Are My Solutions to the High School Probability Exam Correct?

[pretzel1998]

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-resource/exams/2015/91586-exm-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! :D

Question ONE

(a)
(i) I drew this graph, don't 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 similar 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 can't just take the middle value and say its the median. I really don't know though.

Thanks!

 

[jvicens]

Hello! Thank you for sharing your answers to the exam questions. Here are my thoughts on your answers:

Question ONE

(a)
(i) It's great that you have drawn the graph, as it helps to visualize the problem. Just make sure you label the axes and include a title for the graph.
(ii) I agree with your answer of 0.08. The assumption of independence is reasonable, as the arrival times at station A and B are not dependent on each other.

(b)
(i) Your answer of 0.2281453622 is correct. Your explanation of the assumptions is also good.
(ii) Your calculation of 24.07 cars needed to be observed is correct. Just make sure to round to the nearest whole number, so the answer should be 25 cars.

Question TWO

(a)
(i) Your answer of 1.23 attempts is correct.
(ii) Your answer of a fixed charge of $300 is also correct.

(b)
(i) Your answer of 0.87948709 is correct. Your explanation of why the Poisson distribution may not be appropriate is also good.
(ii) Your reasoning is correct. The triangular distribution is not symmetrical, so the median cannot be calculated in the same way as a symmetrical distribution.

Question 3.

(a)
(i) Your answer of 70% is correct.
(ii) Your explanation is good. Just make sure to mention that the outliers are affecting the shape of the distribution and making it non-symmetrical.
(b)
(i) Your answer of 0.8 is correct. The median of the triangular distribution can be calculated by finding the average of the two modes.
(ii) Your reasoning is correct. The non-symmetrical shape of the distribution affects the calculation of the median.

Overall, your answers are well thought out and correct. Just make sure to round to the appropriate number of significant figures and to label your graphs and include titles. Good job!