Continuous Random Variables - IB Mathematics Analysis and Approaches

1. A continuous random variable $X$ has probability density function $f(x) = 3x^2$ for $0 \le x \le 1$. Find the variance of $X$.
   Answer: $\frac{3}{80}$ The variance is given by $Var(X) = E(X^2) - [E(X)]^2$. First, find the mean, $E(X)$: $$E(X) = \int_{0}^{1} x f(x) \,dx = \int_{0}^{1} x(3x^2) \,dx = \int_{0}^{1} 3x^3 \,dx = \left[\frac{3x^4}{4}\right]_{0}^{1} = \frac{3}{4}.$$ Next, find $E(X^2)$: $$E(X^2) = \int_{0}^{1} x^2 f(x) \,dx = \int_{0}^{1} x^2(3x^2) \,dx = \int_{0}^{1} 3x^4 \,dx = \left[\frac{3x^5}{5}\right]_{0}^{1} = \frac{3}{5}.$$ Finally, calculate the variance: $$Var(X) = E(X^2) - [E(X)]^2 = \frac{3}{5} - \left(\frac{3}{4}\right)^2 = \frac{3}{5} - \frac{9}{16} = \frac{48 - 45}{80} = \frac{3}{80}.$$ Distractor B is $[E(X)]^2$. Distractor C is $E(X^2)$. Distractor D is $E(X)$.
2. The probability density function of a continuous random variable $X$ is $f(x) = \frac{4}{\pi(1+x^2)}$ for $0 \le x \le 1$. Find the exact value of $P(0 \le X \le \frac{1}{\sqrt{3}})$.
   Answer: $\frac{2}{3}$ When you see a probability density function and need to find the probability over an interval, you're looking at a fundamental integration problem. The probability $P(a \le X \le b)$ equals the definite integral $\int_a^b f(x) \, dx$. Here, you need to calculate $P(0 \le X \le \frac{1}{\sqrt{3}}) = \int_0^{1/\sqrt{3}} \frac{4}{\pi(1+x^2)} \, dx$. The key insight is recognizing that $\frac{1}{1+x^2}$ is the derivative of $\arctan(x)$. This gives us: $\int_0^{1/\sqrt{3}} \frac{4}{\pi(1+x^2)} \, dx = \frac{4}{\pi} \int_0^{1/\sqrt{3}} \frac{1}{1+x^2} \, dx = \frac{4}{\pi}[\arctan(x)]_0^{1/\sqrt{3}}$ Evaluating: $\frac{4}{\pi}[\arctan(\frac{1}{\sqrt{3}}) - \arctan(0)] = \frac{4}{\pi}[\frac{\pi}{6} - 0] = \frac{4}{\pi} \cdot \frac{\pi}{6} = \frac{2}{3}$ This confirms answer D is correct. Answer A ($\frac{\pi}{6}$) represents just the value of $\arctan(\frac{1}{\sqrt{3}})$ without applying the constant $\frac{4}{\pi}$. Answer B ($\frac{4}{3}$) likely comes from an error in the arctangent evaluation or integration bounds. Answer C ($\frac{3}{\pi}$) suggests a computational mistake, possibly confusing the coefficient manipulation. Remember: when integrating probability density functions, always check if the integrand matches a standard derivative form. The arctangent derivative $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$ appears frequently in these problems, so memorize key arctangent values like $\arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
3. A continuous random variable $X$ has a PDF given by $f(x) = \begin{cases} kx & 0 \le x \le 2 \\ k(4-x) & 2 < x \le 4 \end{cases}$. Find $P(1 < X < 3)$.
   Answer: $\frac{3}{4}$ First, we find the constant $k$. The graph of the PDF is a triangle with base 4 and height (2k). The area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2k = 4k$. For a valid PDF, the area must be 1, so $4k=1 \implies k=\frac{1}{4}$. Now we calculate $P(1 < X < 3)$ by splitting the integral: $$P(1 < X < 3) = \int_{1}^{2} \frac{1}{4}x \,dx + \int_{2}^{3} \frac{1}{4}(4-x) \,dx$$ $$= \frac{1}{4}\left[\frac{x^2}{2}\right]_{1}^{2} + \frac{1}{4}\left[4x - \frac{x^2}{2}\right]_{2}^{3}$$ $$= \frac{1}{4}\left(\frac{4}{2} - \frac{1}{2}\right) + \frac{1}{4}\left( (12 - \frac{9}{2}) - (8 - \frac{4}{2}) \right)$$ $$= \frac{1}{4}\left(\frac{3}{2}\right) + \frac{1}{4}\left( \frac{15}{2} - 6 \right) = \frac{3}{8} + \frac{1}{4}\left(\frac{3}{2}\right) = \frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4}.$$ Distractor A is the value of just one of the integrals, $\int_1^2 f(x) dx$ or $\int_2^3 f(x) dx$. Distractor B is the median of the distribution. By symmetry, the median is 2. This distractor represents $P(X<2)$ or $P(X>2)$.
