Question about HP 33s Calculator

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Programing the hp33s

I need to compute a series of 20 measurements and produce a mean & standard deviation quickly. this is for a timed exam. I have the manual and tried to follow their steps to no satisfaction. Please help

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  • glenne2001 Jan 26, 2009

    refer to the 33s manual; Statistical Operations 11-1 entering one variable data. once you have entered the data using the top left key (Sums+) use the other top keys to select yor answer.


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Check out chapter 11 in the user's guide. This chapter gives you the run down on how to calculate what you are looking for.

Posted on Dec 02, 2010

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What is deviation standard

The standard deviation is a measure of how "tight" the samples are distributed around your mean.
In layman's terms, a small standard deviation indicates that most of your measurements are in the vicinity of the means; a large standard deviation corresponds to readings that are all over the place.
You could also say that the smaller the SD, the more your mean is representative of the data set.
For a better explanation, just look up Standard deviation on Wikipedia!

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How do I calculate z score on TI-36X Pro?

To convert an x value to a z-score, the formula is:
(x value - mean)/standard deviation

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Hope this helps!

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Find the standard deviation of the following data set 5,5,5,5

The standard deviation is a measure on the variability of a number set from its average. Since all numbers in your set are the same, there is no variation from its average, and the standard deviation is 0.

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Question no. 2 For the 50 observation given below. Prepare Frequency Distribution for the the class interval is 10 (lower limit starting is 0) and upper limit is 80. Also plot in graphing. 64 62 52 50 70...

64 62 52 50 70 60 55 20 25 48 35 55 55 47 64 42 36 50 30 53 51 63 58 50 60 54 33 48 40 76 61 46 65 65 15 45 58 40 64 18 33 40 64 18 33 40 48 9 38 41

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Standard deviation from grouped data: 15.40

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If both tests are graed on a curve, who did better and why?

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Maths & statistic

Standard deviation is the average root mean squared deviation from the average of the numbers.

Familiarize yourself with the wikipedia page on standard deviation:

In this case, your question is easy. The standard deviation is 0, so this means the boys ages are all the same.

So they're all 24/3 = 8 years old.

8 + 8 + 8 = 24 <-- satisfies the constraint that the boys ages must total to 24.

Get the standard deviation of these numbers;

8+8+8 / 3 = 8 <-- the average value is 8

The deviation of each boy's age from the average is:
boy 1
8 years old, which deviates from 8 by 0
boy 2
8 years old, which deviates from 8 by 0
boy 3
8 years old, which deviates from 8 by 0

So the deviations are 0, 0 and 0.

To get the standard deviation, you sum the squares of the deviations of each boy, get the average, and square root. So:

0² + 0² + 0² = 0
Average is 0 / 3 = 0
Square root of 0 = 0

So the standard deviation is 0. Which shouldn't be much of a surprise to you. We just had to actually DO the work to show that it was in fact 0.

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