Hey guys sorry for the late post my relative had gone in to the hospital, so i only got home at 11... Also please bare with me on today's class's scribe because i am not understanding the work very well and am still learning. so anything i have posting that is incorrect please leave a comment so i can correct it which possibly some help
Today we worked with Z-scores and the normal curve
Before we got into it MR.k had given us a Littlee quiz on determining the length of arrowheads one standard deviation below and one above the mean of numbers. Also on knowing what percentage of arrowheads are within one standard deviation of one mean length.
We then corrected the quiz and handed it back
Next we talked about curving marks..
The graph above represents the marks of a large number of students with the mean mark of 69.3% and stranded deviation of 7
so this shows that the students with the marks of 62.3 to 76.3 are the mean marks of the class and are about 68-69 percent are in that mean. as the others shows above.
So knowing this then
68% of the marks are between 62.3-76.3 percent
34 percent are between 69.3 and 76.3 percent
50% are below 69.3%
and 16% are above
So since we have been doing non standerd deviations till this point which means the mean and standard deviation of the distribution or problem have been the standard deviation and mean of the data. now we will try a
standard normal deviation
scale on the x-axis on the window of your calculation is the Z-score(standard score) which is the mean. and standard deviation is 1 so since it is a probability distribution the area under the curve=1
which means it is 100% chance of every score being included in this problem(distribution)
WINDOW
(then add your y and x numbers along with the xscl etc..)
QUIT
2nd VARS(dist)
(then move to the right to draw)
then press number 1 which is shadenorm
then in the open bracket put in the standard deviation and mean i think
close bracket
ENTER
your graph should be shown..
I'm trying and hoping to succeed
next scribe is...KATIE
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