4. A continuous random variable $X$ is uniformly distributed on the interval $[1, 3]$. Find $E(X^3)$.
   Answer: 10 For a uniform distribution on $[a, b]$, the PDF is $f(x) = \frac{1}{b-a}$. Here, $f(x) = \frac{1}{3-1} = \frac{1}{2}$ for $1 \le x \le 3$. The expected value of a function of $X$, $g(X)$, is given by $E(g(X)) = \int_a^b g(x)f(x) \,dx$. Here, $g(x) = x^3$. $$E(X^3) = \int_{1}^{3} x^3 \cdot \frac{1}{2} \,dx$$ $$= \frac{1}{2} \left[\frac{x^4}{4}\right]_{1}^{3}$$ $$= \frac{1}{8} [x^4]_{1}^{3} = \frac{1}{8} (3^4 - 1^4) = \frac{1}{8} (81 - 1) = \frac{80}{8} = 10.$$ Distractor A is $[E(X)]^3$. The mean is $E(X) = \frac{1+3}{2} = 2$, so $2^3 = 8$. Distractor C is the result of forgetting the PDF constant $\frac{1}{2}$ in the integration. Distractor D is the result of incorrectly calculating the integral of $x^3$ as $3^3 - 1^3$.
5. A continuous random variable $X$ has a probability density function (PDF) given by $f(x) = kx^2(3-x)$ for $0 \le x \le 3$, and $f(x) = 0$ otherwise. Find the expected value of $X$.
   Answer: $\frac{9}{5}$ First, we must find the value of the constant $k$ by ensuring the total probability is 1. $$\int_{0}^{3} kx^2(3-x) \,dx = 1$$ $$k \int_{0}^{3} (3x^2 - x^3) \,dx = 1$$ $$k \left[x^3 - \frac{x^4}{4}\right]_{0}^{3} = 1$$ $$k \left(3^3 - \frac{3^4}{4}\right) = 1$$ $$k \left(27 - \frac{81}{4}\right) = 1 \implies k \left(\frac{108-81}{4}\right) = 1 \implies k\left(\frac{27}{4}\right)=1 \implies k = \frac{4}{27}.$$ Next, we find the expected value, $E(X)$: $$E(X) = \int_{0}^{3} x \cdot f(x) \,dx = \int_{0}^{3} x \cdot \frac{4}{27}x^2(3-x) \,dx$$ $$E(X) = \frac{4}{27} \int_{0}^{3} (3x^3 - x^4) \,dx$$ $$E(X) = \frac{4}{27} \left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_{0}^{3}$$ $$E(X) = \frac{4}{27} \left(\frac{3(81)}{4} - \frac{243}{5}\right) = \frac{4}{27} \left(\frac{243}{4} - \frac{243}{5}\right)$$ $$E(X) = \frac{4 \cdot 243}{27} \left(\frac{1}{4} - \frac{1}{5}\right) = 36 \left(\frac{1}{20}\right) = \frac{36}{20} = \frac{9}{5}.$$ Distractor A arises from incorrectly defining the PDF as $k(3x-x^2)$. Distractor C is the mode of the distribution, found by maximizing $f(x)$. Distractor D is a plausible calculation error.
6. A continuous random variable $X$ follows a uniform distribution on the interval $[a, b]$. The mean of $X$ is 8 and the variance is 12. Determine the values of $a$ and $b$.
   Answer: $a=2, b=14$ For a uniform distribution on $[a, b]$, the mean is $E(X) = \frac{a+b}{2}$ and the variance is $Var(X) = \frac{(b-a)^2}{12}$. From the given information, we have two equations: 1) $\frac{a+b}{2} = 8 \implies a+b = 16$ 2) $\frac{(b-a)^2}{12} = 12 \implies (b-a)^2 = 144 \implies b-a = 12$ (since $b>a$). We now have a system of linear equations: $a+b = 16$ $b-a = 12$ Adding the two equations gives $2b = 28 \implies b=14$. Substituting $b=14$ into the first equation gives $a+14=16 \implies a=2$. So, $a=2$ and $b=14$. Distractor B has the correct mean but incorrect variance. Distractor C arises from mixing up the formulas for mean and variance. Distractor D arises from incorrectly taking the square root of 12 for $b-a$.
