Question

g True or false: For continuous random variables, the probability of being less than some value, x, IS NOT the same as the probability of being less than or equal to the same value, x. Group of answer choices True False

Answer 1

Answer:  False

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Explanation:

The given statement is claiming something like $P(X < 2)$ produces a different result compared to $P(X \le 2)$. The subtle difference of course is the "or equal to".

When talking about continuous random variables, the two values are the same. Why? Because probability of selecting *exactly* X = 2 is going to be zero. Think of a very very thin strip of height P(X = 2) and the width is approaching zero. The width is doing this because we're trying to pinpoint X = 2 itself. As you can see, this skinny rectangle will have its area approach 0. Ultimately, whether we compute $P(X < 2)$ or $P(X \le 2)$, it doesn't matter. They both produce the same result.

If you are a visual learner, consider the standard normal distribution. Now consider the task if shading to the left of z = 2. This area describes both probabiltiies mentioned above.

If X is a continuous random variable, then $P(X < 2) = P(X \le 2)$ and we can change the "2" to any number we want to have the same outcome.

Therefore, the original claim is false.

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Extra info:

Now if X was discrete (say from the binomial distribution), then the two probabilities would be different. This is because

$P(X < 2) = P(0) + P(1)\\\\P(X \le 2) = P(0) + P(1) + P(2)\\\\P(X < 2) \ne P(X \le 2) \ \text{only if X is discrete}$

We can see in this case that the second line has P(2) while the first one does not.

So the original claim would be true if we changed "continuous" to "discrete".