How do I find if the probability of the sample proportion is greater than something?
I have this problem and I have no clue how to solve it.
In 2012, 31% of the adult population in the US had earned a bachelor’s degree or higher. One hundred people are randomly sampled from the population. What is the probability that the
sample proportion p-hat is greater than 0.40?
statistics probability-distributions
asked Mar 22 '14 at 21:36
Lily H
8124
add a comment |
I have this problem and I have no clue how to solve it.
In 2012, 31% of the adult population in the US had earned a bachelor’s degree or higher. One hundred people are randomly sampled from the population. What is the probability that the
sample proportion p-hat is greater than 0.40?
statistics probability-distributions
asked Mar 22 '14 at 21:36
Lily H
8124
add a comment |
I have this problem and I have no clue how to solve it.
In 2012, 31% of the adult population in the US had earned a bachelor’s degree or higher. One hundred people are randomly sampled from the population. What is the probability that the
sample proportion p-hat is greater than 0.40?
statistics probability-distributions
asked Mar 22 '14 at 21:36
Lily H
8124
I have this problem and I have no clue how to solve it.
In 2012, 31% of the adult population in the US had earned a bachelor’s degree or higher. One hundred people are randomly sampled from the population. What is the probability that the
sample proportion p-hat is greater than 0.40?
statistics probability-distributions
statistics probability-distributions
asked Mar 22 '14 at 21:36
Lily H
8124
asked Mar 22 '14 at 21:36
Lily H
8124
asked Mar 22 '14 at 21:36
Lily H
8124
asked Mar 22 '14 at 21:36
Lily H
8124
asked Mar 22 '14 at 21:36
Lily H
8124
8124
add a comment |
add a comment |
2 Answers
2
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oldest
votes
The mean of $hat{p}$ is equal to p, which is 0.31, and the standard deviation of $hat{p}$ is:
$$sqrt{p(1-p)/n} = (0.31*0.69/100)^{1/2} = 0.0462$$
So
$$P(hat{p} > 0.4) = P(z > (0.4 - 0.31)/0.0462)$$
$$= P(z > 1.948) = 1 - P(z < 1.948) = 1 - .9744 = .0256$$
edited Jul 9 '16 at 7:56
BCLC
1
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
add a comment |
DeepSea's answer is a pretty standard way to do this. But it's important to remember, that treating $hat p$ as Normally distributed is an approximation (albeit a quite good one most of the time).
If you wanted to have a more "exact" answer, you could use the Binomial distribution.
Let $Y$ be the number of people in the sample who have earned a Bachelors or higher. Then $Y sim Binom(100, 0.31)$ and $hat P = frac{Y}{n}$. Therefore:
begin{equation}
P(hat p > .40) = P(Y > 40) = 1 - F_Y(40) = 0.0218
end{equation}
Notice that DeepSea's answer gives a good approximation. But this is answer is more "exact".
answered Mar 1 '17 at 19:25
knrumsey
712310
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The mean of $hat{p}$ is equal to p, which is 0.31, and the standard deviation of $hat{p}$ is:
$$sqrt{p(1-p)/n} = (0.31*0.69/100)^{1/2} = 0.0462$$
So
$$P(hat{p} > 0.4) = P(z > (0.4 - 0.31)/0.0462)$$
$$= P(z > 1.948) = 1 - P(z < 1.948) = 1 - .9744 = .0256$$
edited Jul 9 '16 at 7:56
BCLC
1
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
add a comment |
The mean of $hat{p}$ is equal to p, which is 0.31, and the standard deviation of $hat{p}$ is:
$$sqrt{p(1-p)/n} = (0.31*0.69/100)^{1/2} = 0.0462$$
So
$$P(hat{p} > 0.4) = P(z > (0.4 - 0.31)/0.0462)$$
$$= P(z > 1.948) = 1 - P(z < 1.948) = 1 - .9744 = .0256$$
edited Jul 9 '16 at 7:56
BCLC
1
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
add a comment |
The mean of $hat{p}$ is equal to p, which is 0.31, and the standard deviation of $hat{p}$ is:
$$sqrt{p(1-p)/n} = (0.31*0.69/100)^{1/2} = 0.0462$$
So
$$P(hat{p} > 0.4) = P(z > (0.4 - 0.31)/0.0462)$$
$$= P(z > 1.948) = 1 - P(z < 1.948) = 1 - .9744 = .0256$$
edited Jul 9 '16 at 7:56
BCLC
1
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
The mean of $hat{p}$ is equal to p, which is 0.31, and the standard deviation of $hat{p}$ is:
$$sqrt{p(1-p)/n} = (0.31*0.69/100)^{1/2} = 0.0462$$
So
$$P(hat{p} > 0.4) = P(z > (0.4 - 0.31)/0.0462)$$
$$= P(z > 1.948) = 1 - P(z < 1.948) = 1 - .9744 = .0256$$
edited Jul 9 '16 at 7:56
BCLC
1
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
edited Jul 9 '16 at 7:56
BCLC
1
edited Jul 9 '16 at 7:56
BCLC
1
edited Jul 9 '16 at 7:56
BCLC
1
1
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
answered Mar 22 '14 at 21:59
DeepSea
70.9k54487
70.9k54487
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
add a comment |
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
I know this might sound dumb, but what does the z stand for?