7. A continuous random variable $X$ has a probability density function given by $f(x) = \frac{1}{8}x$ for $0 \le x \le 4$. Find $P(X > 3 | X > 1)$.
   Answer: $\frac{7}{15}$ The conditional probability is given by the formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Let $A$ be the event $X > 3$ and $B$ be the event $X > 1$. The intersection $A \cap B$ is the event $X > 3$. So we need to calculate $\frac{P(X > 3)}{P(X > 1)}$. First, find $P(X > a)$ for a general $a$: $$P(X>a) = \int_{a}^{4} \frac{1}{8}x \,dx = \left[\frac{x^2}{16}\right]_{a}^{4} = \frac{16}{16} - \frac{a^2}{16} = 1 - \frac{a^2}{16}.$$ Now, calculate the specific probabilities: $$P(X > 3) = 1 - \frac{3^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}.$$ $$P(X > 1) = 1 - \frac{1^2}{16} = 1 - \frac{1}{16} = \frac{15}{16}.$$ Finally, the conditional probability is: $$P(X > 3 | X > 1) = \frac{7/16}{15/16} = \frac{7}{15}.$$ Distractor A is the numerator, $P(X>3)$. Distractor C is $P(X>2)$, which might arise from assuming a memoryless property which does not apply here. Distractor D is the denominator, $P(X>1)$.
8. The probability density function of a continuous random variable $X$ is defined by $f(x) = a(2-|x-2|)$ for $0 \le x \le 4$. Find $Var(X)$.
   Answer: $\frac{2}{3}$ The graph of $f(x)$ is an isosceles triangle with vertices at (0,0), (4,0), and (2, 2a). The area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2a = 4a$. Setting the area to 1 gives $4a=1 \implies a=1/4$. The PDF can be written as a piecewise function: $f(x) = \begin{cases} \frac{1}{4}x & 0 \le x \le 2 \\ \frac{1}{4}(4-x) & 2 < x \le 4 \end{cases}$. Due to symmetry about $x=2$, the mean is $E(X)=2$. We need $E(X^2)$ to find the variance. $$E(X^2) = \int_0^4 x^2 f(x) \,dx = \int_0^2 x^2(\frac{x}{4}) \,dx + \int_2^4 x^2(\frac{4-x}{4}) \,dx$$ $$= \frac{1}{4}\left[\frac{x^4}{4}\right]_0^2 + \frac{1}{4}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_2^4$$ $$= \frac{1}{4}(\frac{16}{4}) + \frac{1}{4}\left( (\frac{256}{3}-64) - (\frac{32}{3}-4) \right)$$ $$= 1 + \frac{1}{4}\left( \frac{224}{3} - 60 \right) = 1 + \frac{1}{4}\left(\frac{44}{3}\right) = 1 + \frac{11}{3} = \frac{14}{3}.$$ Finally, $Var(X) = E(X^2) - [E(X)]^2 = \frac{14}{3} - 2^2 = \frac{14}{3} - 4 = \frac{2}{3}$. Distractor B is the mean $E(X)$. Distractor C is a plausible calculation error. Distractor D is the value of $E(X^2)$.
9. A continuous random variable $X$ has probability density function $f(x) = \frac{1}{4}x^3$ for $0 \le x \le 2$. Find the median of $X$.
   Answer: $\sqrt[4]{8}$ The median $m$ is the value such that $P(X \le m) = 0.5$. We solve the integral equation: $$\int_{0}^{m} \frac{1}{4}x^3 \,dx = 0.5$$ $$\left[\frac{1}{4} \cdot \frac{x^4}{4}\right]_{0}^{m} = \frac{1}{2}$$ $$\frac{m^4}{16} - 0 = \frac{1}{2}$$ $$m^4 = 8 \implies m = \sqrt[4]{8}.$$ Distractor A is the mean of the distribution, $E(X) = \int_0^2 x \cdot \frac{1}{4}x^3 dx = \frac{8}{5}$. Distractor C is the mode of the distribution, as $f(x)$ is strictly increasing on its domain. It is also the value obtained by incorrectly setting the integral to 1. Distractor D is the result of incorrectly solving the equation $F(m)=m$, where $F(x)$ is the CDF.
