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

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?

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The density function of X is given by

f(x)={1B(5,2)x5−1(1−x)2−1,   0<x<1,0,elsewhere,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 \text{elsewhere}, \end{cases}

where, from (7.21),

B(α,β)=Γ(α)Γ(β)Γ(α+β).B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.                  (7.21)

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

B(5,2)=Γ(5)Γ(2)Γ(5+2)=4!1!6!=130.B(5,2)=\frac{\Gamma(5)\Gamma(2)}{\Gamma(5+2)}=\frac{4!1!}{6!}=\frac{1}{30}.

Thus, the density takes the form

f(x)={30x4(1−x),   0<x<1,0, elsewhere.f(x)= \begin{cases} 30x^4(1-x),\ \ \ 0\lt x\lt 1, \\ 0, \qquad \qquad \quad ~ \text{elsewhere}. \end{cases}

(i) For the distribution function, we have

F(t)=∫−∞tf(x)dx={0, t<0,6t5−5t6,0≤t<1,1, t≥1.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}

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

E(X)=αα+β=55+2=57E(X)={\frac{\alpha}{\alpha+\beta}}={\frac{5}{5+2}}={\frac{5}{7}}

and

Var(X)=αβ(α+β+1)(α+β)2=5⋅2(5+2+1)(5+2)2=108⋅49=10392≅0.0255,\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,

so that the standard deviation is

σ=Var⁡(X)≅0.0255=0.1597.\sigma={\sqrt{\operatorname{Var}(X)}}\cong{\sqrt{0.0255}}=0.1597.

(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

P(X≤23)=F(23)=6(23)5−5(23)6=256729≅0.3512,P\left(X\leq{\frac{2}{3}}\right)=F\left({\frac{2}{3}}\right)=6{\bigg(}{\frac{2}{3}}{\bigg)}^{5}-5{\bigg(}{\frac{2}{3}}{\bigg)}^{6}={\frac{256}{729}}\cong0.3512,

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

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