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

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH101 C April 2024

• Q1 • Q2 • Q3 • Q4 • Q5 • Q6 • Q7 • Q8 • Q9 • Q10 • Q11 • Q12 • Q13 • Q14 • Q15 • Q16 (a) • Q16 (b) • Q17 (a) • Q17 (b) • Q17 (c) • Q17 (d) • Q18 (a) • Q18 (b) • Q18 (c) • Q19 • Q20 (a) • Q20 (b) • Q20 (c) •

Other MATH101 C Exams

• April 2024 •

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!

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.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution
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 $\left[-{\frac {1}{16}}-{\sqrt {{\frac {5}{96}}-{\frac {1}{16^{2}}}}},-{\frac {1}{16}}+{\sqrt {{\frac {5}{96}}-{\frac {1}{16^{2}}}}}\right]$ .