10. Let $f(x)$ be the probability density function for a continuous random variable $X$. Which of the following statements is not always true?
    Answer: $f(c) \le 1$ for any constant $c$ The value of the probability density function (PDF) $f(x)$ can be greater than 1. The condition for a valid PDF is that the total area under the curve must be equal to 1, i.e., $\int_{-\infty}^{\infty} f(x) \,dx = 1$. For example, consider the uniform distribution on the interval $[0, 0.5]$. The PDF is $f(x) = 2$ for $0 \le x \le 0.5$, and $f(x)=0$ otherwise. Here $f(x) > 1$ on its domain, but $\int_0^{0.5} 2 \,dx = 1$. The other three statements are properties that are always true for any continuous random variable's PDF.
11. The probability density function of a continuous random variable $X$ is given by $f(x) = \frac{1}{2}x^2e^{-x}$ for $x \ge 0$. Determine the mode of $X$.
    Answer: 2 The mode is the value of $x$ for which the PDF $f(x)$ is at its maximum. To find this, we differentiate $f(x)$ and set the derivative to zero. $$f'(x) = \frac{d}{dx} \left(\frac{1}{2}x^2e^{-x}\right)$$ Using the product rule: $$f'(x) = \frac{1}{2} \left( (2x)(e^{-x}) + (x^2)(-e^{-x}) \right)$$ $$f'(x) = \frac{1}{2}xe^{-x}(2-x)$$ Setting $f'(x) = 0$, we get $x=0$ or $x=2$. Since $f(0)=0$ and $f(x) > 0$ for $x>0$, $x=0$ is a minimum. The maximum must occur at $x=2$. Thus, the mode is 2. Distractor A is the value of the constant $k$ in the more general form $kx^2e^{-x}$. Distractor B could be an error in solving $2-x=0$. Distractor D is the mean of the distribution, $E(X) = \int_0^\infty x \cdot \frac{1}{2}x^2e^{-x} dx = 3$.
12. The lifetime $T$ of a component, in years, is modelled by the probability density function $f(t) = \lambda e^{-\lambda t}$ for $t \ge 0$. The mean lifetime is 20 years. What is the probability that a component lasts for more than 30 years?
    Answer: $e^{-3/2}$ For an exponential distribution with PDF $f(t) = \lambda e^{-\lambda t}$, the mean is $E(T) = 1/\lambda$. Given that the mean lifetime is 20 years, we have $1/\lambda = 20 \implies \lambda = 1/20$. The probability that a component lasts for more than 30 years is $P(T > 30)$. The cumulative distribution function (CDF) is $F(t) = P(T \le t) = 1 - e^{-\lambda t}$. Therefore, $P(T > 30) = 1 - P(T \le 30) = 1 - F(30)$. $$P(T > 30) = 1 - (1 - e^{-(1/20)(30)}) = e^{-30/20} = e^{-3/2}.$$ Alternatively, one can integrate: $\int_{30}^{\infty} \frac{1}{20} e^{-t/20} dt = [-e^{-t/20}]_{30}^{\infty} = 0 - (-e^{-30/20}) = e^{-3/2}$. Distractor A arises from incorrectly using $\lambda=20$ and $t=30$. Distractor C arises from swapping the values for mean and time, i.e., using $t=20$ and $1/\lambda=30$. Distractor D is the probability that the component lasts for 30 years or less, $P(T \le 30)$.
13. A continuous random variable $X$ has a probability density function given by $f(x) = \begin{cases} x & 0 \le x \le 1 \\ 2-x & 1 < x \le 2 \end{cases}$. Find the value of $c$ such that $P(X > c) = \frac{1}{8}$.
    Answer: $\frac{3}{2}$ We need to find $c$ such that $\int_c^2 f(x) dx = \frac{1}{8}$. The area of the right half of the triangular distribution (from $x=1$ to $x=2$) is $\int_1^2 (2-x)dx = [2x - x^2/2]_1^2 = (4-2) - (2-1/2) = 2-3/2 = 1/2$. Since $\frac{1}{8} < \frac{1}{2}$, the value of $c$ must be in the interval $(1, 2]$. We solve the equation: $$\int_{c}^{2} (2-x) \,dx = \frac{1}{8}$$ $$\left[2x - \frac{x^2}{2}\right]_{c}^{2} = \frac{1}{8}$$ $$\left(2(2) - \frac{2^2}{2}\right) - \left(2c - \frac{c^2}{2}\right) = \frac{1}{8}$$ $$(4-2) - (2c - \frac{c^2}{2}) = \frac{1}{8}$$ $$2 - 2c + \frac{c^2}{2} = \frac{1}{8}$$ Multiplying by 8 gives: $16 - 16c + 4c^2 = 1$, which simplifies to $4c^2 - 16c + 15 = 0$. Factoring gives $(2c-3)(2c-5) = 0$. The solutions are $c = \frac{3}{2}$ and $c = \frac{5}{2}$. Since $c$ must be in the domain $[0, 2]$, we choose $c = \frac{3}{2}$. Distractor A is the value of $c$ such that $P(X < c) = \frac{1}{8}$. Distractor D is the other, extraneous, solution to the quadratic equation.
