# AP Statistics Curriculum 2007 Limits Bin2HyperG

(Difference between revisions)
 Revision as of 18:55, 14 June 2007 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 01:41, 3 February 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 2: Line 2: === Binomial Approximation to HyperGeometric Distribution=== === Binomial Approximation to HyperGeometric Distribution=== - Example on how to attach images to Wiki documents in included below (this needs to be replaced by an appropriate figure for this section)! + Suppose [[AP_Statistics_Curriculum_2007_Distrib_Dists | $X \sim Hypergeometric(m, N, n)]] and p=m/N. - [[Image:AP_Statistics_Curriculum_2007_IntroVar_Dinov_061407_Fig1.png|500px]] + - ===Approach=== + *If [itex]n=1$ then $X$ has a [[AP_Statistics_Curriculum_2007_Distrib_Binomial  | Bernoulli distribution]] with parameter $p$. - Models & strategies for solving the problem, data understanding & inference. + - * TBD + *If $N and [itex]m$ are large compared to $n$ and $p$ is not close to 0 or 1, then $P(X \le x) \approx P(Y \le x)$, where $Y$ has a [[AP_Statistics_Curriculum_2007_Distrib_Binomial  | Binomial($n$, $p$) distribution]]. - + - ===Model Validation=== + - Checking/affirming underlying assumptions. + - + - * TBD + - + - ===Computational Resources: Internet-based SOCR Tools=== + - * TBD + ===Examples=== ===Examples=== - Computer simulations and real observed data. + An urn contains 50 marbles (35 green and 15 white). Fifteen marbles are selected without replacement.  Find the probability that exactly 10 out of the 15 selected are green marbles.  The answer to this question can be found using the formula: + $P(X=10)=\frac{{35 \choose 10}{15 \choose 5}}{{50 \choose 15}}=0.2449.$  Using SOCR simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below. +
[[Image: SOCR_Activities_ExploreDistributions_Christou_figure2.jpg|600px]]
- * TBD + Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained.  This is equal to - + $P(X=1)=\frac{{35 \choose 1}{15 \choose 1}}{{50 \choose 2}}=0.4286.$  This is also shown on the figure below. - ===Hands-on activities=== +
[[Image: SOCR_Activities_ExploreDistributions_Christou_figure3.jpg|600px]]
- Step-by-step practice problems. + - * TBD + We will approximate the probability of obtaining 1 green marble using Binomial as follows.  Select the SOCR Binomial distribution and choose number of trials 2 and probablity of success $p=\frac{35}{50}=0.7$.  Compare the figure below with the figure above.  They are almost the same!  Why?  Using the Binomial formula we can compute the approximate probability of observing 1 green marble as $P(X=1)={2 \choose 1}0.70^10.30^1=0.42$ (very close to the exact probability, 0.4286). +
[[Image: SOCR_Activities_ExploreDistributions_Christou_figure4.jpg|600px]]

## General Advance-Placement (AP) Statistics Curriculum - Binomial Approximation to HyperGeometric Distribution

### Binomial Approximation to HyperGeometric Distribution

Suppose $X \sim Hypergeometric(m, N, n)$ and p = m / N.

• If N and m are large compared to n and p is not close to 0 or 1, then $P(X \le x) \approx P(Y \le x)$, where Y has a Binomial(n, p) distribution.

### Examples

An urn contains 50 marbles (35 green and 15 white). Fifteen marbles are selected without replacement. Find the probability that exactly 10 out of the 15 selected are green marbles. The answer to this question can be found using the formula: $P(X=10)=\frac{{35 \choose 10}{15 \choose 5}}{{50 \choose 15}}=0.2449.$ Using SOCR simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below.

Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained. This is equal to $P(X=1)=\frac{{35 \choose 1}{15 \choose 1}}{{50 \choose 2}}=0.4286.$ This is also shown on the figure below.

We will approximate the probability of obtaining 1 green marble using Binomial as follows. Select the SOCR Binomial distribution and choose number of trials 2 and probablity of success $p=\frac{35}{50}=0.7$. Compare the figure below with the figure above. They are almost the same! Why? Using the Binomial formula we can compute the approximate probability of observing 1 green marble as $P(X=1)={2 \choose 1}0.70^10.30^1=0.42$ (very close to the exact probability, 0.4286).

