The continuous uniform distribution Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths The continuous uniform distribution questions. See exactly how to solve problems on continuous-uniform-distribution, probability-density-function, probability, mean.

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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.

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 has a continuous uniform distribution over the interval [0,10]\left[0,10\right]. Find P(X<4)P\left(X<4\right).

Worked solution

  1. Write down the probability density function

    f(x)=110−0=110,0≤x≤10f(x)=\frac{1}{10-0}=\frac{1}{10},\quad 0\le x\le 10

    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<4)=∫04110 dxP\left(X<4\right)=\int_{0}^{4}\frac{1}{10}\,dx

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

  3. Evaluate the integral

    P(X<4)=110×(4−0)=25P\left(X<4\right)=\frac{1}{10}\times\left(4-0\right)=\frac{2}{5}

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

  4. State the probability

    P(X<4)=25P\left(X<4\right)=\frac{2}{5}

    This is the exact probability, left as a fraction.

Answer
25\frac{2}{5}
Question 3
2 markseasy
The continuous random variable XX follows a rectangular distribution on the interval [1,9]\left[1,9\right]. Find P(X>6)P\left(X>6\right).

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 the probability as an integral of the density

    P(X>6)=∫6918 dxP\left(X>6\right)=\int_{6}^{9}\frac{1}{8}\,dx

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

  3. Evaluate the integral

    P(X>6)=18×(9−6)=38P\left(X>6\right)=\frac{1}{8}\times\left(9-6\right)=\frac{3}{8}

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

  4. State the probability

    P(X>6)=38P\left(X>6\right)=\frac{3}{8}

    This is the exact probability, left as a fraction.

Answer
38\frac{3}{8}
Question 4
2 markseasy
The continuous random variable XX is modelled by a continuous uniform distribution over the interval [0,5]\left[0,5\right]. Find P(2<X<3)P\left(2<X<3\right).

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(2<X<3)=∫2315 dxP\left(2<X<3\right)=\int_{2}^{3}\frac{1}{5}\,dx

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

  3. Evaluate the integral

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

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

  4. State the probability

    P(2<X<3)=15P\left(2<X<3\right)=\frac{1}{5}

    This is the exact probability, left as a fraction.

Answer
15\frac{1}{5}
Question 5
2 markseasy
The continuous random variable XX is uniformly distributed over the interval [3,11]\left[3,11\right]. Find E(X)E(X).

Worked solution

  1. Write down the probability density function

    f(x)=111−3=18,3≤x≤11f(x)=\frac{1}{11-3}=\frac{1}{8},\quad 3\le x\le 11

    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)=∫311x×18 dxE(X)=\int_{3}^{11}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 and substitute the limits

    E(X)=[x216]311=7E(X)=\left[\frac{x^{2}}{16}\right]_{3}^{11}=7

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

  4. State the mean

    E(X)=7E(X)=7

    This is the exact mean of the distribution.

Answer
77

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