14. The continuous random variable $X$ has probability density function $f(x) = 6x(1-x)$ for $0 \le x \le 1$. Find $Var(2-3X)$.
    Answer: $\frac{9}{20}$ First, we find $Var(X) = E(X^2) - [E(X)]^2$. $$E(X) = \int_0^1 x \cdot 6x(1-x) \,dx = 6 \int_0^1 (x^2-x^3) \,dx = 6[\frac{x^3}{3}-\frac{x^4}{4}]_0^1 = 6(\frac{1}{3}-\frac{1}{4}) = \frac{1}{2}.$$ $$E(X^2) = \int_0^1 x^2 \cdot 6x(1-x) \,dx = 6 \int_0^1 (x^3-x^4) \,dx = 6[\frac{x^4}{4}-\frac{x^5}{5}]_0^1 = 6(\frac{1}{4}-\frac{1}{5}) = \frac{3}{10}.$$ $$Var(X) = \frac{3}{10} - (\frac{1}{2})^2 = \frac{3}{10} - \frac{1}{4} = \frac{6-5}{20} = \frac{1}{20}.$$ Now, we use the property $Var(aX+b) = a^2 Var(X)$. For $Y=2-3X$, we have $a=-3$ and $b=2$. $$Var(2-3X) = (-3)^2 Var(X) = 9 \cdot \frac{1}{20} = \frac{9}{20}.$$ Distractor A is $Var(X)$. Distractor B is the result of using $a=3$ but forgetting to square it, i.e., $3 Var(X)$. Distractor D is the mean of the transformed variable, $E(2-3X) = 2 - 3E(X) = 2 - 3(1/2) = 1/2$.
15. A continuous random variable $X$ has a probability density function $f(x) = k \cos(x)$ for $0 \le x \le \frac{\pi}{2}$. Find $P(X < \frac{\pi}{3})$.
    Answer: $\frac{\sqrt{3}}{2}$ First, find the constant $k$ by integrating the PDF from 0 to $\pi/2$ and setting the result to 1. $$\int_{0}^{\pi/2} k \cos(x) \,dx = 1$$ $$k[\sin(x)]_{0}^{\pi/2} = 1$$ $$k(\sin(\pi/2) - \sin(0)) = 1 \implies k(1-0) = 1 \implies k=1.$$ So the PDF is $f(x) = \cos(x)$. Now, we find the required probability: $$P(X < \frac{\pi}{3}) = \int_{0}^{\pi/3} \cos(x) \,dx$$ $$= [\sin(x)]_{0}^{\pi/3} = \sin(\frac{\pi}{3}) - \sin(0) = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2}.$$ Distractor A is the complementary probability, $P(X > \frac{\pi}{3})$. Distractor B is the value of the PDF at $x=\pi/3$, $\cos(\pi/3)$. Distractor D is the value of the constant $k$.
16. The continuous random variable $X$ has mean $E(X) = 10$ and variance $Var(X) = 4$. A new random variable $Y$ is defined by the transformation $Y = 3X - 5$. Find the variance of $Y$.
    Answer: 36 We use the properties of variance for a linear transformation $Y = aX + b$. The formula is $Var(Y) = a^2 Var(X)$. The additive constant $b$ does not affect the variance. In this case, $a = 3$ and $b = -5$. $$Var(Y) = Var(3X - 5) = 3^2 Var(X) = 9 \times 4 = 36.$$ Distractor A results from forgetting to square the coefficient $a$, calculating $3 \times Var(X)$. Distractor B is the expected value of $Y$, calculated as $E(Y) = 3E(X) - 5 = 3(10) - 5 = 25$. Distractor D results from incorrectly calculating $a^2 Var(X) + b^2$.
17. A continuous random variable $X$ has a probability density function given by $f(x) = \frac{k}{x}$ for $1 \le x \le e^2$. Find $E(X)$.
