Further Maths Discrete random variables Practice Questions

Free Further Maths Discrete random variables practice questions with full step-by-step worked solutions. Covers discrete-random-variables, expectation, probability-distribution, finding-k. Practise exam-style problems and check your method.

discrete-random-variablesexpectationprobability-distributionfinding-kfunction-of-xvariance
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The discrete random variable XX has the probability distribution x1234P(X=x)18143814\begin{array}{c|cccc}x&1&2&3&4\\\hline P(X=x)&\frac{1}{8}&\frac{1}{4}&\frac{3}{8}&\frac{1}{4}\end{array}. Find the exact value of E(X)E(X).
Show worked solution

Worked solution

  1. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  2. Multiply each value of xx by its probability

    E(X)=1×18+2×14+3×38+4×14E(X)=1\times\frac{1}{8}+2\times\frac{1}{4}+3\times\frac{3}{8}+4\times\frac{1}{4}

    This is the sum defining E(X)E(X).

  3. State the value of E(X)E(X)

    E(X)=114E(X)=\frac{11}{4}

    This is the exact value of E(X)E(X).

Answer
E(X)=114E(X)=\frac{11}{4}
Question 2
2 markseasy
The discrete random variable XX has the probability distribution x0123P(X=x)15251515\begin{array}{c|cccc}x&0&1&2&3\\\hline P(X=x)&\frac{1}{5}&\frac{2}{5}&\frac{1}{5}&\frac{1}{5}\end{array}. Which of the following is the exact value of E(X2)E(X^{2})?
Show worked solution

Worked solution

  1. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  2. Multiply each value of x2x^{2} by its probability

    E(X2)=0×15+1×25+4×15+9×15E(X^{2})=0\times\frac{1}{5}+1\times\frac{2}{5}+4\times\frac{1}{5}+9\times\frac{1}{5}

    This is the sum defining E(X2)E(X^{2}).

  3. Evaluate x2x^{2} at each value of xx

    x=0⇒x2=0,x=1⇒x2=1,x=2⇒x2=4,x=3⇒x2=9x=0\Rightarrow x^{2}=0,\quad x=1\Rightarrow x^{2}=1,\quad x=2\Rightarrow x^{2}=4,\quad x=3\Rightarrow x^{2}=9

    Apply the function to each value of xx before weighting.

  4. Select the option equal to this value

    E(X2)=3E(X^{2})=3

    This is the exact value of E(X2)E(X^{2}).

Answer
E(X2)=3E(X^{2})=3
Question 3
4 marksintermediate
The discrete random variable XX has the probability distribution x123P(X=x)161312\begin{array}{c|ccc}x&1&2&3\\\hline P(X=x)&\frac{1}{6}&\frac{1}{3}&\frac{1}{2}\end{array}. Which of the following is the exact value of [E(X)]2\left[E(X)\right]^{2}?
Show worked solution

Worked solution

  1. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  2. Work out E(X)E(X)

    E(X)=1×16+2×13+3×12=73E(X)=1\times\frac{1}{6}+2\times\frac{1}{3}+3\times\frac{1}{2}=\frac{7}{3}

    Multiply each value by its probability and add the results.

  3. Add the contributions to E(X)E(X)

    16+23+32=73\frac{1}{6}+\frac{2}{3}+\frac{3}{2}=\frac{7}{3}

    Adding the exact fractions gives the required expectation.

  4. State the value of E(X)E(X)

    E(X)=73E(X)=\frac{7}{3}

    This is the exact value of E(X)E(X).

  5. Find the contribution to E(X)E(X) from x=1x=1

    1×16=161\times\frac{1}{6}=\frac{1}{6}

    The contribution of a single value of xx to E(X)E(X).

  6. Select the option equal to this value

    [E(X)]2=(73)2=499\left[E(X)\right]^{2}=\left(\frac{7}{3}\right)^{2}=\frac{49}{9}

    The mean is squared AFTER it has been found; this is not the same as E(X2)E(X^{2}).

Answer
[E(X)]2=(73)2=499\left[E(X)\right]^{2}=\left(\frac{7}{3}\right)^{2}=\frac{49}{9}
Question 4
6 markshard
The discrete random variable XX has the probability distribution x102030P(X=x)141214\begin{array}{c|ccc}x&10&20&30\\\hline P(X=x)&\frac{1}{4}&\frac{1}{2}&\frac{1}{4}\end{array}. Which of the following is the exact value of Var(3X+1)\text{Var}(3X+1)?
Show worked solution

Worked solution

  1. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  2. Multiply each value of xx by its probability

    E(X)=10×14+20×12+30×14E(X)=10\times\frac{1}{4}+20\times\frac{1}{2}+30\times\frac{1}{4}

    This is the sum defining E(X)E(X).

  3. Add the contributions to E(X)E(X)

    52+10+152=20\frac{5}{2}+10+\frac{15}{2}=20

    Adding the exact fractions gives the required expectation.

  4. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  5. Evaluate x2x^{2} at each value of xx

    x=10⇒x2=100,x=20⇒x2=400,x=30⇒x2=900x=10\Rightarrow x^{2}=100,\quad x=20\Rightarrow x^{2}=400,\quad x=30\Rightarrow x^{2}=900

    Apply the function to each value of xx before weighting.