– Lily H
Mar 22 '14 at 22:09
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
@LilyH: z is the standard-normal variable with mean = 0, and standard deviation = 1.
– DeepSea
Mar 22 '14 at 22:11
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
how do you know you can use $z$?
– BCLC
Jul 9 '16 at 7:54
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
Because the Binomial distribution (especially when divided by $n$) looks quite Normally distributed. So it makes for a good approximation.
– knrumsey
Mar 1 '17 at 19:28
add a comment |
DeepSea's answer is a pretty standard way to do this. But it's important to remember, that treating $hat p$ as Normally distributed is an approximation (albeit a quite good one most of the time).
If you wanted to have a more "exact" answer, you could use the Binomial distribution.
Let $Y$ be the number of people in the sample who have earned a Bachelors or higher. Then $Y sim Binom(100, 0.31)$ and $hat P = frac{Y}{n}$. Therefore:
begin{equation}
P(hat p > .40) = P(Y > 40) = 1 - F_Y(40) = 0.0218
end{equation}
Notice that DeepSea's answer gives a good approximation. But this is answer is more "exact".
answered Mar 1 '17 at 19:25
knrumsey
712310
add a comment |
DeepSea's answer is a pretty standard way to do this. But it's important to remember, that treating $hat p$ as Normally distributed is an approximation (albeit a quite good one most of the time).
If you wanted to have a more "exact" answer, you could use the Binomial distribution.
Let $Y$ be the number of people in the sample who have earned a Bachelors or higher. Then $Y sim Binom(100, 0.31)$ and $hat P = frac{Y}{n}$. Therefore:
begin{equation}
P(hat p > .40) = P(Y > 40) = 1 - F_Y(40) = 0.0218
end{equation}
Notice that DeepSea's answer gives a good approximation. But this is answer is more "exact".
answered Mar 1 '17 at 19:25
knrumsey
712310
add a comment |
DeepSea's answer is a pretty standard way to do this. But it's important to remember, that treating $hat p$ as Normally distributed is an approximation (albeit a quite good one most of the time).
If you wanted to have a more "exact" answer, you could use the Binomial distribution.
Let $Y$ be the number of people in the sample who have earned a Bachelors or higher. Then $Y sim Binom(100, 0.31)$ and $hat P = frac{Y}{n}$. Therefore:
begin{equation}
P(hat p > .40) = P(Y > 40) = 1 - F_Y(40) = 0.0218
end{equation}
Notice that DeepSea's answer gives a good approximation. But this is answer is more "exact".
answered Mar 1 '17 at 19:25
knrumsey
712310
DeepSea's answer is a pretty standard way to do this. But it's important to remember, that treating $hat p$ as Normally distributed is an approximation (albeit a quite good one most of the time).
If you wanted to have a more "exact" answer, you could use the Binomial distribution.
Let $Y$ be the number of people in the sample who have earned a Bachelors or higher. Then $Y sim Binom(100, 0.31)$ and $hat P = frac{Y}{n}$. Therefore:
begin{equation}
P(hat p > .40) = P(Y > 40) = 1 - F_Y(40) = 0.0218
end{equation}
Notice that DeepSea's answer gives a good approximation. But this is answer is more "exact".
answered Mar 1 '17 at 19:25
knrumsey
712310
answered Mar 1 '17 at 19:25
knrumsey
712310
answered Mar 1 '17 at 19:25
knrumsey
712310
answered Mar 1 '17 at 19:25
knrumsey
712310
712310
add a comment |
add a comment |
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