IB Mathematics: Analysis and Approaches Quiz: Continuous Random Variables

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Continuous Random VariablesQuestion 1 of 20

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 $P(X > c) = \frac{1}{8}$ .

\frac{1}{2}

\frac{3}{2}

\frac{5}{2}

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Continuous Random Variables

Practice Continuous Random Variables in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Continuous Random Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

$\frac{1}{2}$
$\frac{3}{2}$ (correct answer)
2
$\frac{5}{2}$

Explanation: 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

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]

. \nWe solve the equation: \n

\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}

\nMultiplying by 8 gives:

16 - 16c + 4c^2 = 1

, which simplifies to

4c^2 - 16c + 15 = 0

. \nFactoring gives

(2c-3)(2c-5) = 0

. The solutions are

c = \frac{3}{2}

and

c = \frac{5}{2}

. \nSince

c

must be in the domain

[0, 2]

, we choose

c = \frac{3}{2}

. \nDistractor A is the value of

c

such that

P(X < c) = \frac{1}{8}

. \nDistractor D is the other, extraneous, solution to the quadratic equation.

Question 2

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 $X$ .

$\frac{1}{2}$
1
2 (correct answer)
3

Explanation: 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. \n

f'(x) = \frac{d}{dx} \left(\frac{1}{2}x^2e^{-x}\right)

Using the product rule: \n

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

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. \nDistractor A is the value of the constant

k

in the more general form

kx^2e^{-x}

. \nDistractor B could be an error in solving

2-x=0

. \nDistractor D is the mean of the distribution,

E(X) = \int_0^\infty x \cdot \frac{1}{2}x^2e^{-x} dx = 3

Question 3

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?

$e^{-600}$
$e^{-3/2}$ (correct answer)
$e^{-2/3}$
$1 - e^{-3/2}$

Explanation: For an exponential distribution with PDF

f(t) = \lambda e^{-\lambda t}

, the mean is

E(T) = 1/\lambda

. \nGiven that the mean lifetime is 20 years, we have

1/\lambda = 20 \implies \lambda = 1/20

. \nThe probability that a component lasts for more than 30 years is

P(T > 30)

. \nThe cumulative distribution function (CDF) is

F(t) = P(T \le t) = 1 - e^{-\lambda t}

. \nTherefore,

P(T > 30) = 1 - P(T \le 30) = 1 - F(30)

. \n

P(T > 30) = 1 - (1 - e^{-(1/20)(30)}) = e^{-30/20} = e^{-3/2}.

\nAlternatively, 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}

. \nDistractor A arises from incorrectly using

\lambda=20

and

t=30

. \nDistractor C arises from swapping the values for mean and time, i.e., using

t=20

and

1/\lambda=30

. \nDistractor D is the probability that the component lasts for 30 years or less,

P(T \le 30)

Question 4

$\frac{1}{2}$
$\frac{3}{2}$ (correct answer)
2
$\frac{5}{2}$

Explanation: 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

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]

. \nWe solve the equation: \n

\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}

\nMultiplying by 8 gives:

16 - 16c + 4c^2 = 1

, which simplifies to

4c^2 - 16c + 15 = 0

. \nFactoring gives

(2c-3)(2c-5) = 0

. The solutions are

c = \frac{3}{2}

and

c = \frac{5}{2}

. \nSince

c

must be in the domain

[0, 2]

, we choose

c = \frac{3}{2}

. \nDistractor A is the value of

c

such that

P(X < c) = \frac{1}{8}

. \nDistractor D is the other, extraneous, solution to the quadratic equation.

Question 5

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 ( $M$ )?

$\mu = m = M$
$\mu < m < M$
$M < m < \mu$ (correct answer)
$m < \mu < M$

Explanation: 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. \nDistractor A represents a symmetric distribution. \nDistractor B represents a distribution that is skewed to the left (negatively skewed). \nDistractor D does not represent a standard relationship for skewed distributions.

Question 6

$\mu = m = M$
$\mu < m < M$
$M < m < \mu$ (correct answer)
$m < \mu < M$

\mu

Question 7

$e^{-600}$
$e^{-3/2}$ (correct answer)
$e^{-2/3}$
$1 - e^{-3/2}$

Explanation: For an exponential distribution with PDF

f(t) = \lambda e^{-\lambda t}

, the mean is

E(T) = 1/\lambda

. \nGiven that the mean lifetime is 20 years, we have

1/\lambda = 20 \implies \lambda = 1/20

. \nThe probability that a component lasts for more than 30 years is

P(T > 30)

. \nThe cumulative distribution function (CDF) is

F(t) = P(T \le t) = 1 - e^{-\lambda t}

. \nTherefore,

P(T > 30) = 1 - P(T \le 30) = 1 - F(30)

. \n

P(T > 30) = 1 - (1 - e^{-(1/20)(30)}) = e^{-30/20} = e^{-3/2}.

