Mixing
The distribution of sample proportion for given population proportion and sample size
$begingroup$
If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89? (4dp)
work: $n=64$, $hat p=0.89$, so $X=n hat p =56.96$.
$$ P(hat p >0.89)=P( X >56.95)=1- P( X <56.95)$$
and then how to do it?
The answer is $0.6064$
More details of the solutions would be great, thanks
probability-distributions sampling
asked Sep 21 '14 at 4:47
HiverHiver
113
$endgroup$
add a comment |
$begingroup$
If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89? (4dp)
work: $n=64$, $hat p=0.89$, so $X=n hat p =56.96$.
$$ P(hat p >0.89)=P( X >56.95)=1- P( X <56.95)$$
and then how to do it?
The answer is $0.6064$
More details of the solutions would be great, thanks
probability-distributions sampling
asked Sep 21 '14 at 4:47
HiverHiver
113
$endgroup$
add a comment |
$begingroup$
If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89? (4dp)
work: $n=64$, $hat p=0.89$, so $X=n hat p =56.96$.
$$ P(hat p >0.89)=P( X >56.95)=1- P( X <56.95)$$
and then how to do it?
The answer is $0.6064$
More details of the solutions would be great, thanks
probability-distributions sampling
asked Sep 21 '14 at 4:47
HiverHiver
113
$endgroup$
If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89? (4dp)
work: $n=64$, $hat p=0.89$, so $X=n hat p =56.96$.
$$ P(hat p >0.89)=P( X >56.95)=1- P( X <56.95)$$
and then how to do it?
The answer is $0.6064$
More details of the solutions would be great, thanks
probability-distributions sampling
probability-distributions sampling
asked Sep 21 '14 at 4:47
HiverHiver
113
asked Sep 21 '14 at 4:47
HiverHiver
113
edited Sep 21 '14 at 5:47
Hiver
asked Sep 21 '14 at 4:47
HiverHiver
113
asked Sep 21 '14 at 4:47
HiverHiver
113
asked Sep 21 '14 at 4:47
HiverHiver
113
113
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Since $np > 10$ we can apply Central Limit Theorem and use the formula $mathbb{P}!left(p < hat{p}right)= mathbb{P}!left(Z<frac{p-hat{p}}{s}right)$ where $s = sqrt{frac{p(1-p)}{n}}$ and $Z sim N(0,1)$.
My answer came out to be .6052 (I didn't use a standard normal table).
$endgroup$
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $np > 10$ we can apply Central Limit Theorem and use the formula $mathbb{P}!left(p < hat{p}right)= mathbb{P}!left(Z<frac{p-hat{p}}{s}right)$ where $s = sqrt{frac{p(1-p)}{n}}$ and $Z sim N(0,1)$.
My answer came out to be .6052 (I didn't use a standard normal table).
$endgroup$
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
add a comment |
$begingroup$
Since $np > 10$ we can apply Central Limit Theorem and use the formula $mathbb{P}!left(p < hat{p}right)= mathbb{P}!left(Z<frac{p-hat{p}}{s}right)$ where $s = sqrt{frac{p(1-p)}{n}}$ and $Z sim N(0,1)$.
My answer came out to be .6052 (I didn't use a standard normal table).
$endgroup$
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
add a comment |
$begingroup$
Since $np > 10$ we can apply Central Limit Theorem and use the formula $mathbb{P}!left(p < hat{p}right)= mathbb{P}!left(Z<frac{p-hat{p}}{s}right)$ where $s = sqrt{frac{p(1-p)}{n}}$ and $Z sim N(0,1)$.
My answer came out to be .6052 (I didn't use a standard normal table).
$endgroup$
Since $np > 10$ we can apply Central Limit Theorem and use the formula $mathbb{P}!left(p < hat{p}right)= mathbb{P}!left(Z<frac{p-hat{p}}{s}right)$ where $s = sqrt{frac{p(1-p)}{n}}$ and $Z sim N(0,1)$.
My answer came out to be .6052 (I didn't use a standard normal table).
answered Sep 21 '14 at 5:16
user177612
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
add a comment |
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
it should be the np≥5 and then the sampling distribution of the sample proportion p-hat is approximately normally distributed.
$endgroup$
– Hiver
Sep 21 '14 at 5:54
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
thanks. i have figure it out. here: P[Z<(p-p^)/SD]=P[Z<(0.9-0.89)/√(0.9*0.1)/64]=P(Z<0.27)
$endgroup$
– Hiver
Sep 21 '14 at 6:02
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
$begingroup$
@Hiver I think that condition varies depending on how your taught. But my guess is both are right. Anyways, you can still accept my answer, if you wish.
$endgroup$
– user177612
Sep 21 '14 at 14:20
add a comment |
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