  6. Multiply each value of x2x^{2} by its probability

    E(X2)=100×14+400×12+900×14E(X^{2})=100\times\frac{1}{4}+400\times\frac{1}{2}+900\times\frac{1}{4}

    This is the sum defining E(X2)E(X^{2}).

  7. Add the contributions to E(X2)E(X^{2})

    25+200+225=45025+200+225=450

    Adding the exact fractions gives the required expectation.

  8. Recall the formula for Var(X)\text{Var}(X)

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

    The variance is the mean of the squares minus the square of the mean.

  9. Substitute E(X2)E(X^{2}) and E(X)E(X) into the variance formula

    Var(X)=450−(20)2\text{Var}(X)=450-\left(20\right)^{2}

    Both expectations are already known as exact fractions.

  10. Simplify to find Var(X)\text{Var}(X)

    Var(X)=50\text{Var}(X)=50

    Subtracting the exact fractions gives the variance.

  11. Recall the rule for the variance of a linear function of XX

    Var(aX+b)=a2Var(X)\text{Var}(aX+b)=a^{2}\text{Var}(X)

    The multiplier is SQUARED and the constant bb disappears entirely.

  12. Substitute Var(X)\text{Var}(X) and a2=9a^{2}=9

    Var(3X+1)=9×50\text{Var}(3X+1)=9\times50

    Here a=3a=3, so a2=9a^{2}=9.

  13. Select the option equal to this value

    Var(3X+1)=450\text{Var}(3X+1)=450

    The constant term has vanished and the multiplier has been squared.

Answer
Var(3X+1)=450\text{Var}(3X+1)=450
Question 5
9 markschallenging
The discrete random variable XX has probability distribution given by P(X=x)=x10P(X=x)=\frac{x}{10} for x=1,2,3,4x=1,2,3,4. Which of the following is the exact value of Var(3X−1)\text{Var}(3X-1)?
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Worked solution

  1. Evaluate the probability function at each value of xx

    P(X=1)=110,P(X=2)=15,P(X=3)=310,P(X=4)=25P(X=1)=\frac{1}{10},\quad P(X=2)=\frac{1}{5},\quad P(X=3)=\frac{3}{10},\quad P(X=4)=\frac{2}{5}

    Substituting each value of xx into the formula gives its probability.

  2. Write out the completed probability distribution

    x1234P(X=x)1101531025\begin{array}{c|cccc}x&1&2&3&4\\\hline P(X=x)&\frac{1}{10}&\frac{1}{5}&\frac{3}{10}&\frac{2}{5}\end{array}

    With kk known, every probability is a definite fraction.

  3. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  4. Multiply each value of xx by its probability

    E(X)=1×110+2×15+3×310+4×25E(X)=1\times\frac{1}{10}+2\times\frac{1}{5}+3\times\frac{3}{10}+4\times\frac{2}{5}

    This is the sum defining E(X)E(X).

  5. Add the contributions to E(X)E(X)

    110+25+910+85=3\frac{1}{10}+\frac{2}{5}+\frac{9}{10}+\frac{8}{5}=3

    Adding the exact fractions gives the required expectation.

  6. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  7. Evaluate x2x^{2} at each value of xx

    x=1⇒x2=1,x=2⇒x2=4,x=3⇒x2=9,x=4⇒x2=16x=1\Rightarrow x^{2}=1,\quad x=2\Rightarrow x^{2}=4,\quad x=3\Rightarrow x^{2}=9,\quad x=4\Rightarrow x^{2}=16

    Apply the function to each value of xx before weighting.

  8. Multiply each value of x2x^{2} by its probability

    E(X2)=1×110+4×15+9×310+16×25E(X^{2})=1\times\frac{1}{10}+4\times\frac{1}{5}+9\times\frac{3}{10}+16\times\frac{2}{5}

    This is the sum defining E(X2)E(X^{2}).

  9. Add the contributions to E(X2)E(X^{2})

    110+45+2710+325=10\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}=10

    Adding the exact fractions gives the required expectation.

  10. Recall the formula for Var(X)\text{Var}(X)

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

    The variance is the mean of the squares minus the square of the mean.

  11. Substitute E(X2)E(X^{2}) and E(X)E(X) into the variance formula

    Var(X)=10−(3)2\text{Var}(X)=10-\left(3\right)^{2}

    Both expectations are already known as exact fractions.

  12. Simplify to find Var(X)\text{Var}(X)

    Var(X)=1\text{Var}(X)=1

    Subtracting the exact fractions gives the variance.

  13. Recall the rule for the variance of a linear function of XX

    Var(aX+b)=a2Var(X)\text{Var}(aX+b)=a^{2}\text{Var}(X)

    The multiplier is SQUARED and the constant bb disappears entirely.

  14. Substitute Var(X)\text{Var}(X) and a2=9a^{2}=9

    Var(3X−1)=9×1\text{Var}(3X-1)=9\times1

    Here a=3a=3, so a2=9a^{2}=9.

  15. State the value of E(X)E(X)

    E(X)=3E(X)=3

    This is the exact value of E(X)E(X).

  16. Select the option equal to this value

    Var(3X−1)=9\text{Var}(3X-1)=9

    The constant term has vanished and the multiplier has been squared.

Answer
Var(3X−1)=9\text{Var}(3X-1)=9

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