SOLUTION: Students who score in the top 10 percent are recognized publicly for their achievement by the Department of the Treasury. Assuming a normal distribution, how many standard deviati
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Question 1163298: Students who score in the top 10 percent are recognized publicly for their achievement by the Department of the Treasury. Assuming a normal distribution, how many standard deviations above the mean does a student have to score to be publicly recognized? (Round your answer to 2 decimal places.) Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
You'll need to use a calculator or a table. I recommend a calculator. I'm using this one here. You can use any (online) resource that you find preferable.
If you use the calculator I posted, then click the button that says "Value from an area". Type in 0.10 for the area. Leave the mean and standard deviation as they are (0 and 1 respectively). Click the button labeled "above" to have the value 1.282 show up. You may need to click the "recalculate" button.
This means P(Z > 1.282) = 0.10 approximately. So 10% of the population is above z = 1.282; when rounding to two decimal places, we get z = 1.28
We are approximately 1.28 standard deviations above the mean. A positive z score means we are above the mean z = 0, a negative score means we're below the mean. The absolute value of the z score tells us the distance from the mean.
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If you are using a Texas Instruments (TI) calculator, such as a TI83, then press the key labeled "2ND" and then press the VARS key.
Scroll down to option 3.
Type in 0.90 and hit enter
Your TI calculator should show invNorm(0.90) = 1.281551567 approximately. This shows P(Z < 1.281551567) = 0.90 which is equivalent to saying P(Z > 1.281551567) = 0.10
This leads to z = 1.282 and z = 1.28 as found earlier.
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