AP Statistics Curriculum 2007 Distrib Multinomial - Revision history

Tdlee: Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

2020-03-03T16:43:11Z

Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

IvoDinov: /* Examples of Multinomial experiments */

2013-04-17T01:41:53Z

‎Examples of Multinomial experiments

IvoDinov: /* Examples of Multinomial experiments */

2013-04-17T01:40:55Z

‎Examples of Multinomial experiments

IvoDinov: /* Problems */

2011-04-25T15:49:54Z

‎Problems

Jenny: /* Approximate Solution */

2010-06-28T19:51:50Z

‎Approximate Solution

Jenny: /* SOCR Multinomial Examples */

2010-06-28T19:51:25Z

‎SOCR Multinomial Examples

Jenny at 17:00, 28 June 2010

2010-06-28T17:00:06Z

IvoDinov: /* SOCR Multinomial Examples */ clarified the calculations

2010-05-05T17:25:32Z

‎SOCR Multinomial Examples: clarified the calculations

IvoDinov: /* Exact Solution */ typo

2010-05-05T17:17:32Z

‎Exact Solution: typo

IvoDinov: /* SOCR Multinomial Examples */ fixed typos

2010-05-05T17:17:13Z

‎SOCR Multinomial Examples: fixed typos

← Older revision		Revision as of 16:43, 3 March 2020
Line 113:		Line 113:
	* SOCR Home page: http://www.socr.ucla.edu		* SOCR Home page: http://www.socr.ucla.edu

−	{{translate\|pageName=http://wiki.~~stat~~.~~ucla~~.edu~~/socr~~/index.php?title=AP_Statistics_Curriculum_2007_Distrib_Multinomial}}	+	"{{translate\|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_Distrib_Multinomial}}

@@ Line 20: / Line 20: @@
 Plugging these values into the multinomial formula we get the probability of the event of interest to be:
-: <math>P(A) = {5\choose r_1,r_2,r_3}p_1^{r_1}p_2^{r_2}p_3^{r_3}</math>
+: \(P(A) = {5\choose r_1,r_2,r_3}p_1^{r_1}p_2^{r_2}p_3^{r_3}\). In this specific case, \(P(A) = {5\choose 0,2,3}p_1^{0}p_2^{2}p_3^{3}\).
 : <math>P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (4/9)^3=0.0975461.</math>

@@ Line 20: / Line 20: @@
 Plugging these values into the multinomial formula we get the probability of the event of interest to be:
-: <math>P(A) = {5\choose 0, 2, 3}p_1^{r_1}p_2^{r_2}p_3^{r_3}</math>
+: <math>P(A) = {5\choose r_1,r_2,r_3}p_1^{r_1}p_2^{r_2}p_3^{r_3}</math>
 : <math>P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (4/9)^3=0.0975461.</math>

@@ Line 104: / Line 104: @@
 Note that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical compared to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.
-===[[hEBook_Problems_Distrib_Multinomial|Problems]]===
+===[[EBook_Problems_Distrib_Multinomial|Problems]]===
 ===See also===

@@ Line 102: / Line 102: @@
 : <math>P(A) \approx {8 \over 1,000}=0.008</math>.
-Note that that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical compared to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.
+Note that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical compared to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.
 ===[[hEBook_Problems_Distrib_Multinomial|Problems]]===

@@ Line 68: / Line 68: @@
 ===SOCR Multinomial Examples===
-Suppose we row 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (''one'' is the most likely and ''six'' is the least likely outcome).
+Suppose we roll 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (''one'' is the most likely and ''six'' is the least likely outcome).
 <center>
 {| class="wikitable" style="text-align:center; width:75%" border="1"

@@ Line 1: / Line 1: @@
 ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Multinomial Random Variables and Experiments==
-The multinomial experiments (and multinomial distributions) directly extend the their [[AP_Statistics_Curriculum_2007_Distrib_Binomial |bi-nomial counterparts]].
+The multinomial experiments (and multinomial distributions) directly extend their [[AP_Statistics_Curriculum_2007_Distrib_Binomial |bi-nomial counterparts]].
 ===Multinomial experiments===
@@ Line 13: / Line 13: @@
 *  Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue (2+3+4=9). We randomly select 5 marbles from the urn, ''with replacement''. What is the probability (''P(A)'') of the event ''A={selecting 2 green marbles and 3 blue marbles}''?
-*To solve this problem, we apply the multinomial formula. We know the following:
+*To solve this problem, we apply the multinomial formula, we know the following:
 ** The experiment consists of 5 trials, so k = 5.
 ** The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so <math>r_1=r_{red} = 0</math>, <math>r_2=r_{green} = 2</math>, and <math>r_3=r_{blue} = 3</math>.
@@ Line 102: / Line 102: @@
 : <math>P(A) \approx {8 \over 1,000}=0.008</math>.
-Note that that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical approximation to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.
+Note that that this approximation is close to the exact answer above. By the [[AP_Statistics_Curriculum_2007_Limits_LLN | Law of Large Numbers (LLN)]], we know that this SOCR empirical compared to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.
 ===[[hEBook_Problems_Distrib_Multinomial|Problems]]===

@@ Line 88: / Line 88: @@
 :<math>=0.00586690138260962656816896.</math>
-* Using the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multinomial Distribution Calculator]: Enter the above given information in the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multimonial distribution applet] to get the probability density and cumulative distribution values for the given outcome ''{3,3,2,2,0}'', as shown on the image below.
+* Using the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multinomial Distribution Calculator]: Enter the above given information in the [http://socr.ucla.edu/htmls/dist/Multinomial_Distribution.html SOCR Multimonial distribution applet] to get the probability density and cumulative distribution values for the given outcome ''{3,3,2,2,0}'', as shown on the image below. Note that since the event A does not contain the die outcomes of 5 or 6 , we can reduce the case of 6 outcomes <math>\{1,2,3,4,5,6 \}</math> to a case of 5 outcomes <math>\{1,2,3,4,other \}</math> with corresponding probabilities <math>\{0.286, 0.238, 0.19, 0.143, 0.143 \}</math>, where we pool together the probabilities of the die outcomes of 5 and 6, i.e., P(other)=0.095+0.048=0.143.
 <center>[[Image:SOCR_EBook_Dinov_Multinimial_102209_Fig2.png|500px]]</center>

@@ Line 85: / Line 85: @@
 * By-hand calculations:
-: <math>P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 0! \times 0!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.095^0 \times 0.048^0}= </math>
+: <math>P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 0! \times 0!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.095^0 \times 0.048^0= </math>
 :<math>=0.00586690138260962656816896.</math>

@@ Line 79: / Line 79: @@
 : ''P(A)=?'' Note that the complete description of the event of interest is:
-: A={3 ones, 3 twos, 2 threes, 2 fours, and 2 others (5's or 6's!)}
+: A={3 ones, 3 twos, 2 threes, 2 fours, and 0 others (5's or 6's!)}
 ====Exact Solution====
@@ Line 85: / Line 85: @@
 * By-hand calculations:
-: <math>P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 2!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.143^2 = </math>
+: <math>P(A) = {10! \over 3!\times 3! \times 2! \times 2! \times 0! \times 0!} \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 \times 0.095^0 \times 0.048^0}= </math>
 :<math>=0.00586690138260962656816896.</math>