Approximating Probability by Central limit theorem.
$begingroup$
A large number of insects are expected to be attracted to a certain variety of rose plant.
A commercial insecticide is advertised as being $99$%$ $ efective. Suppose $2000$ insects infest
a rose garden where the insecticide has been applied, and let $X=$ number of surviving
insects.
Evaluate an approximation to the probability that fewer than
$100$ insects survive
My attempt
$lambda=np=2000*.01=20$
since $100$ is the large value it tends to normal distribution.
$$P(X<100)=P(frac{X-lambda}{sqrtlambda}<frac{100-20}{sqrt 20})=P(frac{X-lambda}{sqrtlambda}<17.89)$$
But there is no value $17.89$ in normal distribution table.
probability statistics probability-distributions normal-distribution
asked Nov 26 '13 at 9:50
user2983722user2983722
426
$endgroup$
add a comment |
$begingroup$
A large number of insects are expected to be attracted to a certain variety of rose plant.
A commercial insecticide is advertised as being $99$%$ $ efective. Suppose $2000$ insects infest
a rose garden where the insecticide has been applied, and let $X=$ number of surviving
insects.
Evaluate an approximation to the probability that fewer than
$100$ insects survive
My attempt
$lambda=np=2000*.01=20$
since $100$ is the large value it tends to normal distribution.
$$P(X<100)=P(frac{X-lambda}{sqrtlambda}<frac{100-20}{sqrt 20})=P(frac{X-lambda}{sqrtlambda}<17.89)$$
But there is no value $17.89$ in normal distribution table.
probability statistics probability-distributions normal-distribution
asked Nov 26 '13 at 9:50
user2983722user2983722
426
$endgroup$
add a comment |
$begingroup$
A large number of insects are expected to be attracted to a certain variety of rose plant.
A commercial insecticide is advertised as being $99$%$ $ efective. Suppose $2000$ insects infest
a rose garden where the insecticide has been applied, and let $X=$ number of surviving
insects.
Evaluate an approximation to the probability that fewer than
$100$ insects survive
My attempt
$lambda=np=2000*.01=20$
since $100$ is the large value it tends to normal distribution.
$$P(X<100)=P(frac{X-lambda}{sqrtlambda}<frac{100-20}{sqrt 20})=P(frac{X-lambda}{sqrtlambda}<17.89)$$
But there is no value $17.89$ in normal distribution table.
probability statistics probability-distributions normal-distribution
asked Nov 26 '13 at 9:50
user2983722user2983722
426
$endgroup$
A large number of insects are expected to be attracted to a certain variety of rose plant.
A commercial insecticide is advertised as being $99$%$ $ efective. Suppose $2000$ insects infest
a rose garden where the insecticide has been applied, and let $X=$ number of surviving
insects.
Evaluate an approximation to the probability that fewer than
$100$ insects survive
My attempt
$lambda=np=2000*.01=20$
since $100$ is the large value it tends to normal distribution.
$$P(X<100)=P(frac{X-lambda}{sqrtlambda}<frac{100-20}{sqrt 20})=P(frac{X-lambda}{sqrtlambda}<17.89)$$
But there is no value $17.89$ in normal distribution table.
probability statistics probability-distributions normal-distribution
probability statistics probability-distributions normal-distribution
asked Nov 26 '13 at 9:50
user2983722user2983722
426
asked Nov 26 '13 at 9:50
user2983722user2983722
426
edited Nov 26 '13 at 10:50
user2983722
asked Nov 26 '13 at 9:50
user2983722user2983722
426
asked Nov 26 '13 at 9:50
user2983722user2983722
426
asked Nov 26 '13 at 9:50
user2983722user2983722
426
426
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Are you sure that you have calculated the variance of the random varibale correctly?
As far as I correctly understood your task, the variance of the random variable (that repestns the survivness) is the following: $0.99^20.1+0.1^20.9=32,8597%$. The variance of the sum of these random variables is equal to $n*D(X)=2000*32,8597=215,82$. Thus, the standard deviation will be equal to square root of the variance - $107,91$. So, you have to look at the following critical value $p_{value}=80/107,91=0,741359$. Its corresponding probability is equal to $0,30$. Thus, the nullhypothesis should be rejected.
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
$endgroup$
add a comment |
$begingroup$
You cannot find it in a printed table, because it is rather extreme. $Phi^{-1}(17.89)$ is extremely close to $1$, perhaps about $1-7times 10^{-72}$. And this makes sense: you expect $20$ to survive and so the probability fewer than $100$ survive is almost $1$.
You could adjust your calculation slightly using $sqrt{np(1-p)}$ for the standard deviation, or having a continuity correction so checking $P(X le 99.5)$ but in fact these are minor (taking the result to about $1-1 times 10^{-71}$) compared with the poor performance of the normal approximation to the binomial in the tail in relative terms.
You can get a better value from the R code, looking at the probability that up to $1900$ die:
> pbinom(1900,2000,0.99)
[1] 6.886295e-38
so the probability fewer than $100$ survive is about $1-7times 10^{-38}$. This is still extreme, and my suspicion is that there is an error in the question.
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f581736%2fapproximating-probability-by-central-limit-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Are you sure that you have calculated the variance of the random varibale correctly?
As far as I correctly understood your task, the variance of the random variable (that repestns the survivness) is the following: $0.99^20.1+0.1^20.9=32,8597%$. The variance of the sum of these random variables is equal to $n*D(X)=2000*32,8597=215,82$. Thus, the standard deviation will be equal to square root of the variance - $107,91$. So, you have to look at the following critical value $p_{value}=80/107,91=0,741359$. Its corresponding probability is equal to $0,30$. Thus, the nullhypothesis should be rejected.
