Statistics and Standard Deviation?

Question

Can someone please explain how to work this problem out and exactly why that is the answer? I did it, but im not sure if I am right and the data that I have does not make sense to me.

Show which is relatively better, and why: a score of 88 on the first test, or a score of 70 on the second test. Scores on the first test have a mean 80 and a standard deviation of 10. Scores on the second test have a mean 64 and a standard deviation of 6.

I did a little chart, but im not even sure if it is right. (I had to add the ex's cause the data came out weird without it. The chat might still not come out right on here for some reason.)

0 10 20 30 40 50 60 70 80 |90 100 110
xxxxxxxxxxxxxxxxxxxxxxxxx-----68----
xxxxxxxxxxxxxxxxxxxxx----------95----------
xxxxxxxxxxxxxxx ----------------99.7---------------


66 |72 78 64 70 76 82 88
xxxxxxx -----68----
xxxx----------95----------
---------------99.7--------------

Answer 1

Hi,
Well, I think you are on the right track. You could calculate the z-scores, but fortunately the scores can be interpreted in a straightforward way without that. Notice this:
1) The person who scored 70 is exactly one standard deviation above the mean: 64 +6 = 70.
2) Whereas, the person who scored 88 is slightly under one standard deviation above the mean. 80 + 10 = 90 is one SD above the mean. So, the person who scored 70 ranks a little higher in his class than the one who scored 88.
I don’t know if you might have a TI-83 Plus or TI-84 calculator at your disposal, but you can easily find what percentage of the class is equal to or below the scores using the following steps with one of those calculators:
1. normalcdf(: Area under a curve between two points with μ (mean) and σ (std. dev.) given.
a) Press 2nd, DISTR, 2. The term "normalcdf(" will appear on the home screen.
b) Enter the number for the left boundary, right boundary, μ, and σ in that order. You do
not need to close the parentheses, but it's okay if you do.
c) Press ENTER and the value of the area between the two points will be displayed. Notice
that you do not explicitly convert the points to z-values as you would have if you had
done it by hand.
Your Ex. 1: Assume a normal distribution of values for which the mean is 80 and the std.
dev. is 10.
a) Complete item a) above.
b) Enter numbers so that your display is the following: normalcdf(-1E9,88,80,10.
c) Press ENTER and you'll get 0.78814 which is, of course, 78.8 percent.

To do the second problem, just press 2ND, ENTRY (the enter key); then change the values to the second problem. You’ll get .8413… or 84.13 % are below this student. Once you get the hand of this, you can check that out in no more than 30 seconds.

Incidentally, I just copied this procedure from my Website at this URL: www.anglefire.com/pro/fkizer

You’re welcome to go there and make single copies of anything you wish.

Hope this helps.
FE

Answer 2

You want to compare z's. Whichever z-score is greater tells you which score is in the higher percentile and thus is relatively better.

z = (x - μ) / σ

Test 1:
z = (88 - 80) / 10
= 8 / 10
= 4 / 5

Test 2:
z = (70 - 64) / 6
= 6 / 6
= 1

1 > 4/5

The score on the second test was relatively better.

Answer 3

To standardize the scores (based on a normal distribution)
General formula: Z = (x-µ)/σ,
So,
Z1 = (88-80)/10 = 0.80 The percentile is 78.8%
Z2 = (70-64)/6 = 1.0 The percentile is 84.1%
So the 2nd test is in a higher %-tile.

Answer 4

think quite some scholars went into do an examination. The greater the greater advantageous. On a countrywide scale, if each and every of the outcomes have been taken and plotted onto a graph, some scholars will do poorly, some will do very properly and the final public will lie in between the two extremes. by ability of human nature, the graph plotted would be a common distribution graph, it is going to look like a bell shape layout. the utmost ingredient of the graph stands out as the propose, and at this ingredient while a line is drawn down vertically to decrease the x-axis, it is going to divide the graph in 2 equivalent areas. the form of shape is what's undemanding as a common distribution curve and that's symmetrical approximately it extremely is centre line which coincides with the propose (x bar) of the observations. the known deviation is a level of how this archives is unfold out. One known deviation of the two facets of the propose (x bar ±?) will incorporate sixty 8% of the information. 2 known deviation of the two facets of the propose (x bar ±2?) will incorporate ninety 5% of the information. 3 known deviation of the two facets of the propose (x bar ±3?) will incorporate ninety 9?7% of the information. those values observe for a common distribution. So the known distribution is a level of the unfold of archives from the propose.