Suppose 55% of a group a students can pass an exam…Standard deviation of the mean of sample dataGiven a 95% confidence interval why are we using 1.96 and not 1.64?Probability - analyzing “randomness” of datastatistics question (independent sampling)Problem involving normal approximation to the binomial.Rolling a die, probability of the sumFinding Probability on Normal DistributionWhat is the conditional probability distribution for who wrote this particular exam?Among $33$ students in a class 17 of them earned A's on the midterm, $14$ earned A's on the final exam, $11$ of them did not earn $A$ on either exam.Constructing a 95% Confidence Interval (using output from r-studio)
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Suppose 55% of a group a students can pass an exam…
Standard deviation of the mean of sample dataGiven a 95% confidence interval why are we using 1.96 and not 1.64?Probability - analyzing “randomness” of datastatistics question (independent sampling)Problem involving normal approximation to the binomial.Rolling a die, probability of the sumFinding Probability on Normal DistributionWhat is the conditional probability distribution for who wrote this particular exam?Among $33$ students in a class 17 of them earned A's on the midterm, $14$ earned A's on the final exam, $11$ of them did not earn $A$ on either exam.Constructing a 95% Confidence Interval (using output from r-studio)
$begingroup$
Suppose 55% of a grouo of students can pass an exam, if 100 students are randomly selected, what is the probability that at least 50 but less than 60 students pass the exam? Use histogram correction.
This question was given to me as a review for an upcoming exam.
My math:
$overlinex = 100*0.55 = 55$
$sigma = sqrtnp(1-p)= sqrt100*0.55(0.45) = 4.97493$
What we need to calculate: $P( 50 leq x lt 60)$
$=P(X < 60) - P(X < 50)$
$=P(Z< fracoverlinex-correction-Xsigma)-P(Z< fracoverlinex-correction-Xsigma)$
$=P(Z< frac60-0.5-554.97493)-P(Z< frac50-0.5-554.97493)$
$=P(Z < 0.9045) - P(Z < -1.1055)$
$=0.81713-0.13447$
=0.6826 or 68.26%
Is there anything glaringly wrong with my calculations and did I use the histogram correction correct?
probability statistics
asked Mar 20 at 18:15
JoeJoe
596
$endgroup$
add a comment |
$begingroup$
Suppose 55% of a grouo of students can pass an exam, if 100 students are randomly selected, what is the probability that at least 50 but less than 60 students pass the exam? Use histogram correction.
This question was given to me as a review for an upcoming exam.
My math:
$overlinex = 100*0.55 = 55$
$sigma = sqrtnp(1-p)= sqrt100*0.55(0.45) = 4.97493$
What we need to calculate: $P( 50 leq x lt 60)$
$=P(X < 60) - P(X < 50)$
$=P(Z< fracoverlinex-correction-Xsigma)-P(Z< fracoverlinex-correction-Xsigma)$
$=P(Z< frac60-0.5-554.97493)-P(Z< frac50-0.5-554.97493)$
$=P(Z < 0.9045) - P(Z < -1.1055)$
$=0.81713-0.13447$
=0.6826 or 68.26%
Is there anything glaringly wrong with my calculations and did I use the histogram correction correct?
probability statistics
asked Mar 20 at 18:15
JoeJoe
596
$endgroup$
$begingroup$
I am just starting statistics and I am wondering why you subtracted the $text correction$. It would be helpful if anyone could explain it as my brain was already seeing it as $P(X lt 60)=P(Z lt frac 60- musigma)$.
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:26
$begingroup$
We us the histogram correction (AKA continuity correction factor) when we are using normal distribution as an approximation for the binomial distribution.
$endgroup$
– Joe
Mar 20 at 18:40
$begingroup$
Interesting. Is there any specific reason for $0.5?$
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:41
$begingroup$
I believe we use 0.5 because it does not change the bounds of what we are calculating. Since # of students is always going to be an integer, the range of 49.5 to 59.5 is still the domain of what we were originally trying to find with respect to the # of people [50,60)
$endgroup$
– Joe
Mar 20 at 18:46
$begingroup$
Your title need not be the opening sentence of your question; better to state the gist of the actual question.