• TBD
The text passage contains a statistical topic about the binomial approximation to the hypergeometric distribution, including examples with calculations and visual representations using SOCR (Statistical Online Computational Resource) tools.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# AP Statistics Curriculum 2007 Limits Bin2HyperG

(Difference between revisions)
 Revision as of 18:55, 14 June 2007 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 01:41, 3 February 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 2: Line 2: === Binomial Approximation to HyperGeometric Distribution=== === Binomial Approximation to HyperGeometric Distribution=== - Example on how to attach images to Wiki documents in included below (this needs to be replaced by an appropriate figure for this section)! + Suppose [[AP_Statistics_Curriculum_2007_Distrib_Dists | $X \sim Hypergeometric(m, N, n)]] and p=m/N. - [[Image:AP_Statistics_Curriculum_2007_IntroVar_Dinov_061407_Fig1.png|500px]] + - ===Approach=== + *If [itex]n=1$ then $X$ has a [[AP_Statistics_Curriculum_2007_Distrib_Binomial  | Bernoulli distribution]] with parameter $p$. - Models & strategies for solving the problem, data understanding & inference. + - * TBD + *If $N and [itex]m$ are large compared to $n$ and $p$ is not close to 0 or 1, then $P(X \le x) \approx P(Y \le x)$, where $Y$ has a [[AP_Statistics_Curriculum_2007_Distrib_Binomial  | Binomial($n$, $p$) distribution]]. - + - ===Model Validation=== + - Checking/affirming underlying assumptions. + - + - * TBD + - + - ===Computational Resources: Internet-based SOCR Tools=== + - * TBD + ===Examples=== ===Examples=== - Computer simulations and real observed data. + An urn contains 50 marbles (35 green and 15 white). Fifteen marbles are selected without replacement.  Find the probability that exactly 10 out of the 15 selected are green marbles.  The answer to this question can be found using the formula: + $P(X=10)=\frac{{35 \choose 10}{15 \choose 5}}{{50 \choose 15}}=0.2449.$  Using SOCR simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below. +
[[Image: SOCR_Activities_ExploreDistributions_Christou_figure2.jpg|600px]]
- * TBD + Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained.  This is equal to - + $P(X=1)=\frac{{35 \choose 1}{15 \choose 1}}{{50 \choose 2}}=0.4286.$  This is also shown on the figure below. - ===Hands-on activities=== +
[[Image: SOCR_Activities_ExploreDistributions_Christou_figure3.jpg|600px]]
- Step-by-step practice problems. + - * TBD + We will approximate the probability of obtaining 1 green marble using Binomial as follows.  Select the SOCR Binomial distribution and choose number of trials 2 and probablity of success $p=\frac{35}{50}=0.7$.  Compare the figure below with the figure above.  They are almost the same!  Why?  Using the Binomial formula we can compute the approximate probability of observing 1 green marble as $P(X=1)={2 \choose 1}0.70^10.30^1=0.42$ (very close to the exact probability, 0.4286). +
[[Image: SOCR_Activities_ExploreDistributions_Christou_figure4.jpg|600px]]

## General Advance-Placement (AP) Statistics Curriculum - Binomial Approximation to HyperGeometric Distribution

### Binomial Approximation to HyperGeometric Distribution

Suppose $X \sim Hypergeometric(m, N, n)$ and p = m / N.

• If N and m are large compared to n and p is not close to 0 or 1, then $P(X \le x) \approx P(Y \le x)$, where Y has a Binomial(n, p) distribution.

### Examples

An urn contains 50 marbles (35 green and 15 white). Fifteen marbles are selected without replacement. Find the probability that exactly 10 out of the 15 selected are green marbles. The answer to this question can be found using the formula: $P(X=10)=\frac{{35 \choose 10}{15 \choose 5}}{{50 \choose 15}}=0.2449.$ Using SOCR simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below.

Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained. This is equal to $P(X=1)=\frac{{35 \choose 1}{15 \choose 1}}{{50 \choose 2}}=0.4286.$ This is also shown on the figure below.

We will approximate the probability of obtaining 1 green marble using Binomial as follows. Select the SOCR Binomial distribution and choose number of trials 2 and probablity of success $p=\frac{35}{50}=0.7$. Compare the figure below with the figure above. They are almost the same! Why? Using the Binomial formula we can compute the approximate probability of observing 1 green marble as $P(X=1)={2 \choose 1}0.70^10.30^1=0.42$ (very close to the exact probability, 0.4286).

• TBD
user
The text passage contains a statistical topic about the binomial approximation to the hypergeometric distribution, including examples with calculations and visual representations using SOCR (Statistical Online Computational Resource) tools.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS