Science:Math Exam Resources/Courses/MATH101 B/April 2024/Question 17 (d)

MATH101 B April 2024

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[hide] Question 17 (d)

A random variable $X$ has the following probability density function (PDF): $$f(x)={\begin{cases}1,&{\rm {for}}\;-{\frac {1}{2}}<x<0.\\2,&{\rm {for}}\;0<x<c,\\0,&{\rm {otherwise}}.\end{cases}}$$ (d) Find all values of $X$ whose distance from $\mathbb {E} (X)$ is at most one standard deviation. Write your answer as an interval. For this part only, your endpoints may be left in calculator-ready form.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Break the problem down into smaller parts: first find $\mathbb {E} (X^{2})$, then find the variance of $X$, then the standard deviation, then the desired interval. Recall that the variance can be found using the formula $\mathrm {Var} (X)=\mathbb {E} (X^{2})-(\mathbb {E} X)^{2}$, and the standard deviation is just the square root of the variance.

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[show] Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The expected value of $X^{2}$ is $${\begin{aligned}\mathbb {E} (X^{2})&=\int _{-\infty }^{\infty }x^{2}f(x)\;\mathrm {d} x=\int _{-1/2}^{0}x^{2}\;\mathrm {d} x+\int _{0}^{1/4}2x^{2}\;\mathrm {d} x\\&=\left.\left[{\frac {x^{3}}{3}}\right]\right|_{-1/2}^{0}+\left.\left[{\frac {2x^{3}}{3}}\right]\right|_{0}^{1/4}={\frac {1}{3}}\cdot {\frac {1}{8}}+{\frac {2}{3}}\cdot {\frac {1}{64}}={\frac {1}{3}}\left({\frac {1}{8}}+{\frac {1}{32}}\right)={\frac {1}{3}}\cdot {\frac {5}{32}}={\frac {5}{3\cdot 2^{5}}}.\end{aligned}}$$ Recall from part (c) that $\mathbb {E} (X)=-{\frac {1}{16}}=-{\frac {1}{2^{4}}}$. Thus, $${\begin{aligned}\mathrm {Var} (X)&=\mathbb {E} (X^{2})-(\mathbb {E} X)^{2}={\frac {5}{3\cdot 2^{5}}}-{\frac {1}{2^{8}}}={\frac {5\cdot 2^{3}-3}{3\cdot 2^{8}}}={\frac {37}{3\cdot 2^{8}}}.\end{aligned}}$$ Taking the square root of this, the standard deviation of $X$ is $\sigma (X)={\frac {\sqrt {37/3}}{16}}$. Therefore, the interval we are looking for is $[\mathbb {E} (X)-\sigma (X),\mathbb {E} (X)+\sigma (X)]=\left[-{\frac {1}{16}}-{\frac {\sqrt {37/3}}{16}},-{\frac {1}{16}}+{\frac {\sqrt {37/3}}{16}}\right]$. Note that depending on how much or how little one simplified along the way, it is possible to get different (equivalent) answers, such as $\left[-{\frac {1}{16}}-{\sqrt {{\frac {5}{96}}-{\frac {1}{16^{2}}}}},-{\frac {1}{16}}+{\sqrt {{\frac {5}{96}}-{\frac {1}{16^{2}}}}}\right]$.