The maximum time allowed for an exam is three hours. The time required, X, as a proportion of this maximum duration by a student to complete the exam (that is, if a student completes the exam in t hours, X takes the value t∕3) has a Beta distribution with parameters 𝛼 = 5 and 𝛽 = 2.

(i) Find the distribution function of X.

(ii) Obtain the mean and the standard deviation of X.

(iii) What proportion of students complete the exam within two hours?

Question

The maximum time allowed for an exam is three hours. The time required, X, as a proportion of this maximum duration by a student to complete the exam (that is, if a student completes the exam in t hours, X takes the value t∕3) has a Beta distribution with parameters 𝛼 = 5 and 𝛽 = 2.

(i) Find the distribution function of X.

(ii) Obtain the mean and the standard deviation of X.

(iii) What proportion of students complete the exam within two hours?

(i) Find the distribution function of X.

(ii) Obtain the mean and the standard deviation of X.

(iii) What proportion of students complete the exam within two hours?

Accepted Answer

The density function of X is given by

[latex]f(x)= \begin{cases} \frac{1}{B(5,2)}x^{5-1}(1-x)^{2-1},\ \ \ 0\lt x\lt 1, \ 0, \qquad \qquad \qquad \qquad \quad ext{elsewhere}, \end{cases} [/latex]

where, from (7.21),

[latex]B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.[/latex] (7.21)

we see that the value of B(5, 2) is

[latex]B(5,2)=\frac{\Gamma(5)\Gamma(2)}{\Gamma(5+2)}=\frac{4!1!}{6!}=\frac{1}{30}.[/latex]

Thus, the density takes the form

[latex]f(x)= \begin{cases} 30x^4(1-x),\ \ \ 0\lt x\lt 1, \ 0, \qquad \qquad \quad ~ ext{elsewhere}. \end{cases} [/latex]

(i) For the distribution function, we have

[latex]F(t)= \int_{-\infty }^{t}{f(x)dx} = \begin{cases} 0, \qquad \qquad ~ t \lt 0, \ 6t^5-5t^6, \quad 0 \leq t \lt 1,\ 1,\qquad \qquad ~ t \geq 1. \end{cases}[/latex]

(ii) Using the formulas from Proposition 7.10 for the mean and variance, we find

[latex]E(X)={\frac{\alpha}{\alpha+\beta}}={\frac{5}{5+2}}={\frac{5}{7}}[/latex]

and

[latex]\mathrm{Var}(X)={\frac{\alpha\beta}{(\alpha+\beta+1)(\alpha+\beta)^{2}}}={\frac{5\cdot2}{(5+2+1)(5+2)^{2}}}={\frac{10}{8\cdot49}}={\frac{10}{392}}\cong0.0255,[/latex]

so that the standard deviation is

[latex]\sigma={\sqrt{\operatorname{Var}(X)}}\cong{\sqrt{0.0255}}=0.1597.[/latex]

(iii) As explained in the statement, if a student completes the exam in t hours, the associated value of X is t∕3. Thus, in order to find the probability that a student completes the exam within two hours (or, the proportion of students who complete the exam within two hours, which is the same), we need to find P(X ≤ 2∕3). This can be obtained easily upon substituting t = 2∕3 in the distribution function F found in Part (i). Thus, we see that

[latex]P\left(X\leq{\frac{2}{3}} ight)=F\left({\frac{2}{3}} ight)=6{\bigg(}{\frac{2}{3}}{\bigg)}^{5}-5{\bigg(}{\frac{2}{3}}{\bigg)}^{6}={\frac{256}{729}}\cong0.3512,[/latex]

and so about 35% of the students complete the exam within two hours.

Question 7.15: The maximum time allowed for an exam is three hours. The tim......

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