Use technology or the z-distribution table on pages A-1 to A-2 in the reference guide to find the indicated area. The scores on a standardized test are normally distributed with a mean of 500 and a standard deviation of 100. Jake scored 500 on the test. What percent of students scored below Jake?

Ray is a bit off track, there are indeed a few mathematicians around Fixya.

Here you are dealing with the standard error of a proportion. The proportion of students of interest in the sample is 540/970 = 0.5567. This is usually denoted by p.

Now you want the standard error of this proportion, s. It is
SQRT ( p (1-p) / N ) where N is sample size. Note the error is smaller for big samples. Here s = 0.0160.

Now we turn to the Normal Distribution. We want the Z-statistic which corresponds to a confidence level of 92%, or 0.92. Z is the number of Standard Deviations on the Normal Curve which takes in 92% of the area under the curve. You can look this up in any table of the Normal Distribution, or with an online calculator

http://onlinestatbook.com/2/calculators/normal.html

This gives Z = 1.751. So the true result is somewhere in p ± (1.751 * 0.0160)
or 0.5567 ± 0.0279. The margin of error is 0.0279.

Now I don't know here if your tutor wants you to use a Continuity Correction, which is sometimes applied because we are approximating the Normal Distribution here. If so this is
0.5 / N = 0.5 / 970 = 0.0005.
This small amount is added to the Margin of Error, to make 0.0284.

A good reference for you to use is

http://onlinestatbook.com/2/estimation/proportion_ci.html
http://onlinestatbook.com/2/index.html

About 0.002

Check the user guide:

http://www.arrisi.com/support/documentation/user_guides/_docs/tm822_user_guide_std1-2.pdf

Refer to the table on page 43 for what the LEDs are indicating.

Your calculator is giving you the correct answer for a one-tailed distribution. 0.99 of the area is to the left of 2.32635, with 0.01 to the right.

For a two-tailed distribution, 0.99 of the area is between -2.5758 and +2.5758, with 0.01 to the outside.

Midpoints ( x + y)/2 = x47.567.57988.597

Boundaries for lower class -0.5 and +.0.5 for upper.39.5 - 55.559.5 - 75.575.5 - 82.582.5 - 94.594.5 - 99.5
I just don't get how the number of students factor in? It's still an average.

Press 2nd and Distr and find the InvNorm function. Enter InvNorm(2.13).

If you wanted to find the percentage that fall above, you would enter
1-InvNorm(2.13)

The Casio FX-115W/991W/115E/991ES have three functions P(, Q(, R( and t(: the last one calculates the normalized variable noted t.



The Sharp EL-520W ans several similar Sharps do it too. Here is a screen capture.

The Casio FX-115W/991W/115E/991ES have three functions P(, Q(, R( and t(: the last one calculates the normalized variable noted t.



The Sharp EL-520W ans several similar Sharps do it too. Here is a screen capture.

This table uses low voltages for the scoring lever. I agree with fixer1950. Check / disassemble the goal / scoring assembly, and look for the scoring lever. sometimes just 2 metal plates. should see where it's touching the scoring strip and you can test it by moving it with your finger / pen / other device.

Try adjusting it further away, and test again with a puck to ensure proper operation..