Using the normal distribution to find the probability of getting an exact value
$begingroup$
Take the following example. In the US, the average IQ is 100 points with a standard deviation of approx. 18 points. Find the following probability:
Find the probability that a randomly selected individual has an IQ of exactly 100 points.
Here is what I assume needs to be done, after which point we normalize and use tables.
a) $P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
As a further question, would we need to do something similar every time the question asks to find the probability of choosing an individual with a score over or under but also equal to a certain value? For example, find the probability that a randomly selected individual has an IQ of 130 points or more. Would it be done like this??
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
probability statistics normal-distribution
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
$endgroup$
add a comment |
$begingroup$
Take the following example. In the US, the average IQ is 100 points with a standard deviation of approx. 18 points. Find the following probability:
Find the probability that a randomly selected individual has an IQ of exactly 100 points.
Here is what I assume needs to be done, after which point we normalize and use tables.
a) $P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
As a further question, would we need to do something similar every time the question asks to find the probability of choosing an individual with a score over or under but also equal to a certain value? For example, find the probability that a randomly selected individual has an IQ of 130 points or more. Would it be done like this??
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
probability statistics normal-distribution
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
$endgroup$
2
$begingroup$
If $X$ has a continuous distribution, then $P(X=x)=0$ for any $x$. There are no point mass for a continuous random variable.
$endgroup$
– Just_to_Answer
Jan 9 at 3:52
add a comment |
$begingroup$
Take the following example. In the US, the average IQ is 100 points with a standard deviation of approx. 18 points. Find the following probability:
Find the probability that a randomly selected individual has an IQ of exactly 100 points.
Here is what I assume needs to be done, after which point we normalize and use tables.
a) $P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
As a further question, would we need to do something similar every time the question asks to find the probability of choosing an individual with a score over or under but also equal to a certain value? For example, find the probability that a randomly selected individual has an IQ of 130 points or more. Would it be done like this??
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
probability statistics normal-distribution
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
$endgroup$
Take the following example. In the US, the average IQ is 100 points with a standard deviation of approx. 18 points. Find the following probability:
Find the probability that a randomly selected individual has an IQ of exactly 100 points.
Here is what I assume needs to be done, after which point we normalize and use tables.
a) $P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
As a further question, would we need to do something similar every time the question asks to find the probability of choosing an individual with a score over or under but also equal to a certain value? For example, find the probability that a randomly selected individual has an IQ of 130 points or more. Would it be done like this??
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
probability statistics normal-distribution
probability statistics normal-distribution
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
edited Jan 9 at 3:43
Ninosław Brzostowiecki
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
asked Jan 9 at 3:38
Ninosław BrzostowieckiNinosław Brzostowiecki
546414
546414
2
$begingroup$
If $X$ has a continuous distribution, then $P(X=x)=0$ for any $x$. There are no point mass for a continuous random variable.
$endgroup$
– Just_to_Answer
Jan 9 at 3:52
add a comment |
2
$begingroup$
If $X$ has a continuous distribution, then $P(X=x)=0$ for any $x$. There are no point mass for a continuous random variable.
$endgroup$
– Just_to_Answer
Jan 9 at 3:52
2
2
$begingroup$
If $X$ has a continuous distribution, then $P(X=x)=0$ for any $x$. There are no point mass for a continuous random variable.
$endgroup$
– Just_to_Answer
Jan 9 at 3:52
$begingroup$
If $X$ has a continuous distribution, then $P(X=x)=0$ for any $x$. There are no point mass for a continuous random variable.
$endgroup$
– Just_to_Answer
Jan 9 at 3:52
add a comment |
1 Answer
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$begingroup$
$P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
Yes... assuming an IQ values can be any integer (which seems to be the case).
BTW: to avoid confusions you should use different letters for the real (discrete) random variable, and the one that corresponds to the normal approximation.
However, not using the continiuty correction (just evaluating the normal at $100$) is often also fine. That is, if $phi(x)$ is the normal density, we often approximate
$$ int_{X-h/2}^{X+h/2} phi(x) dx approx h phi(X)$$
just because it's simpler.
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
Yes. Of course, in that case the correction is (probably) less important.
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
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add a comment |
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$begingroup$
$P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
Yes... assuming an IQ values can be any integer (which seems to be the case).
BTW: to avoid confusions you should use different letters for the real (discrete) random variable, and the one that corresponds to the normal approximation.
However, not using the continiuty correction (just evaluating the normal at $100$) is often also fine. That is, if $phi(x)$ is the normal density, we often approximate
$$ int_{X-h/2}^{X+h/2} phi(x) dx approx h phi(X)$$
just because it's simpler.
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
Yes. Of course, in that case the correction is (probably) less important.
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
$endgroup$
add a comment |
$begingroup$
$P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
Yes... assuming an IQ values can be any integer (which seems to be the case).
BTW: to avoid confusions you should use different letters for the real (discrete) random variable, and the one that corresponds to the normal approximation.
However, not using the continiuty correction (just evaluating the normal at $100$) is often also fine. That is, if $phi(x)$ is the normal density, we often approximate
$$ int_{X-h/2}^{X+h/2} phi(x) dx approx h phi(X)$$
just because it's simpler.
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
Yes. Of course, in that case the correction is (probably) less important.
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
$endgroup$
add a comment |
$begingroup$
$P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
Yes... assuming an IQ values can be any integer (which seems to be the case).
BTW: to avoid confusions you should use different letters for the real (discrete) random variable, and the one that corresponds to the normal approximation.
However, not using the continiuty correction (just evaluating the normal at $100$) is often also fine. That is, if $phi(x)$ is the normal density, we often approximate
$$ int_{X-h/2}^{X+h/2} phi(x) dx approx h phi(X)$$
just because it's simpler.
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
Yes. Of course, in that case the correction is (probably) less important.
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
$endgroup$
$P(X = 100) rightarrow P(99.5 < X < 100.5)$
Does this look right?
Yes... assuming an IQ values can be any integer (which seems to be the case).
BTW: to avoid confusions you should use different letters for the real (discrete) random variable, and the one that corresponds to the normal approximation.
However, not using the continiuty correction (just evaluating the normal at $100$) is often also fine. That is, if $phi(x)$ is the normal density, we often approximate
$$ int_{X-h/2}^{X+h/2} phi(x) dx approx h phi(X)$$
just because it's simpler.
$P(X geq 130) rightarrow P(X > 129.5)$
In other words, do we correct for continuity every time equality is involved?
Yes. Of course, in that case the correction is (probably) less important.
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
answered Jan 9 at 3:52
leonbloyleonbloy
41.3k645107
41.3k645107
add a comment |
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$begingroup$
If $X$ has a continuous distribution, then $P(X=x)=0$ for any $x$. There are no point mass for a continuous random variable.
$endgroup$
– Just_to_Answer
Jan 9 at 3:52