\nAlternatively, 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}

. \nDistractor A arises from incorrectly using

\lambda=20

and

t=30

. \nDistractor C arises from swapping the values for mean and time, i.e., using

t=20

and

1/\lambda=30

. \nDistractor D is the probability that the component lasts for 30 years or less,

P(T \le 30)

Question 8

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 $P(0 \le X \le \frac{1}{\sqrt{3}})$ .

$\frac{\pi}{6}$
$\frac{4}{3}$
$\frac{3}{\pi}$
$\frac{2}{3}$ (correct answer)

Explanation: 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}

Question 9

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 $E(X)$ .

$\frac{1}{2}$
1
2
$\frac{1}{2}(e^2-1)$ (correct answer)

Explanation: First, find the constant

k

by ensuring the total probability is 1. \n

\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}.

\nSo the PDF is

f(x) = \frac{1}{2x}

. Now, find the expected value,

E(X)

: \n

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).

\nDistractor A is the value of the constant

k

. \nDistractor B could arise from a calculation error, or it is the value of

E(\ln X)

. \nDistractor C is the value of the integral

\int_1^{e^2} \frac{1}{x} dx

Question 10

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 $b$ .

$a=2, b=14$ (correct answer)
$a=4, b=12$
$a=8, b=16$
$a=8-\sqrt{3}, b=8+\sqrt{3}$

Explanation: 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}

. \nFrom the given information, we have two equations: \n1)

\frac{a+b}{2} = 8 \implies a+b = 16

\n2)

\frac{(b-a)^2}{12} = 12 \implies (b-a)^2 = 144 \implies b-a = 12

(since

b>a

). \nWe now have a system of linear equations: \n

a+b = 16

b-a = 12

\nAdding the two equations gives

2b = 28 \implies b=14

. \nSubstituting

b=14

into the first equation gives

a+14=16 \implies a=2

. \nSo,

a=2

and

b=14

. \nDistractor B has the correct mean but incorrect variance. \nDistractor C arises from mixing up the formulas for mean and variance. \nDistractor D arises from incorrectly taking the square root of 12 for

b-a

Question 11

The continuous random variable $X$ has probability density function $f(x) = 6x(1-x)$ for $0 \le x \le 1$ . Find $Var(2-3X)$ .

$\frac{1}{20}$
$\frac{3}{20}$
$\frac{9}{20}$ (correct answer)
$\frac{1}{2}$

Explanation: First, we find

Var(X) = E(X^2) - [E(X)]^2

. \n

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}.

\nNow, we use the property

Var(aX+b) = a^2 Var(X)

. For

Y=2-3X

, we have

a=-3

and

b=2

. \n

Var(2-3X) = (-3)^2 Var(X) = 9 \cdot \frac{1}{20} = \frac{9}{20}.

\nDistractor A is

Var(X)

. \nDistractor B is the result of using

a=3

but forgetting to square it, i.e.,

3 Var(X)

. \nDistractor D is the mean of the transformed variable,

E(2-3X) = 2 - 3E(X) = 2 - 3(1/2) = 1/2

Question 12

Let $f(x)$ be the probability density function for a continuous random variable $X$ . Which of the following statements is not always true?

$\int_{-\infty}^{\infty} f(x) \,dx = 1$
$f(x) \ge 0$ for all $x$
$P(X=c) = 0$ for any constant $c$
$f(c) \le 1$ for any constant $c$ (correct answer)

Explanation: 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.

Question 13

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 $X$ .

$\sqrt{5}$
$\sqrt{7} + \sqrt{3}$
4
$\sqrt{7} - \sqrt{3}$ (correct answer)

Explanation: 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.

Question 14

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 $Var(X)$ .

$\frac{2}{3}$ (correct answer)
2
$\frac{7}{3}$
$\frac{14}{3}$

Explanation: 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

. \nThe 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}

. \nDue to symmetry about

x=2

, the mean is

E(X)=2

. \nWe need

E(X^2)

to find the variance. \n

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}.

