Probability Distributions: 91586 Exam paper

pretzel1998

New member
Joined
Jan 4, 2015
Messages
4
Hi, I am trying to solve the problems in the exam paper posted below, this is a HIGH SCHOOL probability exam paper.
http://www.nzqa.govt.nz/nqfdocs/ncea-re ... m-2015.pdf

I have put down my answers to these questions. Could you guys do it as well and take a look at my answers and whether or not you agree with them? Thanks!
icon_biggrin.gif


Question ONE

(a)
(i) I drew this graph, dont need to worry about this one.
(ii) I got 0.08, and the assumption was independence between the amount of time it takes for a train to arrive at station A and B.

(b)
(i) I got 0.2281453622
(ii) There are fixed number of trials = 7 cars, fixed probability = red is a 0.13 chance, independence between the colour of each car - we assume, and there is a success or failure (Red car or not red car).
(iii) I got 24.07 cars needed to be observed to have a 0.965 chance of seeing at least one car.

Question TWO

(a)
(i) I got a mean of 1.23 attempts.
(ii) I got a fixed charge of $300.

(b)
(i) I got 0.87948709. The assumption is independence between the probability that each bus breaks down.
(ii) I used a Possion with a mean of 3.6 and plotted the actual values on the graph (Which did not match the experimental results and were off by a long shot). I calculated the variance to be 6.6 and this was not close to the mean of 3.6 (They should be relatively similiar if a Poisson was appropriate). Hence because the theoretical values from the poisson distribution do not not close at all to the results of the experiment and because the variance is not close to the mean, it indicates that the Possion distribution is not a appropriate model to model the situation. This could be because that buses can technically break down simultaneously (Invaliding one of the conditions of the poisson distribution) and independence could be questionable (If buses in the local area are all from one company that has maintenance issues, then independence might not be valid (invalidating one of the conditions of the Poisson).

Question 3.

(a)
(i) I got 70%
(ii) I said a sample size of 49 is a little small and unlikely going to form a nearly symmetrical normal distribution. Also I commented that the 2 crazy low outliers were dragging the tail of the normal distribution curve to the left heavily, and that if they were cleaned out of the data - the data would look more normally distributed.

(b)
(i) I got 0.8
(ii) I actually have no idea for this question. I just said that it was because the triangular distribution wasn't symmetrical, hence you cant just take the middle value and say its the median. I really don't know though.

Thanks!
 
Top