Further Maths Continuous random variables Practice Questions

Free Further Maths Continuous random variables practice questions with full step-by-step worked solutions. Covers continuous-random-variables, probability-density-function, find-k, probability. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The continuous random variable XX has probability density function f(x)={kx0≤x≤40otherwisef(x)=\begin{cases}k x&0\le x\le 4\\0&\text{otherwise}\end{cases}, where kk is a constant. Find the value of kk.
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Worked solution

  1. Use the fact that the total area under a density is 11

    ∫04kx dx=1\int_{0}^{4}k x\,dx=1

    The density is zero outside the support, so this integral fixes kk.

  2. Integrate and evaluate the limits

    [kx22]04=8k\left[\frac{k x^{2}}{2}\right]_{0}^{4}=8 k

    The total area comes out as a multiple of kk.

  3. Solve for kk

    8k=1 ⇒ k=188 k=1\ \Rightarrow\ k=\frac{1}{8}

    Dividing through gives the constant immediately.

  4. State the value of kk

    k=18k=\frac{1}{8}

    This is the only value of kk for which the total area is 11.

Answer
k=18k=\frac{1}{8}
Question 2
2 markseasy
Each of the following functions is defined to be zero outside the interval shown. Which one is a valid probability density function?
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Worked solution

  1. Recall the two conditions for a probability density function

    f(x)≥0and∫−∞∞f(x) dx=1f(x)\ge0\quad\text{and}\quad\int_{-\infty}^{\infty}f(x)\,dx=1

    Both conditions must hold; either one on its own is not enough.

  2. Integrate the candidate over its stated interval

    ∫023x28 dx=1\int_{0}^{2}\frac{3 x^{2}}{8}\,dx=1

    The total area under a valid density is exactly 11.

  3. Check the sign of the candidate on that interval

    3x28≥0 for 0≤x≤2\frac{3 x^{2}}{8}\ge0\ \text{for}\ 0\le x\le 2

    A density can never take a negative value.

  4. Select the valid density

    f(x)=3x28, 0≤x≤2f(x)=\frac{3 x^{2}}{8},\ 0\le x\le 2

    This is the only option that is non-negative and encloses unit area.

Answer
f(x)=3x28, 0≤x≤2f(x)=\frac{3 x^{2}}{8},\ 0\le x\le 2
Question 3
4 marksintermediate
The continuous random variable XX has probability density function f(x)={x290≤x≤30otherwisef(x)=\begin{cases}\frac{x^{2}}{9}&0\le x\le 3\\0&\text{otherwise}\end{cases}. Which of the following is the median of XX?
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Worked solution

  1. Write down the equation satisfied by the median

    F(m)=12F\left(m\right)=\frac{1}{2}

    The median is the value with an area of 12\frac{1}{2} to its left.

  2. Use the branch of FF that contains the required value

    F(x)=x327 on 0≤x≤3F(x)=\frac{x^{3}}{27}\ \text{on}\ 0\le x\le 3

    The cumulative function reaches 12\frac{1}{2} somewhere on this interval.

  3. Form the equation

    m327=12\frac{m^{3}}{27}=\frac{1}{2}

    Replace xx by the unknown quantile and solve.

  4. Solve for the quantile

    m=3⋅2232m=\frac{3 \cdot 2^{\frac{2}{3}}}{2}

    Only the root that lies inside the support is admissible.

  5. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  6. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  7. Select the matching option

    m=3⋅2232m=\frac{3 \cdot 2^{\frac{2}{3}}}{2}

    Exactly half of the area under the density lies to the left of it.

Answer
m=3⋅2232m=\frac{3 \cdot 2^{\frac{2}{3}}}{2}
Question 4
6 markshard
The continuous random variable XX has probability density function f(x)={3x280≤x≤20otherwisef(x)=\begin{cases}\frac{3 x^{2}}{8}&0\le x\le 2\\0&\text{otherwise}\end{cases}. Which of the following is the value of Var(4X+5)\text{Var}\left(4 X + 5\right)?
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Worked solution

  1. Write down the integral for the mean

    E(X)=∫−∞∞xf(x) dx=∫023x38 dxE(X)=\int_{-\infty}^{\infty}xf(x)\,dx=\int_{0}^{2}\frac{3 x^{3}}{8}\,dx

    Multiply the density by xx and integrate over the support.

  2. Integrate

    =[3x432]02=\left[\frac{3 x^{4}}{32}\right]_{0}^{2}

    Antidifferentiate each piece of xf(x)xf(x).

  3. Evaluate the limits

    E(X)=32E(X)=\frac{3}{2}

    Substituting the limits gives the exact mean.

  4. Write down the integral for E(X2)E\left(X^{2}\right)

    E(X2)=∫−∞∞x2f(x) dx=∫023x48 dxE\left(X^{2}\right)=\int_{-\infty}^{\infty}x^{2}f(x)\,dx=\int_{0}^{2}\frac{3 x^{4}}{8}\,dx

    The same integral as for the mean, but with x2x^{2} in place of xx.

  5. Integrate and evaluate the limits

    E(X2)=125E\left(X^{2}\right)=\frac{12}{5}

    This is the second moment of XX about the origin.

