Solve the problem that involves Normal distribution

In summary: I made a mistake ...find the corrected version here;In summary,The interest is on part (c), and the take is that when z =-0.66666, the area subtended between the two x points would be equal to 0.3546.
  • #1
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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  • #2
What do you mean with the probability of -2/3? The density of the distribution there?
 
  • #3
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...
 
  • #4
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##
 
  • #5
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.
 
  • #6
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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  • #7
It's OK, making mistakes is how we learn ;)
 
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  • #8
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...
 
  • Like
Likes Maarten Havinga

What is Normal distribution?

Normal distribution, also known as Gaussian distribution, is a probability distribution that is symmetric and bell-shaped. It is often used to model natural phenomena such as height, weight, and IQ scores.

How do you calculate the mean and standard deviation of a Normal distribution?

The mean of a Normal distribution can be calculated by taking the sum of all the data points and dividing it by the total number of data points. The standard deviation can be calculated by taking the square root of the variance, which is the average squared difference from the mean.

What is the 68-95-99.7 rule in Normal distribution?

The 68-95-99.7 rule states that approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations in a Normal distribution.

How do you use the Normal distribution to solve real-world problems?

The Normal distribution can be used to solve real-world problems by determining the probability of a certain event occurring. This can be done by finding the area under the curve using the mean and standard deviation, and then converting it to a percentage.

What are some common applications of Normal distribution in science?

Normal distribution is commonly used in science to analyze and interpret data, such as in medical research to study the effectiveness of a new drug, or in environmental science to study the distribution of pollutants in a population. It is also used in quality control to monitor the consistency of a product or process.

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