\nFinally,

Var(X) = E(X^2) - [E(X)]^2 = \frac{14}{3} - 2^2 = \frac{14}{3} - 4 = \frac{2}{3}

. \nDistractor B is the mean

E(X)

. \nDistractor C is a plausible calculation error. \nDistractor D is the value of

E(X^2)

Question 15

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 $X$ .

$\frac{3}{2}$
$\frac{9}{5}$ (correct answer)
2
$\frac{12}{5}$

Explanation: First, we must find the value of the constant

k

by ensuring the total probability is 1. \n

\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}.

\nNext, we find the expected value,

E(X)

: \n

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}.

\nDistractor A arises from incorrectly defining the PDF as

k(3x-x^2)

. \nDistractor C is the mode of the distribution, found by maximizing

f(x)

. \nDistractor D is a plausible calculation error.

Question 16

A continuous random variable $X$ has a probability density function $f(x) = k \cos(x)$ for $0 \le x \le \frac{\pi}{2}$ . Find $P(X < \frac{\pi}{3})$ .

$1 - \frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$ (correct answer)
1

Explanation: First, find the constant

k

by integrating the PDF from 0 to

\pi/2

and setting the result to 1. \n

\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.

\nSo the PDF is

f(x) = \cos(x)

. Now, we find the required probability: \n

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}.

\nDistractor A is the complementary probability,

P(X > \frac{\pi}{3})

. \nDistractor B is the value of the PDF at

x=\pi/3

\cos(\pi/3)

. \nDistractor D is the value of the constant

k

Question 17

A continuous random variable $X$ is uniformly distributed on the interval $[1, 3]$ . Find $E(X^3)$ .

8
10 (correct answer)
20
26

Explanation: 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

. \nThe expected value of a function of

X

g(X)

, is given by

E(g(X)) = \int_a^b g(x)f(x) \,dx

. \nHere,

g(x) = x^3

. \n

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.

\nDistractor A is

[E(X)]^3

. The mean is

E(X) = \frac{1+3}{2} = 2

, so

2^3 = 8

. \nDistractor C is the result of forgetting the PDF constant

\frac{1}{2}

in the integration. \nDistractor D is the result of incorrectly calculating the integral of

x^3

3^3 - 1^3

Question 18

The continuous random variable $X$ has probability density function $f(x) = 6x(1-x)$ for $0 \le x \le 1$ . Find $Var(2-3X)$ .

$\frac{1}{20}$
$\frac{3}{20}$
$\frac{9}{20}$ (correct answer)
$\frac{1}{2}$

Explanation: First, we find

Var(X) = E(X^2) - [E(X)]^2

. \n

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}.

\nNow, we use the property

Var(aX+b) = a^2 Var(X)

. For

Y=2-3X

, we have

a=-3

and

b=2

. \n

Var(2-3X) = (-3)^2 Var(X) = 9 \cdot \frac{1}{20} = \frac{9}{20}.

\nDistractor A is

Var(X)

. \nDistractor B is the result of using

a=3

but forgetting to square it, i.e.,

3 Var(X)

. \nDistractor D is the mean of the transformed variable,

E(2-3X) = 2 - 3E(X) = 2 - 3(1/2) = 1/2

Question 19

A continuous random variable $X$ has probability density function $f(x) = 3x^2$ for $0 \le x \le 1$ . Find the variance of $X$ .

$\frac{3}{80}$ (correct answer)
$\frac{9}{16}$
$\frac{3}{5}$
$\frac{3}{4}$

Explanation: The variance is given by

Var(X) = E(X^2) - [E(X)]^2

. \nFirst, find the mean,

E(X)

: \n

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}.

\nNext, find

E(X^2)

: \n

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}.

\nFinally, calculate the variance: \n

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}.

\nDistractor B is

[E(X)]^2

. \nDistractor C is

E(X^2)

. \nDistractor D is

E(X)

Question 20

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 $X$ .

$\frac{3}{2}$
$\frac{9}{5}$ (correct answer)
2
$\frac{12}{5}$

Explanation: First, we must find the value of the constant

k

by ensuring the total probability is 1. \n

\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}.

\nNext, we find the expected value,

E(X)

: \n

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}.

\nDistractor A arises from incorrectly defining the PDF as

k(3x-x^2)

. \nDistractor C is the mode of the distribution, found by maximizing

f(x)

. \nDistractor D is a plausible calculation error.