Further Maths The continuous uniform distribution Practice Questions

Free Further Maths The continuous uniform distribution practice questions with full step-by-step worked solutions. Covers continuous-uniform-distribution, probability-density-function, probability, mean. 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 is uniformly distributed over the interval [2,8]\left[2,8\right]. Write down the value of the probability density function f(x)f(x) for 2≤x≤82\le x\le 8.
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Worked solution

  1. Find the width of the interval

    b−a=8−2=6b-a=8-2=6

    Every result for a rectangular distribution is built from this width.

  2. Recall that the density is the reciprocal of the width

    f(x)=1b−af(x)=\frac{1}{b-a}

    The rectangle must have area 11, so its height is 1b−a\frac{1}{b-a}.

  3. State the value of the density

    f(x)=16,2≤x≤8f(x)=\frac{1}{6},\quad 2\le x\le 8

    This constant value is the density everywhere inside the interval.

Answer
16\frac{1}{6}
Question 2
2 markseasy
The continuous random variable XX is uniformly distributed over the interval [0,5]\left[0,5\right]. Which of the following is the value of P(X>2)P\left(X>2\right)?
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Worked solution

  1. Write down the probability density function

    f(x)=15−0=15,0≤x≤5f(x)=\frac{1}{5-0}=\frac{1}{5},\quad 0\le x\le 5

    For a rectangular distribution the height of the density is one over the width of the interval.

  2. Write the probability as an integral of the density

    P(X>2)=∫2515 dxP\left(X>2\right)=\int_{2}^{5}\frac{1}{5}\,dx

    A probability for a continuous variable is an area under the density.

  3. Evaluate the integral

    P(X>2)=15×(5−2)=35P\left(X>2\right)=\frac{1}{5}\times\left(5-2\right)=\frac{3}{5}

    The density is constant, so the area is height ×\times width.

  4. State the probability

    P(X>2)=35P\left(X>2\right)=\frac{3}{5}

    This is the exact probability, left as a fraction.

Answer
35\frac{3}{5}
Question 3
4 marksintermediate
The continuous random variable XX follows a rectangular distribution on the interval [−1,7]\left[-1,7\right]. Which of the following is the value of P(X<3)P\left(X<3\right)?
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Worked solution

  1. Write down the probability density function

    f(x)=17−(−1)=18,−1≤x≤7f(x)=\frac{1}{7-\left(-1\right)}=\frac{1}{8},\quad -1\le x\le 7

    For a rectangular distribution the height of the density is one over the width of the interval.

  2. Write the probability as an integral of the density

    P(X<3)=∫−1318 dxP\left(X<3\right)=\int_{-1}^{3}\frac{1}{8}\,dx

    A probability for a continuous variable is an area under the density.

  3. Evaluate the integral

    P(X<3)=18×(3−(−1))=12P\left(X<3\right)=\frac{1}{8}\times\left(3-\left(-1\right)\right)=\frac{1}{2}

    The density is constant, so the area is height ×\times width.

  4. Sketch the density and shade the required region

    rectangle of height 18 over −1≤x≤7\text{rectangle of height }\frac{1}{8}\text{ over }-1\le x\le 7

    The probability is the shaded fraction of the rectangle.

  5. Express the probability as a fraction of the interval

    P=length of the eventlength of the intervalP=\frac{\text{length of the event}}{\text{length of the interval}}

    For a uniform distribution, probability is proportional to length.

  6. State the probability

    P(X<3)=12P\left(X<3\right)=\frac{1}{2}

    This is the exact probability, left as a fraction.

Answer
12\frac{1}{2}
Question 4
6 markshard
The continuous random variable XX follows a rectangular distribution on the interval [1,9]\left[1,9\right]. Which of the following is the value of Var(−3X+4)\text{Var}\left(-3X+4\right)?
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Worked solution

  1. Write down the probability density function

    f(x)=19−1=18,1≤x≤9f(x)=\frac{1}{9-1}=\frac{1}{8},\quad 1\le x\le 9

    For a rectangular distribution the height of the density is one over the width of the interval.

  2. Write down the integral defining the mean

    E(X)=∫19x×18 dxE(X)=\int_{1}^{9}x\times\frac{1}{8}\,dx

    The mean of a continuous random variable is ∫xf(x) dx\int xf(x)\,dx over its range.

  3. Integrate

    E(X)=[x216]19E(X)=\left[\frac{x^{2}}{16}\right]_{1}^{9}

    The density is a constant, so only xx has to be integrated.

  4. Substitute the limits

    E(X)=8116−116=5E(X)=\frac{81}{16}-\frac{1}{16}=5

    Upper limit minus lower limit gives the value of the definite integral.

  5. Write down the integral defining E(X2)E(X^{2})

    E(X2)=∫19x2×18 dxE(X^{2})=\int_{1}^{9}x^{2}\times\frac{1}{8}\,dx

    For any function of XX the rule is E(g(X))=∫g(x)f(x) dxE\left(g(X)\right)=\int g(x)f(x)\,dx.

  6. Integrate the squared term

    E(X2)=[x324]19E(X^{2})=\left[\frac{x^{3}}{24}\right]_{1}^{9}

    Raise the power by one and divide by the new power.

  7. Substitute the limits to find E(X2)E(X^{2})

    E(X2)=2438−124=913E(X^{2})=\frac{243}{8}-\frac{1}{24}=\frac{91}{3}

    This is the second moment of the distribution about the origin.