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
$endgroup$
add a comment |
$begingroup$
Are you sure that you have calculated the variance of the random varibale correctly?
As far as I correctly understood your task, the variance of the random variable (that repestns the survivness) is the following: $0.99^20.1+0.1^20.9=32,8597%$. The variance of the sum of these random variables is equal to $n*D(X)=2000*32,8597=215,82$. Thus, the standard deviation will be equal to square root of the variance - $107,91$. So, you have to look at the following critical value $p_{value}=80/107,91=0,741359$. Its corresponding probability is equal to $0,30$. Thus, the nullhypothesis should be rejected.
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
$endgroup$
add a comment |
$begingroup$
Are you sure that you have calculated the variance of the random varibale correctly?
As far as I correctly understood your task, the variance of the random variable (that repestns the survivness) is the following: $0.99^20.1+0.1^20.9=32,8597%$. The variance of the sum of these random variables is equal to $n*D(X)=2000*32,8597=215,82$. Thus, the standard deviation will be equal to square root of the variance - $107,91$. So, you have to look at the following critical value $p_{value}=80/107,91=0,741359$. Its corresponding probability is equal to $0,30$. Thus, the nullhypothesis should be rejected.
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
$endgroup$
Are you sure that you have calculated the variance of the random varibale correctly?
As far as I correctly understood your task, the variance of the random variable (that repestns the survivness) is the following: $0.99^20.1+0.1^20.9=32,8597%$. The variance of the sum of these random variables is equal to $n*D(X)=2000*32,8597=215,82$. Thus, the standard deviation will be equal to square root of the variance - $107,91$. So, you have to look at the following critical value $p_{value}=80/107,91=0,741359$. Its corresponding probability is equal to $0,30$. Thus, the nullhypothesis should be rejected.
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
answered Nov 26 '13 at 15:37
Taras MurzenkovTaras Murzenkov
1363
1363
add a comment |
add a comment |
$begingroup$
You cannot find it in a printed table, because it is rather extreme. $Phi^{-1}(17.89)$ is extremely close to $1$, perhaps about $1-7times 10^{-72}$. And this makes sense: you expect $20$ to survive and so the probability fewer than $100$ survive is almost $1$.
You could adjust your calculation slightly using $sqrt{np(1-p)}$ for the standard deviation, or having a continuity correction so checking $P(X le 99.5)$ but in fact these are minor (taking the result to about $1-1 times 10^{-71}$) compared with the poor performance of the normal approximation to the binomial in the tail in relative terms.
You can get a better value from the R code, looking at the probability that up to $1900$ die:
> pbinom(1900,2000,0.99)
[1] 6.886295e-38
so the probability fewer than $100$ survive is about $1-7times 10^{-38}$. This is still extreme, and my suspicion is that there is an error in the question.
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
$endgroup$
add a comment |
$begingroup$
You cannot find it in a printed table, because it is rather extreme. $Phi^{-1}(17.89)$ is extremely close to $1$, perhaps about $1-7times 10^{-72}$. And this makes sense: you expect $20$ to survive and so the probability fewer than $100$ survive is almost $1$.
You could adjust your calculation slightly using $sqrt{np(1-p)}$ for the standard deviation, or having a continuity correction so checking $P(X le 99.5)$ but in fact these are minor (taking the result to about $1-1 times 10^{-71}$) compared with the poor performance of the normal approximation to the binomial in the tail in relative terms.
You can get a better value from the R code, looking at the probability that up to $1900$ die:
> pbinom(1900,2000,0.99)
[1] 6.886295e-38
so the probability fewer than $100$ survive is about $1-7times 10^{-38}$. This is still extreme, and my suspicion is that there is an error in the question.
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
$endgroup$
add a comment |
$begingroup$
You cannot find it in a printed table, because it is rather extreme. $Phi^{-1}(17.89)$ is extremely close to $1$, perhaps about $1-7times 10^{-72}$. And this makes sense: you expect $20$ to survive and so the probability fewer than $100$ survive is almost $1$.
You could adjust your calculation slightly using $sqrt{np(1-p)}$ for the standard deviation, or having a continuity correction so checking $P(X le 99.5)$ but in fact these are minor (taking the result to about $1-1 times 10^{-71}$) compared with the poor performance of the normal approximation to the binomial in the tail in relative terms.
You can get a better value from the R code, looking at the probability that up to $1900$ die:
> pbinom(1900,2000,0.99)
[1] 6.886295e-38
so the probability fewer than $100$ survive is about $1-7times 10^{-38}$. This is still extreme, and my suspicion is that there is an error in the question.
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
$endgroup$
You cannot find it in a printed table, because it is rather extreme. $Phi^{-1}(17.89)$ is extremely close to $1$, perhaps about $1-7times 10^{-72}$. And this makes sense: you expect $20$ to survive and so the probability fewer than $100$ survive is almost $1$.
You could adjust your calculation slightly using $sqrt{np(1-p)}$ for the standard deviation, or having a continuity correction so checking $P(X le 99.5)$ but in fact these are minor (taking the result to about $1-1 times 10^{-71}$) compared with the poor performance of the normal approximation to the binomial in the tail in relative terms.
You can get a better value from the R code, looking at the probability that up to $1900$ die:
> pbinom(1900,2000,0.99)
[1] 6.886295e-38
so the probability fewer than $100$ survive is about $1-7times 10^{-38}$. This is still extreme, and my suspicion is that there is an error in the question.
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
answered Dec 5 '15 at 1:11
HenryHenry
100k480166
100k480166
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f581736%2fapproximating-probability-by-central-limit-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
