Further holiday homework

[Albeno69]

I think im doing:
UNIT 3 - No sacs til term 2
Data Analysis - univariate and bivariate
Trigonometry and geometry
UNIT 4
Matrices


Mine the same as you 
Networks

[mandy]

well school wants to do
trig
matrices
networks
Through i wanna change for the exam
business math
sequences and patterns
wbu

my school's doing trig and geometry, business math and relations and graphs.
i think i'm just gonna stick with what the school does for the exams too, lol.
so i don't need to learn anything extra :]

MANDYYYY!!!
SAME SAME SAME

GREAT. NOW WE CAN WORK TOGETHER SOMETIME :]

[mandy]

BELL SHAPED DATA SETS.
I have a question that I don't understand.

The heights climbed by a group of rock climbers are normally distributed. A certain rock climber, Jack, has a standardised height climbed of 2 (i.e. z = 2).
a. Where does the height Jack climbed sit in terms of standard deviations from the mean?
The height Jack climbed is 2 standard deviations above the mean.

b. Approximately what percentage of climbers did not climb as high as Jack?
c. If there are 400 climbers climbing that year, how many of them will have climbed higher than Jack?

I don't know what to do to get the percentage in b.

[mandy]

Helppp anyone?

[amirite?]

Is there a graph/picture that goes with this question?

[mandy]

Is there a graph/picture that goes with this question?

Nah, nothing. Just the info they give in words :S

[amirite?]

The question is pretty vague too. Do you count the people who climbed higher because they didnt climb as high as Jack?

I suck at standard deviation, I try to do it in my head rather than on paper or with a calculator. :\ No wonder. If you cant get help here just make sure youve got and understanding of standard deviation and perhaps highlight the question so you can ask someone at a later date.

[Aden]

Since it asks for the 'approximate' percentage, I'm guessing it just wants you to use general facts regarding standard normal distributions. Basically it goes like this:

68% lie within 1 standard deviation from the mean
95% lie within 2 standard deviations from the mean
99.7% lie within 3 standard deviations from the mean

Since Jack is 2 deviations above the mean, then the percentage of people who didn't climb as high as him would be:
%

Sorry if it's wrong

[mandy]

Since it asks for the 'approximate' percentage, I'm guessing it just wants you to use general facts regarding standard normal distributions. Basically it goes like this:

68% lie within 1 standard deviation from the mean
95% lie within 2 standard deviations from the mean
99.7% lie within 3 standard deviations from the mean

Since Jack is 2 deviations above the mean, then the percentage of people who didn't climb as high as him would be:
%

Sorry if it's wrong

Thats the right answer! But how did you get the and the ?

[Aden]

I think I failed at making the curve in paint, but I hope it helps your understanding:

[stonecold]

Make sure you guys know know normal distributions and z scores very well.  I can guarentee you there will a question on it in the exams, most likely multi choice.

You guys seem to get it which is good.    And you won't be asked to calculate any values other than the percentages 68, 95 or 99.7 so don't worry about that.  It wont come up.

I suggest everyone take a diagram similar to Aden's one into the exam as well.

[mandy]

Thanks so much Aden, but I still don't get where you got those values from :S
Am I meant to see it somewhere on the curve you drew?

Thanks stonecold, you did so well in Further this year! I'll be remembering all your tips ;]

[Aden]

Since the question asks for the percentage of climbers who did not climb as high as Jack, you want to find the total percentage of people who are below 2 standard deviations above the mean. If you look at the attached image, we want to find the total percentage that is inside that red rectangle thing. Since a standard normal distribution is symmetrical at the mean (at 0, where there is 50% above and 50% below), then you know that the total percentage below 0 is 50%. Then there is the bit from 0 to 2, which is 47.5% (half of 95%).

[mandy]

Since the question asks for the percentage of climbers who did not climb as high as Jack, you want to find the total percentage of people who are below 2 standard deviations above the mean. If you look at the attached image, we want to find the total percentage that is inside that red rectangle thing. Since a standard normal distribution is symmetrical at the mean (at 0, where there is 50% above and 50% below), then you know that the total percentage below 0 is 50%. Then there is the bit from 0 to 2, which is 47.5% (half of 95%).

Omg, I see it now. The fool I am, lol. Thanks so muchhhhh.
Wait, you're not doing further, how do you know all this :O

[Aden]

You cover this in Methods as well.