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

Let 𝑋 be a continuous random variable with probability density function 𝑓(π‘₯) = (1/8) (6π‘₯ βˆ’ 7) if 2 ≀ π‘₯ ≀ 3 and 𝑓(π‘₯) = 0, otherwise. Find 𝑃(2 ≀ 𝑋 ≀ 2.5).

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Video Transcript

Let 𝑋 be a continuous random variable with probability density function 𝑓 of π‘₯ equals one-eighth multiplied by six π‘₯ minus seven if π‘₯ is between two and three inclusively or zero otherwise. Find the probability that 𝑋 is between two and 2.5.

A continuous random variable is characterized by its probability density function. That is a nonnegative function whose area under the curve is one, and that represents the probability of the whole sample space. So, we often use integration when we’re finding a probability from a probability density function of a continuous random variable. This probability density function has been given to us as a formula. So, we can use an integral to find the probability that we’re looking for. That is the probability of 𝑋 being between two and 2.5 is the integral between two and 2.5 of 𝑓 of π‘₯ with respect to π‘₯.

Over the interval that we’re looking at between two and 2.5, 𝑓 of π‘₯ is equal to one-eighth multiplied by six π‘₯ minus seven because we were told in the question that this is 𝑓 of π‘₯ when π‘₯ is between two and three. So, we’re going to find the integral between two and 2.5 of one-eighth multiplied by six π‘₯ minus seven with respect to π‘₯. Let’s begin this question by bringing one-eighth to the front of the integral. So, we now have one-eighth multiplied by the integral between two and 2.5 of six π‘₯ minus seven with respect to π‘₯. We can now use the usual rules of integration to integrate six π‘₯ minus seven.

We know that to integrate a function of π‘₯, we can add one to the power and divide by the new power. And we do this term by term. So, six π‘₯ integrates to six π‘₯ squared over two, but that’s just the same as three π‘₯ squared. And negative seven is a constant, so that integrates to negative seven π‘₯. Okay, so we’re nearly there. We just need to apply our limits of integration. We do this by firstly substituting in our upper limit of integration, which is 2.5. And then we subtract our integral with the lower limit substituted, and that’s two.

So, now, we just need to simplify. 2.5 squared is 25 over four. So, when we multiply that by three, we get 75 over four. And seven multiplied by 2.5 is 35 over two. Notice that I’m keeping everything in fraction form here. Three multiplied by two squared is three multiplied by four. So, that’s 12. And seven multiplied by two gives us 14.

Okay, to simplify the first bracket, let’s rewrite 35 over two as something over four. That’s going to be 70 over four. So, simplifying what we’ve got, 75 over four minus 70 over four is five over four. So, we have one-eighth multiplied by five over four. And then subtract this bracket, which is 12 minus 14. So, that’s negative two. As negative two is the same as negative eight over four, we have one-eighth multiplied by five over four minus negative eight over four. But when we subtract a negative, we can just add instead. This is one-eighth multiplied by 13 over four. And multiplying the numerators and the denominators gives us 13 over 32.

So that’s our final answer. The probability of 𝑋 being between two and 2.5 inclusively is 13 over 32.

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