Thread: Probability calculation using Standard Normal Distribution Table

Hi, I think this will be a simple question to answer but I can't seem to find a definitive answer anywhere, other than by observing examples and working out what I think is correct.

For Z~N(0,1), looking up P(z<1.62) for example, I get 0.5 + 0.4474 = 0.9474. Would P(z<=1.62) be the same?

It appears that all examples I've seen make no distinction between < and <= but I have an old question sheet from college that contains questions with some containing < and others containing <=, so I'd like to know if there is a difference, and if so how it affects obtaining a result from the table.

(If is doesn't make a difference can you give me a quick hint as to why? My (possibly faulty) logic tells me that the probability including 1.62, from the example above, should be slightly higher than that not including it. Or is it due to rounding, being that the table is to 2dp?)

Thanks!

Hello !

Would P(z<=1.62) be the same?

Yes !

The difference in counting 1.62 or not comes with the fact that you're working on continuous or discrete variables.

If it's continuous, we rather work on intervals, that is to say an uncountable amount of values. Imagine you want the probability of X being between 0 & 1, where X can take any value between 0 & 1.
The probability of taking one value is 0.

If it's discrete, it means that the variable takes a countable amount of values, that is to say it takes specific values. So the probability that it takes one value can't be neglected.

A way to remember is that when working with discrete random variables, you work with a probability distribution, that is to say with the probability that the variable takes one value.
When working with continuous random variables, you work with a probability density function, that is to say that the probability will be representing an area. Does it matter including one point or not ?

I'm not sure whether it's clear or not

Originally Posted by Moo

If it's continuous, we rather work on intervals, that is to say an uncountable amount of values. Imagine you want the probability of X being between 0 & 1, where X can take any value between 0 & 1.
The probability of taking one value is 0.

Is this because, being basically infinite other values X could take between 0 and 1, it is essentially the case that the probability of it being any one approaches zero? Or am I thinking of it wrongly?

I guess in my original thinking (that P(z<=1.62) would be greater than P(z<1.62)) I was stuck in the trap of thinking of a finite number of values between the limits, rather than infinite?

That seems to make sense. Thanks for the reply.

Originally Posted by Wonkihead

Hi, I think this will be a simple question to answer but I can't seem to find a definitive answer anywhere, other than by observing examples and working out what I think is correct.

For Z~N(0,1), looking up P(z<1.62) for example, I get 0.5 + 0.4474 = 0.9474. Would P(z<=1.62) be the same?

It appears that all examples I've seen make no distinction between < and <= but I have an old question sheet from college that contains questions with some containing < and others containing <=, so I'd like to know if there is a difference, and if so how it affects obtaining a result from the table.

(If is doesn't make a difference can you give me a quick hint as to why? My (possibly faulty) logic tells me that the probability including 1.62, from the example above, should be slightly higher than that not including it. Or is it due to rounding, being that the table is to 2dp?)

Thanks!

Although there is a technical difference between using strict inequalities and not, this difference is not something you should worry about. It has absolutely no practical affect on calculated probability values. The reason for this is because the calculated probability is the area under the probability density function.

To make it crystal clear, there is no difference between using strict inequalities < and >, and 'less than or equal to' and 'greater than or equal to' when calculating the value of a probability.

You should have been taught that Pr(z = a) = 0 for any value of a. And in fact, Pr(X = a) for any value of a when X is any continuous random variable.

Originally Posted by mr fantastic

You should have been taught that Pr(z = a) = 0 for any value of a. And in fact, Pr(X = a) for any value of a when X is any continuous random variable.

I don't recall being taught that and I can't find reference to it in my notes, but it could be because some areas of the notes are a little spotty. It makes going through them a couple of years later a little difficult

If I understand this correctly, then Pr(z = a) would be zero because the integral would have limits from a to a, thus yielding a zero result always. If that's correct, then I'm happy that I understand it properly.

Originally Posted by Wonkihead

[snip]
If I understand this correctly, then Pr(z = a) would be zero because the integral would have limits from a to a, thus yielding a zero result always. If that's correct, then I'm happy that I understand it properly.

Be happy.