$endgroup$
– StubbornAtom
Mar 20 at 19:33
add a comment |
$begingroup$
Suppose 55% of a grouo of students can pass an exam, if 100 students are randomly selected, what is the probability that at least 50 but less than 60 students pass the exam? Use histogram correction.
This question was given to me as a review for an upcoming exam.
My math:
$overlinex = 100*0.55 = 55$
$sigma = sqrtnp(1-p)= sqrt100*0.55(0.45) = 4.97493$
What we need to calculate: $P( 50 leq x lt 60)$
$=P(X < 60) - P(X < 50)$
$=P(Z< fracoverlinex-correction-Xsigma)-P(Z< fracoverlinex-correction-Xsigma)$
$=P(Z< frac60-0.5-554.97493)-P(Z< frac50-0.5-554.97493)$
$=P(Z < 0.9045) - P(Z < -1.1055)$
$=0.81713-0.13447$
=0.6826 or 68.26%
Is there anything glaringly wrong with my calculations and did I use the histogram correction correct?
probability statistics
asked Mar 20 at 18:15
JoeJoe
596
$endgroup$
Suppose 55% of a grouo of students can pass an exam, if 100 students are randomly selected, what is the probability that at least 50 but less than 60 students pass the exam? Use histogram correction.
This question was given to me as a review for an upcoming exam.
My math:
$overlinex = 100*0.55 = 55$
$sigma = sqrtnp(1-p)= sqrt100*0.55(0.45) = 4.97493$
What we need to calculate: $P( 50 leq x lt 60)$
$=P(X < 60) - P(X < 50)$
$=P(Z< fracoverlinex-correction-Xsigma)-P(Z< fracoverlinex-correction-Xsigma)$
$=P(Z< frac60-0.5-554.97493)-P(Z< frac50-0.5-554.97493)$
$=P(Z < 0.9045) - P(Z < -1.1055)$
$=0.81713-0.13447$
=0.6826 or 68.26%
Is there anything glaringly wrong with my calculations and did I use the histogram correction correct?
probability statistics
probability statistics
asked Mar 20 at 18:15
JoeJoe
596
asked Mar 20 at 18:15
JoeJoe
596
asked Mar 20 at 18:15
JoeJoe
596
asked Mar 20 at 18:15
JoeJoe
596
asked Mar 20 at 18:15
JoeJoe
596
596
$begingroup$
I am just starting statistics and I am wondering why you subtracted the $text correction$. It would be helpful if anyone could explain it as my brain was already seeing it as $P(X lt 60)=P(Z lt frac 60- musigma)$.
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:26
$begingroup$
We us the histogram correction (AKA continuity correction factor) when we are using normal distribution as an approximation for the binomial distribution.
$endgroup$
– Joe
Mar 20 at 18:40
$begingroup$
Interesting. Is there any specific reason for $0.5?$
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:41
$begingroup$
I believe we use 0.5 because it does not change the bounds of what we are calculating. Since # of students is always going to be an integer, the range of 49.5 to 59.5 is still the domain of what we were originally trying to find with respect to the # of people [50,60)
$endgroup$
– Joe
Mar 20 at 18:46
$begingroup$
Your title need not be the opening sentence of your question; better to state the gist of the actual question.
$endgroup$
– StubbornAtom
Mar 20 at 19:33
add a comment |
$begingroup$
I am just starting statistics and I am wondering why you subtracted the $text correction$. It would be helpful if anyone could explain it as my brain was already seeing it as $P(X lt 60)=P(Z lt frac 60- musigma)$.
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:26
$begingroup$
We us the histogram correction (AKA continuity correction factor) when we are using normal distribution as an approximation for the binomial distribution.
$endgroup$
– Joe
Mar 20 at 18:40
$begingroup$
Interesting. Is there any specific reason for $0.5?$
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:41
$begingroup$
I believe we use 0.5 because it does not change the bounds of what we are calculating. Since # of students is always going to be an integer, the range of 49.5 to 59.5 is still the domain of what we were originally trying to find with respect to the # of people [50,60)
$endgroup$
– Joe
Mar 20 at 18:46
$begingroup$
Your title need not be the opening sentence of your question; better to state the gist of the actual question.