    Answer: $\frac{1}{2}(e^2-1)$ First, find the constant $k$ by ensuring the total probability is 1. $$\int_1^{e^2} \frac{k}{x} \,dx = 1$$ $$k[\ln|x|]_1^{e^2} = 1$$ $$k(\ln(e^2) - \ln(1)) = 1 \implies k(2-0) = 1 \implies k = \frac{1}{2}.$$ So the PDF is $f(x) = \frac{1}{2x}$. Now, find the expected value, $E(X)$: $$E(X) = \int_1^{e^2} x \cdot f(x) \,dx = \int_1^{e^2} x \cdot \frac{1}{2x} \,dx = \int_1^{e^2} \frac{1}{2} \,dx$$ $$= \left[\frac{1}{2}x\right]_1^{e^2} = \frac{1}{2}(e^2 - 1).$$ Distractor A is the value of the constant $k$. Distractor B could arise from a calculation error, or it is the value of $E(\ln X)$. Distractor C is the value of the integral $\int_1^{e^2} \frac{1}{x} dx$.
18. The probability density function of a continuous random variable $X$ corresponds to a distribution that is skewed to the right (positively skewed). Which of the following shows the most likely relationship between the mean ($\mu$), median ($m$), and mode ($M$)?
    Answer: $M < m < \mu$ For a unimodal distribution that is skewed to the right (positively skewed), the 'tail' of the distribution is on the right. The mode (M) is the peak of the distribution. The median (m) is the value that splits the area in half. The mean ($\mu$) is pulled towards the long tail by the larger values. Therefore, the typical order is Mode < Median < Mean. Distractor A represents a symmetric distribution. Distractor B represents a distribution that is skewed to the left (negatively skewed). Distractor D does not represent a standard relationship for skewed distributions.
19. A continuous random variable $X$ has a cumulative distribution function $F(x) = \frac{x^2-1}{8}$ for $1 \le x \le 3$. Find the interquartile range of $X$.
    Answer: $\sqrt{7} - \sqrt{3}$ When you encounter a cumulative distribution function (CDF) problem asking for the interquartile range, you need to find the first quartile (Q₁) and third quartile (Q₃), then calculate Q₃ - Q₁. The quartiles are found by solving $F(Q_1) = 0.25$ and $F(Q_3) = 0.75$. For Q₁: Set $\frac{x^2-1}{8} = 0.25$ $x^2 - 1 = 2$ $x^2 = 3$ $x = \sqrt{3}$ (taking the positive root since $x \geq 1$) For Q₃: Set $\frac{x^2-1}{8} = 0.75$ $x^2 - 1 = 6$ $x^2 = 7$ $x = \sqrt{7}$ Therefore, the interquartile range is $Q_3 - Q_1 = \sqrt{7} - \sqrt{3}$, which is answer D. Let's examine why the other options are wrong: A) $\sqrt{5}$ likely comes from incorrectly calculating $\sqrt{7-2} = \sqrt{5}$ instead of $\sqrt{7} - \sqrt{3}$. B) $\sqrt{7} + \sqrt{3}$ represents adding the quartiles instead of finding their difference—a sign error. C) 4 might result from using the range of the domain (3-1=2) and doubling it, or from arithmetic errors in the quartile calculations. Study tip: Always remember that interquartile range equals Q₃ minus Q₁, not their sum. When working with CDFs, double-check your algebra when solving $F(x) = p$ for specific probability values, as small errors compound quickly in these calculations.
20. The cumulative distribution function (CDF) of a continuous random variable $X$ is given by $F(x) = \frac{x^3}{27}$ for $0 \le x \le 3$. Find $P(X > 2)$.
    Answer: $\frac{19}{27}$ The probability $P(X > 2)$ can be calculated using the complement rule and the cumulative distribution function (CDF), $F(x)$. $$P(X > 2) = 1 - P(X \le 2)$$ By definition, $P(X \le 2) = F(2)$. $$F(2) = \frac{2^3}{27} = \frac{8}{27}$$ Therefore, $P(X > 2) = 1 - \frac{8}{27} = \frac{19}{27}$. Distractor A is $P(X \le 2)$. Distractor B is the value of the probability density function $f(x) = F'(x) = \frac{3x^2}{27} = \frac{x^2}{9}$ evaluated at $x=2$, i.e., $f(2) = \frac{4}{9}$. This is a common error of confusing the PDF value with probability. Distractor D is $P(X>1)$, a misreading of the question.

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