Stats Question

For my stats exam we need to write probabilities in the form P(X≤k). For instance, I know that P(X<k)=P(X≤k-1); P(X>k)=1-P(X≤k); P(X≥k)=1-P(X≤k-1). This enables me to put these inequalities in the form I need to get the values (as I have all values for P(X≤k) available).

However, if we get something like P(a<X<b), or P(a≤X≤b), what do I do with that? I tried to say that this is the same as P(X>a)*P(X<b) and then convert each one individually as above, but this doesn't bring the correct answers.

So how should I separate my inequalities to rewrite them in terms of P(X≤k) (k can vary, I can use multiple values of k, but my final inequalities need to read X≤ rather than X> or X<)?

Reply 1

CrazyKid123

5

P(a<X<b) = P(X<b) - P(X \le a)

?

Reply 2

Jkn

11

Original post by Big-Daddy

For my stats exam we need to write probabilities in the form P(X≤k). For instance, I know that P(X<k)=P(X≤k-1); P(X>k)=1-P(X≤k); P(X≥k)=1-P(X≤k-1). This enables me to put these inequalities in the form I need to get the values (as I have all values for P(X≤k) available).

However, if we get something like P(a<X<b), or P(a≤X≤b), what do I do with that? I tried to say that this is the same as P(X>a)*P(X<b) and then convert each one individually as above, but this doesn't bring the correct answers.

So how should I separate my inequalities to rewrite them in terms of P(X≤k) (k can vary, I can use multiple values of k, but my final inequalities need to read X≤ rather than X> or X<)?

Note that this is only difficult in the case of discrete probability distributions (where k are integers). For continuous distributions, like the normal distribution, you need not consider such things. You simply have:

P(x<k)=P(x\le k)

For discrete distributions, it's best to see the numbers as little "blocks" if you like. Here is how I visualise it:

Possible values of x:

1, 2, 3, 4, 5, ... , (a-2), (a-1), a, (a+1), ... , (b-2), (b-1), b, (b+1), ...

If a<x<b, we can "group" the values like so:

1, 2, 3, 4, 5, ... , (a-2), (a-1), a, (a+1), ... , (b-2), (b-1), b, (b+1), ... by knowing that x is between them but not equal.

So now we can instantly see that x<b would be this:

1, 2, 3, 4, 5, ... , (a-2), (a-1), a, (a+1), ... , (b-2), (b-1), b, (b+1), ...

and x<(a+1) would be this:

1, 2, 3, 4, 5, ... , (a-2), (a-1), a, (a+1), ... , (b-2), (b-1), b, (b+1), ...

Looking at the difference between these two sets gives us:

P(a<x<b)=P(x<b)-P(x<(a+1))

To get them in the "less than or equal to" form, simply visualise the same thing. :smile:

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