$endgroup$
– StubbornAtom
Mar 20 at 19:33
$begingroup$
I am just starting statistics and I am wondering why you subtracted the $text correction$. It would be helpful if anyone could explain it as my brain was already seeing it as $P(X lt 60)=P(Z lt frac 60- musigma)$.
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:26
$begingroup$
I am just starting statistics and I am wondering why you subtracted the $text correction$. It would be helpful if anyone could explain it as my brain was already seeing it as $P(X lt 60)=P(Z lt frac 60- musigma)$.
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:26
$begingroup$
We us the histogram correction (AKA continuity correction factor) when we are using normal distribution as an approximation for the binomial distribution.
$endgroup$
– Joe
Mar 20 at 18:40
$begingroup$
We us the histogram correction (AKA continuity correction factor) when we are using normal distribution as an approximation for the binomial distribution.
$endgroup$
– Joe
Mar 20 at 18:40
$begingroup$
Interesting. Is there any specific reason for $0.5?$
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:41
$begingroup$
Interesting. Is there any specific reason for $0.5?$
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:41
$begingroup$
I believe we use 0.5 because it does not change the bounds of what we are calculating. Since # of students is always going to be an integer, the range of 49.5 to 59.5 is still the domain of what we were originally trying to find with respect to the # of people [50,60)
$endgroup$
– Joe
Mar 20 at 18:46
$begingroup$
I believe we use 0.5 because it does not change the bounds of what we are calculating. Since # of students is always going to be an integer, the range of 49.5 to 59.5 is still the domain of what we were originally trying to find with respect to the # of people [50,60)
$endgroup$
– Joe
Mar 20 at 18:46
$begingroup$
Your title need not be the opening sentence of your question; better to state the gist of the actual question.
$endgroup$
– StubbornAtom
Mar 20 at 19:33
$begingroup$
Your title need not be the opening sentence of your question; better to state the gist of the actual question.
$endgroup$
– StubbornAtom
Mar 20 at 19:33
add a comment |
1 Answer
1
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$begingroup$
Your math seems to be correct. However, I would have written $60 - mboxcorrection$ as
$59 + mboxcorrection$ to demonstrate that you saw that the upper limit is $59$ and the right-side histogram correction adds half a bin.
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Your math seems to be correct. However, I would have written $60 - mboxcorrection$ as
$59 + mboxcorrection$ to demonstrate that you saw that the upper limit is $59$ and the right-side histogram correction adds half a bin.
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
Your math seems to be correct. However, I would have written $60 - mboxcorrection$ as
$59 + mboxcorrection$ to demonstrate that you saw that the upper limit is $59$ and the right-side histogram correction adds half a bin.
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
Your math seems to be correct. However, I would have written $60 - mboxcorrection$ as
$59 + mboxcorrection$ to demonstrate that you saw that the upper limit is $59$ and the right-side histogram correction adds half a bin.
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
$endgroup$
Your math seems to be correct. However, I would have written $60 - mboxcorrection$ as
$59 + mboxcorrection$ to demonstrate that you saw that the upper limit is $59$ and the right-side histogram correction adds half a bin.
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 18:23
Mark FischlerMark Fischler
33.9k12552
33.9k12552
add a comment |
add a comment |
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$begingroup$
I am just starting statistics and I am wondering why you subtracted the $text correction$. It would be helpful if anyone could explain it as my brain was already seeing it as $P(X lt 60)=P(Z lt frac 60- musigma)$.
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:26
$begingroup$
We us the histogram correction (AKA continuity correction factor) when we are using normal distribution as an approximation for the binomial distribution.
$endgroup$
– Joe
Mar 20 at 18:40
$begingroup$
Interesting. Is there any specific reason for $0.5?$
$endgroup$
– Mohammad Zuhair Khan
Mar 20 at 18:41
$begingroup$
I believe we use 0.5 because it does not change the bounds of what we are calculating. Since # of students is always going to be an integer, the range of 49.5 to 59.5 is still the domain of what we were originally trying to find with respect to the # of people [50,60)
$endgroup$
– Joe
Mar 20 at 18:46
$begingroup$
Your title need not be the opening sentence of your question; better to state the gist of the actual question.
$endgroup$
– StubbornAtom
Mar 20 at 19:33