  6. Apply the variance formula

    Var(X)=E(X2)−[E(X)]2=125−(32)2=320\text{Var}(X)=E\left(X^{2}\right)-\left[E(X)\right]^{2}=\frac{12}{5}-\left(\frac{3}{2}\right)^{2}=\frac{3}{20}

    Subtracting the square of the mean gives the variance.

  7. Apply the rule for a linear change

    Var(4X+5)=(4)2Var(X)=125\text{Var}\left(4 X + 5\right)=\left(4\right)^{2}\text{Var}(X)=\frac{12}{5}

    The multiplier is squared; the constant has no effect.

  8. Reject the option that forgets to square the multiplier

    4 Var(X)=354\,\text{Var}(X)=\frac{3}{5}

    This is the classic error in this topic.

  9. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  10. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  11. Select the matching option

    Var(4X+5)=125\text{Var}\left(4 X + 5\right)=\frac{12}{5}

    This is the only option consistent with the variance rule.

Answer
Var(4X+5)=125\text{Var}\left(4 X + 5\right)=\frac{12}{5}
Question 5
9 markschallenging
The continuous random variable XX has probability density function f(x)={6x250≤x≤1125−6x51≤x≤20otherwisef(x)=\begin{cases}\frac{6 x^{2}}{5}&0\le x\le 1\\\frac{12}{5} - \frac{6 x}{5}&1\le x\le 2\\0&\text{otherwise}\end{cases}. Which of the following is the cumulative distribution function of XX?
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Worked solution

  1. Build FF as the accumulated area under the density

    F(x)=∫0xf(u) duF(x)=\int_{0}^{x}f(u)\,du

    Integrate the density from the left-hand end of the support up to xx.

  2. Integrate the density on 0≤x≤10\le x\le 1

    F(x)=∫0x6t25 dt=2x35F(x)=\int_{0}^{x}\frac{6 t^{2}}{5}\,dt=\frac{2 x^{3}}{5}

    This is the area under the density from 00 up to xx.

  3. Integrate the next piece and add the area already accumulated

    F(x)=25+∫1x125−6t5 dt=−3x25+12x5−75F(x)=\frac{2}{5}+\int_{1}^{x}\frac{12}{5} - \frac{6 t}{5}\,dt=- \frac{3 x^{2}}{5} + \frac{12 x}{5} - \frac{7}{5}

    The area 25\frac{2}{5} under the earlier piece must be carried forward.

  4. Check the value at the top of the support

    F(2)=1F\left(2\right)=1

    A cumulative distribution function must finish at 11.

  5. Reject any option that does not start at 00

    F(0)=0F\left(0\right)=0

    No area has accumulated at the left-hand end of the support.

  6. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  7. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  8. Recall that a probability is an area

    P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_{a}^{b}f(x)\,dx

    For a continuous variable, probability is the area under the density.

  9. Note that single values carry no probability

    P(X=a)=0P(X=a)=0

    This is why P(a<X<b)P(a<X<b) and P(a≤X≤b)P(a\le X\le b) are the same number.

  10. Recall the definition of the cumulative distribution function

    F(x)=P(X≤x)=∫−∞xf(u) duF(x)=P(X\le x)=\int_{-\infty}^{x}f(u)\,du

    FF accumulates the area under the density from the left.

  11. Recall how to recover the density from FF

    f(x)=dFdxf(x)=\frac{dF}{dx}

    Differentiation undoes the integration that produced FF.

  12. Recall the end conditions on FF

    F(−∞)=0,F(∞)=1F(-\infty)=0,\qquad F(\infty)=1

    FF increases from 00 to 11 across the support.

  13. Recall the definition of the mean

    E(X)=∫−∞∞xf(x) dxE(X)=\int_{-\infty}^{\infty}xf(x)\,dx

    The mean weights each value of xx by the density at that point.

  14. Recall the expectation of a function of XX

    E(g(X))=∫−∞∞g(x)f(x) dxE\left(g(X)\right)=\int_{-\infty}^{\infty}g(x)f(x)\,dx

    Apply gg to xx inside the integral; never apply it to E(X)E(X).

  15. Recall the working formula for the variance

    Var(X)=E(X2)−[E(X)]2\text{Var}(X)=E\left(X^{2}\right)-\left[E(X)\right]^{2}

    This is the form quoted in the formula book.

  16. Select the matching option

    F(x)={0x<02x350≤x≤1−3x25+12x5−751≤x≤21x>2F(x)=\begin{cases}0&x<0\\\frac{2 x^{3}}{5}&0\le x\le 1\\- \frac{3 x^{2}}{5} + \frac{12 x}{5} - \frac{7}{5}&1\le x\le 2\\1&x>2\end{cases}

    This is the only option that starts at 00, ends at 11 and differentiates back to ff.

Answer
F(x)={0x<02x350≤x≤1−3x25+12x5−751≤x≤21x>2F(x)=\begin{cases}0&x<0\\\frac{2 x^{3}}{5}&0\le x\le 1\\- \frac{3 x^{2}}{5} + \frac{12 x}{5} - \frac{7}{5}&1\le x\le 2\\1&x>2\end{cases}

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