Central limit theorem. Calculating probability P(N≤49)
$begingroup$
Here is the problem:
Apples are being packed in a box. One apple weight is expected to be 200 g with a dispersion of 20 g.
Packing is stopped as soon as the total weight is 10 kg or more. Calculate the probability $$P(N≤49)$$
when $N$ is the number of apples in the box.
I assume that I have to use central limit theorem somehow. I have done similar exercises before but this one is a little bit different and I just can't get it started.
central-limit-theorem
asked Dec 19 '18 at 13:45
jtejte
336
$endgroup$
add a comment |
$begingroup$
Here is the problem:
Apples are being packed in a box. One apple weight is expected to be 200 g with a dispersion of 20 g.
Packing is stopped as soon as the total weight is 10 kg or more. Calculate the probability $$P(N≤49)$$
when $N$ is the number of apples in the box.
I assume that I have to use central limit theorem somehow. I have done similar exercises before but this one is a little bit different and I just can't get it started.
central-limit-theorem
asked Dec 19 '18 at 13:45
jtejte
336
$endgroup$
add a comment |
$begingroup$
Here is the problem:
Apples are being packed in a box. One apple weight is expected to be 200 g with a dispersion of 20 g.
Packing is stopped as soon as the total weight is 10 kg or more. Calculate the probability $$P(N≤49)$$
when $N$ is the number of apples in the box.
I assume that I have to use central limit theorem somehow. I have done similar exercises before but this one is a little bit different and I just can't get it started.
central-limit-theorem
asked Dec 19 '18 at 13:45
jtejte
336
$endgroup$
Here is the problem:
Apples are being packed in a box. One apple weight is expected to be 200 g with a dispersion of 20 g.
Packing is stopped as soon as the total weight is 10 kg or more. Calculate the probability $$P(N≤49)$$
when $N$ is the number of apples in the box.
I assume that I have to use central limit theorem somehow. I have done similar exercises before but this one is a little bit different and I just can't get it started.
central-limit-theorem
central-limit-theorem
asked Dec 19 '18 at 13:45
jtejte
336
asked Dec 19 '18 at 13:45
jtejte
336
asked Dec 19 '18 at 13:45
jtejte
336
asked Dec 19 '18 at 13:45
jtejte
336
asked Dec 19 '18 at 13:45
jtejte
336
336
add a comment |
add a comment |
1 Answer
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$X_i$ be the weight of $i$-th apple. $P(Nleq 49)=P(sum_{i=1}^{49}X_igeq 10000)=P(bar Xgeq 10000/49)=P(sqrt{49}(bar X-200)/20geq sqrt{49}(10000/49-200)/20)approxPhi(7(10000/49-200)/20)$
where $Phi$ is the cdf of $N(0,1).$
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
$endgroup$
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
1
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
$X_i$ be the weight of $i$-th apple. $P(Nleq 49)=P(sum_{i=1}^{49}X_igeq 10000)=P(bar Xgeq 10000/49)=P(sqrt{49}(bar X-200)/20geq sqrt{49}(10000/49-200)/20)approxPhi(7(10000/49-200)/20)$
where $Phi$ is the cdf of $N(0,1).$
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
$endgroup$
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
1
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
add a comment |
$begingroup$
$X_i$ be the weight of $i$-th apple. $P(Nleq 49)=P(sum_{i=1}^{49}X_igeq 10000)=P(bar Xgeq 10000/49)=P(sqrt{49}(bar X-200)/20geq sqrt{49}(10000/49-200)/20)approxPhi(7(10000/49-200)/20)$
where $Phi$ is the cdf of $N(0,1).$
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
$endgroup$
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
1
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
add a comment |
$begingroup$
$X_i$ be the weight of $i$-th apple. $P(Nleq 49)=P(sum_{i=1}^{49}X_igeq 10000)=P(bar Xgeq 10000/49)=P(sqrt{49}(bar X-200)/20geq sqrt{49}(10000/49-200)/20)approxPhi(7(10000/49-200)/20)$
where $Phi$ is the cdf of $N(0,1).$
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
$endgroup$
$X_i$ be the weight of $i$-th apple. $P(Nleq 49)=P(sum_{i=1}^{49}X_igeq 10000)=P(bar Xgeq 10000/49)=P(sqrt{49}(bar X-200)/20geq sqrt{49}(10000/49-200)/20)approxPhi(7(10000/49-200)/20)$
where $Phi$ is the cdf of $N(0,1).$
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
edited Dec 19 '18 at 14:07
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
answered Dec 19 '18 at 13:59
John_WickJohn_Wick
1,616111
1,616111
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
1
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
add a comment |
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
1
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
$begingroup$
Last = sign should only be an approximation. Also, the "dispersion" is the standard deviation, not the variance.
$endgroup$
– Did
Dec 19 '18 at 14:06
1
1
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
A level-2 remark, if I may, now that the obvious points are corrected: usually, one checks at the end that the deviation considered indeed belongs to the CLT regime. Here the normalized deviation is $frac{10}7$, around $1$, hence everything is fine -- but, to fail to even mention the fact encourages people to apply the CLT in regimes where it is not even relevant.
$endgroup$
– Did
Dec 19 '18 at 14:18
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
I have some questions about this. How does the answer take into account the fact that $N$ could be for example 48 or 47? Also I don't understand the first equality. Thank you very much for the answer.
$endgroup$
– jte
Dec 19 '18 at 17:29
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
$begingroup$
Never mind, I figured it out. Also I noticed that the answer should be $1-Φ(10/7)$
$endgroup$
– jte
Dec 19 '18 at 18:05
add a comment |
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