Given a set of values of a variable x= 40, 25, 20, 15, 12, and 5:
The probability of exceedance of a value x=20 written as P (x>=20) can be got by arranging data is descending order thus giving a value of 0.5
The probability of non-exceedance of a value x=20 written as P (x=1) = 1 - P(x
This is because you have 6 values and therefore no central point. Add in one more value - say 10 - and try again.
George is right. What Philip has done is to include 20 in both groups. The probabilities of getting >20, =20, & 20=2/6=0.333333...
=20=1/6=0.16666666..
One more thing. You should write the equation as
P(x>20)+Px=(=20)+Px(
When the random variables are discrete, equality sign can not be ignored. A number can not be simultaneously in a given set and in its complement. The correct fact is, when X is discrete r.v. P(X>=1) = 1 - P(X
The actual fact is that your assumption is wrong
its actually P(X>=x) = 1 - P(X
Agreeing with the above answers, it is worth to add, that there is difference between X
Dear all, thank you for the answers. I now see where the problem came from.
So all these following expressions below are wrong:
- P(x < X) = 1 - P(x > X)
- P(x =X)
The correct expression should be:
- P(x >= X) = 1 - P(x < X)
However, for a continuous variable, P(x = X) = 0 implying P(x>=X) = P(x>X), which makes it still confusing. Note that standard normal tables for reading probabilities always give P(X=X).
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