# High school Normal distribution two questions

• synkk
In summary: Never mind trying to understand whether (45 - mu) / sigma is 0.67, just tell me: what do YOU think the value of...
synkk
TITLE: Normal distributions (sorry)

Question 1:

Working:

Now the answer for sigma i got correct, but μ i got incorrect for some reason. Could anyone explain where I've gone wrong for μ?

Question 2:

Workings:

All i need help is with part c and d:

for c) i done 1 - 0.6463 to get the probability of the two shaded bits, but where do i go to get the part they are asking for?

for d) i have no idea, any help would be great.
thanks.

Well, one obvious error is that you have both $(25-\mu)/\sigma$ and $(64-\mu)/\sigma$ equal to -0.67. One of them should be positive. In any case, the normal distribution is symmertric so it should be clear that the mean will be exactly between the two "quartiles".

The two shaded bits? I assume you mean the one shaded bit and the other outlieing area. That, of course, is 1- .6463. To find the two parts separately, use the fact, as you are given, that upper piece has twice the area of the lower piece.

HallsofIvy said:
Well, one obvious error is that you have both $(25-\mu)/\sigma$ and $(64-\mu)/\sigma$ equal to -0.67. One of them should be positive. In any case, the normal distribution is symmertric so it should be clear that the mean will be exactly between the two "quartiles".

The two shaded bits? I assume you mean the one shaded bit and the other outlieing area. That, of course, is 1- .6463. To find the two parts separately, use the fact, as you are given, that upper piece has twice the area of the lower piece.

for question 1) Why is one of the positive (the second one) if they are symmetrical. Also how can i use use symmetry to find the mean? Is it 25+45 / 2?

for question 2) I still don't get how to get part C.

synkk said:
for question 1) Why is one of the positive (the second one) if they are symmetrical. Also how can i use use symmetry to find the mean? Is it 25+45 / 2?

for question 2) I still don't get how to get part C.

Is it 25+45 / 2? Close, but not quite: it IS (25 + 45)/2---what you wrote is 25 + 22.5. Brackets matter!

RGV

Ray Vickson said:
Is it 25+45 / 2? Close, but not quite: it IS (25 + 45)/2---what you wrote is 25 + 22.5. Brackets matter!

RGV

Could you explain why it is (25 + 45) / 2? I don't understand how the symmetry makes this true.

synkk said:
Could you explain why it is (25 + 45) / 2? I don't understand how the symmetry makes this true.

The distance between the mean and the 25th percentile is the same as the distance between the mean and the 75th percentile, because of symmetry. Therefore, the mean is the average of the 25th and 75th percentiles.

RGV

Ray Vickson said:
The distance between the mean and the 25th percentile is the same as the distance between the mean and the 75th percentile, because of symmetry. Therefore, the mean is the average of the 25th and 75th percentiles.

RGV

Alright thank you. Do you have any idea on how to do question 2 part c and d?

synkk said:
Alright thank you. Do you have any idea on how to do question 2 part c and d?

Yes, but I think you have already been given enough hints by others. Just look at the diagram, sit down, relax, and _think_. Don't try to write down the answer right away: approach it systematically, piece-by-piece, and don't worry if it takes longer than you think it should. After you see what is happening you can go back and clean it up.

RGV

Ray Vickson said:
Yes, but I think you have already been given enough hints by others. Just look at the diagram, sit down, relax, and _think_. Don't try to write down the answer right away: approach it systematically, piece-by-piece, and don't worry if it takes longer than you think it should. After you see what is happening you can go back and clean it up.

RGV

still on question 1) i don't understand why (45 - mu) / sigma is positive 0.67, could you explain please?

for c i done (1-0.6463)/3 as there is 3 parts, with the shaded region being the smallest, which gets me the correct value.

For D I've standardized it but I'm not sure what value i should put it against...

synkk said:
still on question 1) i don't understand why (45 - mu) / sigma is positive 0.67, could you explain please?

for c i done (1-0.6463)/3 as there is 3 parts, with the shaded region being the smallest, which gets me the correct value.

For D I've standardized it but I'm not sure what value i should put it against...

Never mind trying to understand whether (45 - mu) / sigma is 0.67, just tell me: what do YOU think the value of (45 - mu)/sigma should be? Remember, read the whole question carefully before answering.

RGV

Ray Vickson said:
Never mind trying to understand whether (45 - mu) / sigma is 0.67, just tell me: what do YOU think the value of (45 - mu)/sigma should be? Remember, read the whole question carefully before answering.

RGV

I think it should equal -0.67 because P(z>(45-mu)/sigma) = 0.75 meaning it is in the left hand side of the distribution, so we reflect it and we are trying to find P(z<(45-mu)/sigma) which is 0.67, but because we reflected it doesn't it mean it is negative i.e that originally z was in the left hand side so negative.

thanks for continuing to help.

synkk said:
I think it should equal -0.67 because P(z>(45-mu)/sigma) = 0.75 meaning it is in the left hand side of the distribution, so we reflect it and we are trying to find P(z<(45-mu)/sigma) which is 0.67, but because we reflected it doesn't it mean it is negative i.e that originally z was in the left hand side so negative.

thanks for continuing to help.

You should avoid just blindly using formulas you do not really understand. Instead: think! If 45 is the 75th percentile, and μ is the 50th percentile (also equal to the mean), we MUST HAVE 45 > μ, so (45 - μ)/σ must be > 0. It cannot possibly be -0.67, which is a negative number.
Your "equation" P(z>(45-mu)/sigma) = 0.75 is wrong, but would be OK if you replaced the 0.75 by ___ ? (I'm leaving it to you to fill in the blank.) Always, always, draw a picture.

RGV

by 0.25? so P(Z>(45-mu)/sigma) = 0.25 ? Is this correct?

synkk said:
by 0.25? so P(Z>(45-mu)/sigma) = 0.25 ? Is this correct?

Yes.

RGV

Ray Vickson said:
Yes.

RGV

Thank you for all the help, i done question 2 part d also.

## What is a normal distribution in high school?

A normal distribution in high school refers to a statistical concept that is commonly seen in many aspects of high school education, such as test scores, grades, and other performance measures. It is a type of probability distribution that is symmetrical and bell-shaped, with the majority of data points falling near the mean or average, and fewer points towards the extremes.

## Why is normal distribution important in high school?

Normal distribution is important in high school because it allows us to understand and analyze large sets of data that are typically seen in high school settings. It helps us identify trends, patterns, and outliers in student performance, which can be useful for improving teaching methods and identifying areas of improvement for students.

## How is normal distribution used in high school math?

Normal distribution is used in high school math to calculate probabilities and make predictions about a range of outcomes, such as test scores or grades. It is also used in data analysis and statistics courses to help students understand and interpret data sets.

## What are the characteristics of a normal distribution in high school?

The main characteristics of a normal distribution in high school include a symmetrical bell-shaped curve, with the majority of data points falling near the mean or average, and fewer points towards the extremes. It is also characterized by the mean, median, and mode being equal, and the standard deviation being a measure of spread around the mean.

## How can normal distribution be applied in high school science?

Normal distribution can be applied in high school science to analyze and interpret data collected from experiments or research. It can help determine the probability of a certain outcome or the likelihood of a particular trend seen in the data. It can also be used to compare data sets and determine if there are any significant differences between them.

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