Binominal distribution “fewer than”












0














I have this exercise:




We learned in Exercise 3.26 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults.



(a) What is the mean and standard deviation.



(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?




I have used Z score to calculate it, since both $np$ and $n(1-p)$ is above 10, we can use normal distribution. First I calculated mean and SD from exercise (a) for a binominal distribution, then the Z score:



Mean:
$μ=np$



$μ=120*0.90$



$μ=108$



Then the Standard Deviation:
$σ=sqrt{np(1-p)}$



$σ=sqrt{120*0.90(1-0.90)}$



$σ=sqrt{10.8}$



$σ=3.28$



Now I calculate the Z-score:



$Z=(observation-mean)/SD$



$Z= (105-108)/3.28$



$Z=-0.91$



We will look up that number on the Z-table: 0.1814 = 18.14%



Is this correctly done? I tried looking it up on the internet, but I see people getting 21% when they use pbinom(105,120,.9) from the R language = 0.2181634 = 21%
The same website calculates it this way, without R and still gets 21%
enter image description here






probability statistics binomial-distribution







share|cite|improve this question






















  • The image at the end doesn't match what you are trying to say (how to arrive at $21%$ without using R).
    – Lee David Chung Lin
    Dec 1 at 13:40












  • Here are the calculations by using the binomial distribution and the approximation by using the normal distribution (with continuity correction factor): Bin, Norm. The upper bound at the normal distribution is $frac{105+0.5-108}{sqrt{0.1cdot 0.9cdot 120}}$. Using the continuity correction factor is the right way like in the answer of rzch.
    – callculus
    Dec 1 at 14:34










  • When you round $21.81634%$ to the the nearest integer it is $color{red}{22%}$ and not 21%
    – callculus
    Dec 1 at 14:40


















0














I have this exercise:




We learned in Exercise 3.26 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults.



(a) What is the mean and standard deviation.



(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?




I have used Z score to calculate it, since both $np$ and $n(1-p)$ is above 10, we can use normal distribution. First I calculated mean and SD from exercise (a) for a binominal distribution, then the Z score:



Mean:
$μ=np$



$μ=120*0.90$



$μ=108$



Then the Standard Deviation:
$σ=sqrt{np(1-p)}$



$σ=sqrt{120*0.90(1-0.90)}$



$σ=sqrt{10.8}$



$σ=3.28$



Now I calculate the Z-score:



$Z=(observation-mean)/SD$



$Z= (105-108)/3.28$



$Z=-0.91$



We will look up that number on the Z-table: 0.1814 = 18.14%



Is this correctly done? I tried looking it up on the internet, but I see people getting 21% when they use pbinom(105,120,.9) from the R language = 0.2181634 = 21%
The same website calculates it this way, without R and still gets 21%
enter image description here






probability statistics binomial-distribution







share|cite|improve this question






















  • The image at the end doesn't match what you are trying to say (how to arrive at $21%$ without using R).
    – Lee David Chung Lin
    Dec 1 at 13:40












  • Here are the calculations by using the binomial distribution and the approximation by using the normal distribution (with continuity correction factor): Bin, Norm. The upper bound at the normal distribution is $frac{105+0.5-108}{sqrt{0.1cdot 0.9cdot 120}}$. Using the continuity correction factor is the right way like in the answer of rzch.
    – callculus
    Dec 1 at 14:34










  • When you round $21.81634%$ to the the nearest integer it is $color{red}{22%}$ and not 21%
    – callculus
    Dec 1 at 14:40
















0












0








0







I have this exercise:




We learned in Exercise 3.26 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults.



(a) What is the mean and standard deviation.



(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?




I have used Z score to calculate it, since both $np$ and $n(1-p)$ is above 10, we can use normal distribution. First I calculated mean and SD from exercise (a) for a binominal distribution, then the Z score:



Mean:
$μ=np$



$μ=120*0.90$



$μ=108$



Then the Standard Deviation:
$σ=sqrt{np(1-p)}$



$σ=sqrt{120*0.90(1-0.90)}$



$σ=sqrt{10.8}$



$σ=3.28$



Now I calculate the Z-score:



$Z=(observation-mean)/SD$



$Z= (105-108)/3.28$



$Z=-0.91$



We will look up that number on the Z-table: 0.1814 = 18.14%



Is this correctly done? I tried looking it up on the internet, but I see people getting 21% when they use pbinom(105,120,.9) from the R language = 0.2181634 = 21%
The same website calculates it this way, without R and still gets 21%
enter image description here






probability statistics binomial-distribution







share|cite|improve this question













I have this exercise:




We learned in Exercise 3.26 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults.



(a) What is the mean and standard deviation.



(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?




