106.png
My interest is on part (c),
My take,

##Z=\dfrac{160−200}{60}=−0.666666##

##Pr(−0.66666)=0.3546##

##⇒\dfrac{x_1-200}{60}=1.05##

##x_1=63+200=263##

Yes, i am aware that they want the answer to ##5## significant figures...i just wanted to check the alternative method...
Appreciate your insight...
 
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What do you mean with the probability of -2/3? The density of the distribution there?
 
chwala said:
Homework Statement:: See attached
Relevant Equations:: stats

View attachment 302276View attachment 302277My interest is on part (c),
My take,

##Z=\dfrac{160−200}{60}=−0.666666##

##Pr(−0.66666)=0.3546##
One approach is to calculate P(X > 160). Your work should include the probabilities that some random variable is greater than or less than some number. Pr(-.66666) is meaningless as a probability, a pointed hinted at by the previous poster.

To get you started, calculate Pr(X > 160) = Pr(##\frac{X - 160}{60} > \frac{160 - 200}{60}##) = Pr(Z > -0.66667)##. That should give you a number that is larger than 0.6. You should then be able to find the Z score (and hence the X value) that gives you the correct probability, using a table of z-scores.
chwala said:
##⇒\dfrac{x_1-200}{60}=1.05##

##x_1=63+200=263##

Yes, i am aware that they want the answer to ##5## significant figures...i just wanted to check the alternative method...
Appreciate your insight...
 
Mark44 said:
One approach is to calculate P(X > 160). Your work should include the probabilities that some random variable is greater than or less than some number. Pr(-.66666) is meaningless as a probability, a pointed hinted at by the previous poster.

To get you started, calculate Pr(X > 160) = Pr(##\frac{X - 160}{60} > \frac{160 - 200}{60}##) = Pr(Z > -0.66667)##. That should give you a number that is larger than 0.6. You should then be able to find the Z score (and hence the X value) that gives you the correct probability, using a table of z-scores.
My point was that when ##z =-0.66666## Then the area subtended between the two x points would be equal to ##0.3546##
 
Maarten Havinga said:
What do you mean with the probability of -2/3? The density of the distribution there?
Yes I meant the density...I shouldn't have used term probability...I hope that makes sense...otherwise, I need to check again...and I should have used inequalities instead of equal sign...Will amend that later.
 
chwala said:
My point was that when ##z =-0.66666##
Then that is incorrect since Pr(Z = -0.66666) = 0. For a continuous probability, any probability statement should involve a random variable in an inequality.
 
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It's OK, making mistakes is how we learn ;)
 
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Maarten Havinga said:
It's OK, making mistakes is how we learn ;)

Yes, in my approach i first found the area bound by ##160## and ##X.## I found that to be; ##0.2454## then i took ##0.6-0.2454=0.3542## ...
chwala said:
Homework Statement:: See attached
Relevant Equations:: stats

View attachment 302276View attachment 302277My interest is on part (c),
My take,

##Z=\dfrac{160−200}{60}=−0.666666##

##Pr(−0.66666)=0.2454##

chwala said:
##⇒\dfrac{x_1-200}{60}=1.05##

##x_1=63+200=263##

Yes, i am aware that they want the answer to ##5## significant figures...i just wanted to check the alternative method...
Appreciate your insight...

chwala said:
Homework Statement:: See attached
Relevant Equations:: stats

View attachment 302276View attachment 302277My interest is on part (c),
My take,

##Z=\dfrac{160−200}{60}=−0.666666##

##Pr(−0.66666)=0.3546##

##⇒\dfrac{x_1-200}{60}=1.05##

##x_1=63+200=263##

Yes, i am aware that they want the answer to ##5## significant figures...i just wanted to check the alternative method...
Appreciate your insight...
I made a mistake ...find the corrected version here;

My interest is on part (c),
My take,

##Z=\dfrac{160−200}{60}=−0.666666##

##Pr(−0.66666)=0.2454##

##0.6-0.2454=0.3542##

##0.3542= Z_{1.05}## Therefore

##⇒\dfrac{x_1-200}{60}=1.05##

##x_1=63+200=263##

Yes, i am aware that they want the answer to ##5## significant figures...i just wanted to check the alternative method...
Appreciate your insight...
 
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