# High school Normal distribution two questions

1. Mar 2, 2012

### 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.

2. Mar 2, 2012

### HallsofIvy

Staff Emeritus
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.

3. Mar 3, 2012

### synkk

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 dont get how to get part C.

4. Mar 3, 2012

### Ray Vickson

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

RGV

5. Mar 3, 2012

### synkk

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

6. Mar 3, 2012

### Ray Vickson

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

7. Mar 3, 2012

### synkk

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

8. Mar 3, 2012

### Ray Vickson

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

9. Mar 3, 2012

### synkk

still on question 1) i dont 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...

10. Mar 4, 2012

### Ray Vickson

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

11. Mar 5, 2012

### synkk

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 doesnt it mean it is negative i.e that originally z was in the left hand side so negative.

thanks for continuing to help.

12. Mar 6, 2012

### Ray Vickson

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

13. Mar 6, 2012

### synkk

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

14. Mar 6, 2012

### Ray Vickson

Yes.

RGV

15. Mar 6, 2012

### synkk

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