Discrete random variables Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Discrete random variables questions. See exactly how to solve problems on discrete-random-variables, expectation, probability-distribution, finding-k.

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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).

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}. Find the exact value of E(X)E(X).

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)=0×15+1×25+2×15+3×15E(X)=0\times\frac{1}{5}+1\times\frac{2}{5}+2\times\frac{1}{5}+3\times\frac{1}{5}

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

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

    E(X)=75E(X)=\frac{7}{5}

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

Answer
E(X)=75E(X)=\frac{7}{5}
Question 3
2 markseasy
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}. Find the exact value of E(X)E(X).

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

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

  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).

Answer
E(X)=73E(X)=\frac{7}{3}
Question 4
2 markseasy
The discrete random variable XX has the probability distribution x2468P(X=x)1103102515\begin{array}{c|cccc}x&2&4&6&8\\\hline P(X=x)&\frac{1}{10}&\frac{3}{10}&\frac{2}{5}&\frac{1}{5}\end{array}. Find the exact value of E(X)E(X).

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)=2×110+4×310+6×25+8×15E(X)=2\times\frac{1}{10}+4\times\frac{3}{10}+6\times\frac{2}{5}+8\times\frac{1}{5}

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

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

    E(X)=275E(X)=\frac{27}{5}

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

Answer
E(X)=275E(X)=\frac{27}{5}
Question 5
2 markseasy
The discrete random variable XX has the probability distribution x1234P(X=x)3101511025\begin{array}{c|cccc}x&1&2&3&4\\\hline P(X=x)&\frac{3}{10}&\frac{1}{5}&\frac{1}{10}&\frac{2}{5}\end{array}. Find the exact value of E(X)E(X).

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

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

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

    E(X)=135E(X)=\frac{13}{5}

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

Answer
E(X)=135E(X)=\frac{13}{5}

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