Video: Using Probability Density Function of Continuous Random Variable to Find Probabilities

Let π‘₯ be a continuous random variable with the probability density function 𝑓(π‘₯) = π‘₯/8 if 2 < π‘₯ < 3, 𝑓(π‘₯) = 1/48 if 3 < π‘₯ < 36, 𝑓(π‘₯) = 0 otherwise . Find 𝑃(11 ≀ π‘₯ ≀ 24).

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Video Transcript

Let π‘₯ be a continuous random variable with the probability density function 𝑓 of π‘₯ is equal to π‘₯ over eight if π‘₯ is greater than two and less than three, one over 48 if π‘₯ is greater than three and less than 36, and zero otherwise. Find the probability that π‘₯ is greater than or equal to 11 and less than or equal to 24.

Remember, a continuous random variable is one that takes on an unaccountably infinite number of possible values. To find the probability that π‘₯ lies in some interval, we use a probability density function. Now, a probability density function satisfies a few criteria. The first is 𝑓 of π‘₯ must be greater than or equal to zero for all values of π‘₯. The second is that the definite integral between negative ∞ and ∞ of 𝑓 of π‘₯ with respect to π‘₯ is equal to one. And the third β€” and this is the one we’re interested in β€” says that the probability that π‘₯ takes values greater than π‘Ž and less than 𝑏 is equal to the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯.

Now, in fact, it’s important to realise that since this is a continuous function, the same holds for the probability that π‘₯ is greater than or equal to π‘Ž and less than or equal to 𝑏. And that’s great because we’re looking to find the probability that π‘₯ is greater than or equal to 11 and less than or equal to 24. So, the probability that π‘₯ is greater than or equal to 11 and less than or equal to 24 must be equal to the definite integral between 11 and 24 of 𝑓 of π‘₯ with respect to π‘₯. But of course, our 𝑓 of π‘₯ is defined using a piecewise function. So, which part of the function do we use?

Well, the part of the function we’re interested in is for values of π‘₯ greater than three and less than 36. Remember, we’re looking for π‘₯ is greater than or equal to 11 and less than or equal to 24. So we need to find the definite integral between 11 and 24 of one over 48 with respect to π‘₯. Well, when we integrate one over 48, we get π‘₯ over 48. So the probability that π‘₯ is greater than or equal to 11 and less than or equal to 24 is 24 over 48 minus 11 over 48 which is equal to 13 over 48. Given that π‘₯ is the continuous random variable with the probability density function shown, the probability that π‘₯ is greater than or equal to 11 and less than or equal to 24 is 13 over 48.

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