# AP Statistics Curriculum 2007 Normal Std

(Difference between revisions)
 Revision as of 20:16, 31 January 2008 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 20:18, 31 January 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 1: Line 1: - ====A Game of Chance==== + ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Standard Normal Variables and Experiments== - * Suppose we are offered to play a game of chance under these conditions: it costs us to play $1.50 and the awarded prices are {$1, $2,$3}. Assume the probabilities of winning each price are {0.6, 0.3, 0.1}, respectively. Should we play the game? What are our chances of winning/loosing? Let's let X=awarded price. Then X={1, 2, 3}. + - + -
+ - {| class="wikitable" style="text-align:center; width:75%" border="1" + - |- + - | x  || 1 || 2 || 3 + - |- + - | P(X=x) || 0.6 || 0.3 || 0.1 + - |- + - | x*P(X=x) || 0.6 || 0.6 || 0.3 + - |} + -
+ - Then the mean of this game (i.e., expected return or expectation) is computed as the weighted (by the outcome probabilities) average of all the outcome prices: $E[X] = x_1P(X=x_1) + x_2P(X=x_2)+x_3P(X=x_3) = 1\times 0.6 + 2\times 0.3 + 3\times 0.1 = 1.5$. In other words, the expected return of this came is 1.5, which equals the entry fee, and hence the game is fair - neither the player nor the house has an advantage in this game (on the long run!) Of course, each streak of n games will produce different outcomes and may give small advantage to one side, however, on the long run, no one will make money. + - + - The variance for this game is computed by $VAR[X] = (x_1-1.5)^2P(X=x_1) + (x_2-1.5)^2P(X=x_2)+(x_3-1.5)^2P(X=x_3) =$ + - $=0.25\times 0.6 + 0.25\times 0.3 + 2.25\times 0.1 = 0.45$. Thus, the standard deviation is $SD[X] = \sqrt{VAR[X]}=0.67$. + - + - * Suppose now we ''alter the rules for the game of chance'' and the new pay-off is as follows: + - + - {| class="wikitable" style="text-align:center; width:75%" border="1" + - |- + - | x || 0 || 1.5 || 3 + - |- + - | P(X=x) || 0.6 || 0.3 || 0.1 + - |- + - | x*P(X=x) || 0 || 0.45 || 0.3 + - |} + - + - + - ** What is the ''new expected return'' of the game? Remember, the old expectation was equal to the entrance fee of1.50, and the game was fair! + - ** The change in the pay-off of the game may be represented by this linear transformation $Y = {3(X-1)\over 2}$. Therefore, by our rules for computing expectations of linear functions, $E[Y] = 3/2 E(X) –3/2 = 3/4 = 0.75$, and the game became clearly biased. Note how easy it is to compute ''E[Y]'', using this formula. At the same time, we could have computed the expectation of ''Y'' using first-principles (adding the values of the last row in the revised table above)! + - + - * You can play similar games under different conditions for the probability distribution of the prices using the SOCR [[SOCR_EduMaterials_Activities_BinomialCoinExperiment | Binomial Coin]] or [[SOCR_EduMaterials_Activities_DiceExperiment | Die]] experiments. ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Standard Normal Variables and Experiments== + === Standard Normal Distribution=== === Standard Normal Distribution===

## General Advance-Placement (AP) Statistics Curriculum - Standard Normal Variables and Experiments

### Standard Normal Distribution

The standard normal distribution is a continuous distribution where the following exact areas are bound between the Standard Normal Density function and the x-axis on the symmetric intervals around the origin:

• The area: -1 < z < 1 = 0.8413 - 0.1587 = 0.6826
• The area: -2.0 < z < 2.0 = 0.9772 - 0.0228 = 0.9544
• The area: -3.0 < z < 3.0 = 0.9987 - 0.0013 = 0.9974
• Standard Normal density function $f(x)= {e^{-x^2} \over \sqrt{2 \pi}}.$
• The Standard Normal distribution is also a special case of the more general normal distribution where the mean is set to zero and a variance to one. The Standard Normal distribution is often called the bell curve because the graph of its probability density resembles a bell.

### Experiments

Suppose we decide to test the state of 100 used batteries. To do that, we connect each battery to a volt-meter by randomly attaching the positive (+) and negative (-) battery terminals to the corresponding volt-meter's connections. Electrical current always flows from + to -, i.e., the current goes in the direction of the voltage drop. Depending upon which way the battery is connected to the volt-meter we can observe positive or negative voltage recordings (voltage is just a difference, which forces current to flow from higher to the lower voltage.) Denote Xi={measured voltage for battery i} - this is random variable 0 and assume the distribution of all Xi is Standard Normal, $X_i \sim N(0,1)$. Use the Normal Distribution (with mean=0 and variance=1) in the SOCR Distribution applet to address the following questions. This Distributions help-page may be useful in understanding SOCR Distribution Applet. How many batteries, from the sample of 100, can we expect to have?

• Absolute Voltage > 1? P(X>1) = 0.1586, thus we expect 15-16 batteries to have voltage exceeding 1.
• |Absolute Voltage| > 1? P(|X|>1) = 1- 0.682689=0.3173, thus we expect 31-32 batteries to have absolute voltage exceeding 1.
• Voltage < -2? P(X<-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than -2.
• Voltage <= -2? P(X<=-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than or equal to -2.
• -1.7537 < Voltage < 0.8465? P(-1.7537 < X < 0.8465) = 0.761622, thus we expect 76 batteries to have voltage in this range.