Continuous random variables Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Continuous random variables questions. See exactly how to solve problems on continuous-random-variables, probability-density-function, find-k, probability.

continuous-random-variablesprobability-density-functionfind-kprobabilitycumulative-distribution-functionexpectation
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.

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
The continuous random variable XX has probability density function f(x)={kx20≤x≤20otherwisef(x)=\begin{cases}k x^{2}&0\le x\le 2\\0&\text{otherwise}\end{cases}, where kk is a constant. Find the value of kk.

Worked solution

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

    ∫02kx2 dx=1\int_{0}^{2}k x^{2}\,dx=1

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

  2. Integrate and evaluate the limits

    [kx33]02=8k3\left[\frac{k x^{3}}{3}\right]_{0}^{2}=\frac{8 k}{3}

    The total area comes out as a multiple of kk.

  3. Solve for kk

    8k3=1 ⇒ k=38\frac{8 k}{3}=1\ \Rightarrow\ k=\frac{3}{8}

    Dividing through gives the constant immediately.

  4. State the value of kk

    k=38k=\frac{3}{8}

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

Answer
k=38k=\frac{3}{8}
Question 3
2 markseasy
The continuous random variable XX has probability density function f(x)={k(4−x)0≤x≤40otherwisef(x)=\begin{cases}k \left(4 - x\right)&0\le x\le 4\\0&\text{otherwise}\end{cases}, where kk is a constant. Find the value of kk.

Worked solution

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

    ∫04k(4−x) dx=1\int_{0}^{4}k \left(4 - x\right)\,dx=1

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

  2. Integrate and evaluate the limits

    [−kx22+4kx]04=8k\left[- \frac{k x^{2}}{2} + 4 k x\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 4
2 markseasy
The continuous random variable XX has probability density function f(x)={k(x2+1)0≤x≤30otherwisef(x)=\begin{cases}k \left(x^{2} + 1\right)&0\le x\le 3\\0&\text{otherwise}\end{cases}, where kk is a constant. Find the value of kk.

Worked solution

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

    ∫03k(x2+1) dx=1\int_{0}^{3}k \left(x^{2} + 1\right)\,dx=1

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

  2. Integrate and evaluate the limits

    [kx33+kx]03=12k\left[\frac{k x^{3}}{3} + k x\right]_{0}^{3}=12 k

    The total area comes out as a multiple of kk.

  3. Solve for kk

    12k=1 ⇒ k=11212 k=1\ \Rightarrow\ k=\frac{1}{12}

    Dividing through gives the constant immediately.

  4. State the value of kk

    k=112k=\frac{1}{12}

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

Answer
k=112k=\frac{1}{12}
Question 5
2 markseasy
The continuous random variable XX has probability density function f(x)={ksin⁡(x)0≤x≤π0otherwisef(x)=\begin{cases}k \sin{\left(x \right)}&0\le x\le \pi\\0&\text{otherwise}\end{cases}, where kk is a constant. Find the value of kk.

Worked solution

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

    ∫0πksin⁡(x) dx=1\int_{0}^{\pi}k \sin{\left(x \right)}\,dx=1

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

  2. Integrate and evaluate the limits

    [−kcos⁡(x)]0π=2k\left[- k \cos{\left(x \right)}\right]_{0}^{\pi}=2 k

    The total area comes out as a multiple of kk.

  3. Solve for kk

    2k=1 ⇒ k=122 k=1\ \Rightarrow\ k=\frac{1}{2}

    Dividing through gives the constant immediately.

  4. State the value of kk

    k=12k=\frac{1}{2}

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

Answer
k=12k=\frac{1}{2}

Unlock 65 more Continuous random variables questions

Create a free account to work through every Further Maths Continuous random variables question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Continuous random variables practice

Hard QuestionsPractice Questions

Related Statistics topics

Central limit theoremChi-squared testsCombinations of random variablesThe continuous uniform distributionDiscrete random variablesEstimation and confidence intervals