Question

Question 3: 27 points The picture below shows a probability distribution function for the number of minutes before class begins that a randomly selected BC student arrives at class for Monday, Wednesday, Friday classes. The mean is 5 minutes early and the variance is 12.5 (minutes-squared). I will denote the random variable ej for the number of minutes early that a randomly selected student i arrives to class. f(e) (0 10 e i. Is e a Normally distributed random variable? (no explanation required) 12 pointsl ii. Find f(e) when e = 0. Show clearly how you found this value. [10 points] iii. If you randomly selected 100 BC students, what is the probability that e would fall between 2.5 and 5? Explain what theorem helped you to determine the distribution of e and hence the probability. [15 points

Answer 1

i) From the figure we can see that $\textbf{e}$ is not a normally distributed random variable (Normal distribution has bell shape).

ii) Let be the maximum height of the distribution. The area under a valid PDF is always 1. So we have

5a +(10 -5)a -1

Thus

f(O) a = 0.2

iii) The mean of $E\left (\overline{e} \right )=5$ . The Variance of . Use Central Limit Theorem.

Var (e) 12.5%/100 1.25

According to Central Limit Theorem (CLT), . The probability,

e~N(5,1.25)

P (2.5 <-< 5) = P P (2.5 < ē < 5) = P (-2.236068〈 Z 〈 0) P(2.5 < е < 5)-0.4873 V1.25 V1.25