Question

A computerized driving test held ten sessions in a day. The overall passing rate of the driving test is 87.5%. In each session, there are 14 candidates.
(a) Most likely, how many candidates would pass the test in a session? What is the probability of its
happening?
(b) What is the probability that more than 10 candidates in a session can pass the test?
(c) What are the (i) expectation, (ii) variance, and (iii) standard deviation of number of candidates can pass
the driving test in a day.
(d) Suppose each candidate who fail the driving test will immediately join the tutorial class for the 2nd test.
The charge of the tutorial class is $400. What are the (i) expectation and (ii) standard deviation of the daily income earned from the tutorial class?

Answer 1

Let $X=$ the number of candidates passing the test: $X\sim Bin(n,p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given that $p=0.875, n=14$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & P(X=x) \\ \hline 0 & 2.27\times10^{-13} \\ \hline 1 & 2.23\times10^{-11} \\ \hline 2 & 1.01\times10^{-9} \\ \hline 3 & 2.84\times10^{-8} \\ \hline 4 & 5.47\times10^{-7} \\ \hline 5 & 7.65\times10^{-6} \\ \hline 6 & 8.03\times10^{-5} \\ \hline 7 & 6.43\times10^{-4} \\ \hline 8 & 0.003936 \\ \hline 9 & 0.018369\\ \hline 10 & 0.064292\\ \hline 11 & 0.163652\\ \hline 12 & 0.286390 \\ \hline 13 & 0.308420 \\ \hline 14 & 0.154210 \\ \hline \end{array}$$

(a) Most likely, 13 candidates would pass the test in a session. The probability is $0.308420$

(b) The probability that more than 10 candidates in a session can pass the test is

$$P(X>10)=P(X=11)+P(X=12)+$$ $$+P(X=13)+P(X=14)\approx$$ $$\approx0.163652+0.286390+0.308420+0.154210\approx$$ $$\approx0.912672$$

(c)

$$E(X)=np=14(0.875)=125.25$$

$$Var(X)=\sigma^2=np(1-p)=14(0.875)(1-0.875)=1.53125$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{1.53125}\approx1.23744$$

(d)

(i)

$$E(Y)=400(14)(1-0.875)=\$700$$

(ii)

$$Var(Y)=\sigma_Y^2=(400)^2(14)(1-0.875)(0.875)=245000$$

$$\sigma_Y=\sqrt{245000}=\$494.97$$