Question

A large university will begin a 13-day period during which students may register for that semester’s courses. Of those 13 days, the number of elapsed days before a randomly selected student registers has a continuous distribution with density function f (t) that is symmetric about t = 6.5 and proportional to 1/(t + 1) between days 0 and 6.5.A student registers at the 60th percentile of this distribution.Calculate the number of elapsed days in the registration period for this student.(A) 4.01
(B) 7.80
(C) 8.99
(D) 10.22
(E) 10.51

Answer 1

Answer:

8.99 days elapsed. Option (C) is correct

Step-by-step explanation:

The distribution  has density function k/t+1 for a constant k and t between 0 and 6.5 . Since the distribution is symmetrical in 6.5, the area it forms between 0 and 6.5 should be 1/2, thus

$\frac{1}{2} = \int\limits_0^{6.5} \frac{k}{t+1} \, dt = k *(ln(t+1) \, |_0^{6.5}) = k * (ln(7.5)-ln(1)) = k*ln(7.5)$

Hence k = 1/(2ln(7.5)), approx 1/4.

We need to find the percentil 0.6, since the integral of the random variable is 1/2 over the first half, we need to find t such that the integral of the random variable between o and 6.5 + t is 0.6. This is equivalent to find t such that the integral between 6.5 and 6.5+t is 0.1. Due to the  over 6.5, this t should satisfy that the integral between 6.5-t and 6.5 is also 0.1. Lets compute the integral and find t

$\int\limits^{6.5}_{6.5-t} {\frac{k}{t+1}} \, dx = \frac{1}{2ln(7.5)}*(ln(t+1) \, |_{6.5-t}^{6.5} \, ) = \frac{1}{2ln(7.5)} * (ln(7.5)-ln(7.5-t)) = \\\frac{1}{2} - \frac{ln(7.5-t)}{2ln(7.5)} = 0.1$

Therefore,

$\frac{ln(7.5-t)}{2ln(7.5)} = 0.4\\\\ln(7.5-t) = 0.8*ln(7.5)\\\\7.5-t = e^{0.8*ln(7.5)}\\\\t = 7.5-e^{0.8*ln(7.5)} = 2.49$

As a result, the student sould have registered 2.49 days after the day 6.5, thus it should have registeredd at day 8.99. Option (C) is correct.