S2: Continuous Random Variables

When Random Gets Smooth 📈

Introduction

1. Introduction

Hey there! So far in statistics, we've mostly dealt with discrete random variables—things you can count, like the number of heads in a coin toss. But what about things you measure ? Think height, weight, or the time it takes to run a race. These can take on any value in a range, and that's where Continuous Random Variables come in. It's a whole new ball game! 🤩 In this chapter, we're going to dive into the world of smooth, continuous data. First, we'll meet the Probability Density Function (PDF), which is like the rulebook for a continuous variable. Then, we'll learn how to calculate the mean and variance using our old friend, integration. Finally, we'll put it all together to find probabilities and percentiles, answering questions like 'What's the chance the bus arrives in the next 5 minutes?' It might sound tricky, but it's just about seeing probability as an area under a curve. Let's get started!

f(x)

. This rule is super important because it allows us to find unknown constants in the function, often called 'k', by setting up an integral and making it equal to 1. We call this finding the 'normalising constant'.

Worked example

Solving for 'k' - The Unlockable Character 🔓

X

X

m

Worked example

When is my pizza getting here?! 🍕

T

1
$P(5 < T < 7)$
$P(5 < T < 7) = \int_{5}^{7} \frac{1}{32}(t-2) \,dt$
2
$\frac{1}{32}$
$\frac{1}{32} \left[ \frac{t^2}{2} - 2t \right]_{5}^{7} = \frac{1}{32} \left( (\frac{7^2}{2} - 2(7)) - (\frac{5^2}{2} - 2(5)) \right) = \frac{1}{32} ((24.5 - 14) - (12.5 - 10)) = \frac{1}{32}(10.5 - 2.5) = \frac{8}{32} = 0.25$
3
$m$
$\int_{2}^{m} \frac{1}{32}(t-2) \,dt = 0.5$
4
$m$
$\frac{1}{32} \left[ \frac{t^2}{2} - 2t \right]_{2}^{m} = 0.5$
5
$m$
$\frac{1}{32} \left( (\frac{m^2}{2} - 2m) - (\frac{2^2}{2} - 2(2)) \right) = 0.5 \\ (\frac{m^2}{2} - 2m) - (2 - 4) = 16 \\ \frac{m^2}{2} - 2m + 2 = 16 \\ m^2 - 4m + 4 = 32 \\ (m-2)^2 = 32 \\ m-2 = \pm\sqrt{32} = \pm 4\sqrt{2}$
6
$m$
$m = 2 + 4\sqrt{2} \approx 7.66 \text{ minutes (3 s.f.)}$

Answer

m = 2 + 4\sqrt{2} \approx 7.66 \text{ minutes (3 s.f.)}

E(X)

Var(X)

Worked example

Let's Get This Bread: A Part-Time Job Problem 🍞

X

1
$E(X)$
$E(X) = \int_{0}^{4} x f(x) \,dx = \int_{0}^{4} x \left[\frac{3}{32}x(4-x)\right] \,dx = \frac{3}{32} \int_{0}^{4} (4x^2 - x^3) \,dx$
2
$Now we just do the integration, applying the power rule. Then we'll substitute the limits (4 and 0) to get our final value for the mean.$
$\frac{3}{32} \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_{0}^{4} = \frac{3}{32} \left[ \left( \frac{4(4)^3}{3} - \frac{(4)^4}{4} \right) - (0) \right] = \frac{3}{32} \left[ \frac{256}{3} - 64 \right] = \frac{3}{32} \left[ \frac{64}{3} \right] = 2$
3
$E(X^2)$
$E(X^2) = \int_{0}^{4} x^2 f(x) \,dx = \int_{0}^{4} x^2 \left[\frac{3}{32}x(4-x)\right] \,dx = \frac{3}{32} \int_{0}^{4} (4x^3 - x^4) \,dx$
4
$Time to integrate this new expression and substitute the limits, just like we did for the mean.$
$\frac{3}{32} \left[ \frac{4x^4}{4} - \frac{x^5}{5} \right]_{0}^{4} = \frac{3}{32} \left[ x^4 - \frac{x^5}{5} \right]_{0}^{4} = \frac{3}{32} \left[ (4^4 - \frac{4^5}{5}) - (0) \right] = \frac{3}{32} \left[ 256 - \frac{1024}{5} \right] = \frac{3}{32} \left[ \frac{256}{5} \right] = \frac{24}{5} = 4.8$
5
$Var(X) = E(X^2) - [E(X)]^2$
$Var(X) = E(X^2) - [E(X)]^2 = 4.8 - (2)^2 = 4.8 - 4 = 0.8$

Answer

Var(X) = E(X^2) - [E(X)]^2 = 4.8 - (2)^2 = 4.8 - 4 = 0.8

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