I have used Z score to calculate it, since both $np$ and $n(1-p)$ is above 10, we can use normal distribution. First I calculated mean and SD from exercise (a) for a binominal distribution, then the Z score:



Mean:
$μ=np$



$μ=120*0.90$



$μ=108$



Then the Standard Deviation:
$σ=sqrt{np(1-p)}$



$σ=sqrt{120*0.90(1-0.90)}$



$σ=sqrt{10.8}$



$σ=3.28$



Now I calculate the Z-score:



$Z=(observation-mean)/SD$



$Z= (105-108)/3.28$



$Z=-0.91$



We will look up that number on the Z-table: 0.1814 = 18.14%



Is this correctly done? I tried looking it up on the internet, but I see people getting 21% when they use pbinom(105,120,.9) from the R language = 0.2181634 = 21%
The same website calculates it this way, without R and still gets 21%
enter image description here






probability statistics binomial-distribution




probability statistics binomial-distribution






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share|cite|improve this question











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asked Dec 1 at 11:50









StudentCoderJava

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  • The image at the end doesn't match what you are trying to say (how to arrive at $21%$ without using R).
    – Lee David Chung Lin
    Dec 1 at 13:40












  • Here are the calculations by using the binomial distribution and the approximation by using the normal distribution (with continuity correction factor): Bin, Norm. The upper bound at the normal distribution is $frac{105+0.5-108}{sqrt{0.1cdot 0.9cdot 120}}$. Using the continuity correction factor is the right way like in the answer of rzch.
    – callculus
    Dec 1 at 14:34










  • When you round $21.81634%$ to the the nearest integer it is $color{red}{22%}$ and not 21%
    – callculus
    Dec 1 at 14:40




















  • The image at the end doesn't match what you are trying to say (how to arrive at $21%$ without using R).
    – Lee David Chung Lin
    Dec 1 at 13:40












  • Here are the calculations by using the binomial distribution and the approximation by using the normal distribution (with continuity correction factor): Bin, Norm. The upper bound at the normal distribution is $frac{105+0.5-108}{sqrt{0.1cdot 0.9cdot 120}}$. Using the continuity correction factor is the right way like in the answer of rzch.
    – callculus
    Dec 1 at 14:34










  • When you round $21.81634%$ to the the nearest integer it is $color{red}{22%}$ and not 21%
    – callculus
    Dec 1 at 14:40


















The image at the end doesn't match what you are trying to say (how to arrive at $21%$ without using R).
– Lee David Chung Lin
Dec 1 at 13:40






The image at the end doesn't match what you are trying to say (how to arrive at $21%$ without using R).
– Lee David Chung Lin
Dec 1 at 13:40














Here are the calculations by using the binomial distribution and the approximation by using the normal distribution (with continuity correction factor): Bin, Norm. The upper bound at the normal distribution is $frac{105+0.5-108}{sqrt{0.1cdot 0.9cdot 120}}$. Using the continuity correction factor is the right way like in the answer of rzch.
– callculus
Dec 1 at 14:34




Here are the calculations by using the binomial distribution and the approximation by using the normal distribution (with continuity correction factor): Bin, Norm. The upper bound at the normal distribution is $frac{105+0.5-108}{sqrt{0.1cdot 0.9cdot 120}}$. Using the continuity correction factor is the right way like in the answer of rzch.
– callculus
Dec 1 at 14:34












When you round $21.81634%$ to the the nearest integer it is $color{red}{22%}$ and not 21%
– callculus
Dec 1 at 14:40






When you round $21.81634%$ to the the nearest integer it is $color{red}{22%}$ and not 21%
– callculus
Dec 1 at 14:40












1 Answer
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oldest

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I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).



To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $left(105.5 - 108right)/3.28$ instead of $left(105 - 108right)/3.28$. Using this continuity correction gives an answer of $22.3%$ which is a bit closer to the true probability compared to your answer.



Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.






share|cite|improve this answer





















  • Thank you very much :)
    – StudentCoderJava
    Dec 1 at 16:41











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I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).



To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $left(105.5 - 108right)/3.28$ instead of $left(105 - 108right)/3.28$. Using this continuity correction gives an answer of $22.3%$ which is a bit closer to the true probability compared to your answer.



Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.






share|cite|improve this answer





















  • Thank you very much :)
    – StudentCoderJava
    Dec 1 at 16:41
















1














I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).



To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $left(105.5 - 108right)/3.28$ instead of $left(105 - 108right)/3.28$. Using this continuity correction gives an answer of $22.3%$ which is a bit closer to the true probability compared to your answer.



Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.






share|cite|improve this answer





















  • Thank you very much :)
    – StudentCoderJava
    Dec 1 at 16:41














1












1








1






I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).



To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $left(105.5 - 108right)/3.28$ instead of $left(105 - 108right)/3.28$. Using this continuity correction gives an answer of $22.3%$ which is a bit closer to the true probability compared to your answer.



Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.






share|cite|improve this answer












I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).



To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $left(105.5 - 108right)/3.28$ instead of $left(105 - 108right)/3.28$. Using this continuity correction gives an answer of $22.3%$ which is a bit closer to the true probability compared to your answer.



Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 14:12









rzch

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  • Thank you very much :)
    – StudentCoderJava
    Dec 1 at 16:41


















  • Thank you very much :)
    – StudentCoderJava
    Dec 1 at 16:41
















Thank you very much :)
– StudentCoderJava
Dec 1 at 16:41




Thank you very much :)
– StudentCoderJava
Dec 1 at 16:41


















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