Question Video: Using the Probability Density Function of a Continuous Random Variable to Find Probabilities | Nagwa Question Video: Using the Probability Density Function of a Continuous Random Variable to Find Probabilities | Nagwa

Question Video: Using the Probability Density Function of a Continuous Random Variable to Find Probabilities Mathematics • Third Year of Secondary School

Let 𝑋 be a continuous random variable with probability density function 𝑓(𝑥) = 1/6 (𝑥 − 5), 7 ≤ 𝑥 ≤ 9 and 𝑓(𝑥) = 0 otherwise. Find 𝑃(𝑋 ≤ 8).

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Video Transcript

Let 𝑋 be a continuous random variable with probability density function 𝑓 of 𝑥 equals one-sixth 𝑥 minus five when 𝑥 is between seven and nine inclusive and zero otherwise. Find the probability that 𝑋 is less than or equal to eight.

Continuous random variables are characterized by their probability density function, and this is a nonnegative function which we can graph, and we find that the area under the curve is one. This probability density function is 𝑓 of 𝑥 equals one-sixth multiplied by 𝑥 minus five when 𝑥 is between seven and nine inclusive or zero when it’s out of range. We’ve been asked to calculate the probability of 𝑋 being less than or equal to eight.

One thing to remember with continuous random variables is that less than or equal to and less than are interchangeable. So, to find this probability, what we’re doing is calculating the area under the curve defined by this function between zero and eight. But we know that this function is equal to zero when 𝑥 is from zero up to seven. So, we actually only really need to calculate the area when 𝑥 is between seven and eight. That is the integral between seven and eight of 𝑓 of 𝑥 with respect to 𝑥.

To begin calculating this integral, let’s distribute the parentheses. That gives us the integral between seven and eight of one over six 𝑥 minus five over six with respect to 𝑥. We then recall that to integrate, we add one to the power and divide by the new power. So, we begin by integrating one over six 𝑥 to be one over six 𝑥 squared divided by two. Then, because five over six is a constant, this integrates to five over six 𝑥. We can actually simplify what we’ve got here because we know that one over six divided by two is one over 12.

So now, we’ve just got to apply these limits of integration. Remember we do this by substituting our upper limit into the integral and then subtracting the integral with the lower limit substituted for 𝑥. To calculate this, we could then use a calculator, or we could write each of these fractions with the same denominator of 12. That would look like this. Then simplifying each parenthesis gives us negative 16 over 12 minus negative 21 over 12, which is the same as negative 16 over 12 add 21 over 12. And that gives us five over 12.

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