  8. Apply the working formula for the variance

    Var(X)=E(X2)−[E(X)]2=913−(5)2\text{Var}(X)=E(X^{2})-\left[E(X)\right]^{2}=\frac{91}{3}-\left(5\right)^{2}

    Both moments are now known, so the variance follows immediately.

  9. Simplify the variance

    Var(X)=913−25=163\text{Var}(X)=\frac{91}{3}-25=\frac{16}{3}

    Subtracting the exact fractions gives the exact variance.

  10. Recall the rule for the variance of a linear function

    Var(cX+d)=c2Var(X)\text{Var}(cX+d)=c^{2}\text{Var}(X)

    The multiplier is squared and the constant disappears entirely.

  11. Substitute c=−3c=-3 and Var(X)=163\text{Var}(X)=\frac{16}{3}

    Var(−3X+4)=(−3)2×163=48\text{Var}\left(-3X+4\right)=\left(-3\right)^{2}\times\frac{16}{3}=48

    Here c2=9c^{2}=9.

  12. State the value of Var(−3X+4)\text{Var}\left(-3X+4\right)

    Var(−3X+4)=48\text{Var}\left(-3X+4\right)=48

    This is the exact value of Var(−3X+4)\text{Var}\left(-3X+4\right).

Answer
4848
Question 5
9 markschallenging
The continuous random variable XX is uniformly distributed over the interval [0,3]\left[0,3\right]. Which of the following is the exact value of E((X+1)2)E\left(\left(X+1\right)^{2}\right)?
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Worked solution

  1. Write down the probability density function

    f(x)=13−0=13,0≤x≤3f(x)=\frac{1}{3-0}=\frac{1}{3},\quad 0\le x\le 3

    For a rectangular distribution the height of the density is one over the width of the interval.

  2. Write down the integral defining E((X+1)2)E\left(\left(X+1\right)^{2}\right)

    E((X+1)2)=∫03(x+1)2×13 dxE\left(\left(X+1\right)^{2}\right)=\int_{0}^{3}\left(x+1\right)^{2}\times\frac{1}{3}\,dx

    The expectation of a function of XX weights g(x)g(x) by the density.

  3. Integrate

    E((X+1)2)=[x(x2+3x+3)9]03E\left(\left(X+1\right)^{2}\right)=\left[\frac{x \left(x^{2} + 3 x + 3\right)}{9}\right]_{0}^{3}

    The constant density 13\frac{1}{3} can be taken outside the integral.

  4. Substitute the limits

    E((X+1)2)=7−0E\left(\left(X+1\right)^{2}\right)=7-0

    Upper limit minus lower limit.

  5. Note that this is not the same as substituting the mean

    E((X+1)2)≠g(E(X))E\left(\left(X+1\right)^{2}\right)\neq g\left(E(X)\right)

    The expectation of a function is not the function of the expectation unless gg is linear.

  6. Compare with the value of gg at the mean

    g(E(X))=g(32)=254≠7g\left(E(X)\right)=g\left(\frac{3}{2}\right)=\frac{25}{4}\neq7

    Substituting the mean into gg is a common error; the two values differ.

  7. Identify the antiderivative used

    ∫(x+1)2×13 dx=x(x2+3x+3)9+c\int \left(x+1\right)^{2}\times\frac{1}{3}\,dx=\frac{x \left(x^{2} + 3 x + 3\right)}{9}+c

    Only a single power rule is needed once the density is taken outside.

  8. Interpret the answer as a weighted average

    E((X+1)2)=13∫03(x+1)2 dxE\left(\left(X+1\right)^{2}\right)=\frac{1}{3}\int_{0}^{3}\left(x+1\right)^{2}\,dx

    Because the density is constant, E((X+1)2)E\left(\left(X+1\right)^{2}\right) is just the average value of g(x)g(x) across the interval.

  9. Check the answer numerically

    E((X+1)2)≈7.0E\left(\left(X+1\right)^{2}\right)\approx 7.0

    A decimal check confirms the exact value is of a sensible size.

  10. Find the width of the interval

    b−a=3−0=3b-a=3-0=3

    Every result for a rectangular distribution is built from this width.

  11. Note the danger of leaving out the density

    ∫03(x+1)2 dx=21≠7\int_{0}^{3}\left(x+1\right)^{2}\,dx=21\neq7

    Integrating g(x)g(x) without the factor f(x)f(x) gives the wrong answer.

  12. Note the mean and the variance of XX

    E(X)=32,Var(X)=34E(X)=\frac{3}{2},\qquad \text{Var}(X)=\frac{3}{4}

    These are the standard results for the interval in the question.

  13. Integrate the density to obtain the cumulative distribution function

    F(x)=∫0x13 dt=x3F(x)=\int_{0}^{x}\frac{1}{3}\,dt=\frac{x}{3}

    The cdf is the accumulated area under the density up to xx.

  14. Recall the probability density function of a rectangular distribution

    f(x)=1b−a,a≤x≤b,f(x)=0 otherwisef(x)=\frac{1}{b-a},\quad a\le x\le b,\qquad f(x)=0\ \text{otherwise}

    The density is constant on the interval, which is what makes the distribution rectangular.

  15. Recall that the total area under the density must be 11

    ∫ab1b−a dx=b−ab−a=1\int_{a}^{b}\frac{1}{b-a}\,dx=\frac{b-a}{b-a}=1

    The graph is a rectangle of width b−ab-a and height 1b−a\frac{1}{b-a}, so its area is 11.

  16. State the exact value

    E((X+1)2)=7E\left(\left(X+1\right)^{2}\right)=7

    This is the exact value of E((X+1)2)E\left(\left(X+1\right)^{2}\right).

Answer
77

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