Help needed on a probability question from the 2005 VCAA paper please

[strawberries321]

Hey everyone, i'm quite weak on probability and i don't get how to do this question(question 2) from the VCAA 05 paper 2 at all : http://www.vcaa.vic.edu.au/vce/studies/mathematics/methods/pastexams/2005mm2.pdf


Could someone please explain it, even just the first few parts would be awesome.

Thanks in advance so much! :laugh:

[curry_in_a_hurry69]

hey hey....the 1st few q's r all done on calc.....to complete the table u need 2 use normalcdf on ur calc..so..

A)
greater than A standard: they tell u the A standard is 81.8 and u want the prob of greater than the A standard...which is normalcdf(lower, upper, mean standard dev) ---> normalcfd(81.8, infintiy, 80.8, 4.5) = .412

greater than A but less than olympic recored: lower is 81.8, upper is now 90.17 ---> nrmalcfd(81.8, 90.17, 80.8, 4.5) = .393

greater than olympic record: normalcdf(90.17, infinity, 80.8 4.5) = .019

B)it tells u 90% of his throws r atleast at point M...which means that the porb that x is greater than or equal to M is 90 which means the prob that x is leass than M is 10%...now we can use inversenorm as we have a prob from the x value 2 neg infintity.....invnorm(.1, 80.8, 4.5) = 75.03

hope thats enough 2 get u started

ps. these r not 100%....some1 pls correct me if im wrong

[bec]

I'll explain the first few parts for you

Mean: 80.8
Standard deviation: 4.5

So for part (a), you just plug the relevant things into your calculator:
TIStat.normCdf(81.8, , 80.8,4.5) = 0.412 is the first row of the table

b) 90% of his throws are at least M metres. That means that 10% are less than that.
You want to find the inverse of 0.1, (tistat.invnorm(.1) on your calc), which is -1.2816.
This number is the value in the STANDARD normal distribution, so you need to convert it back to the normal distribution using this formula:








c) This is conditional probability. They want you to find Pr(he throws A standard given that it isn't an olympic record)

If you let X be the length of his throw, this is:



Use your calculator to find out these values (same method as in part a)

(but when you do this on your calculator, keep it to as many decimal places as the calculator has stored - not just the 4 i've shown here)

[tex] =0.401


d) Expected reward is another way of saying E(X). If you work out the probability for all the things in the left column, it will start to look like a regular discrete probability table and you will probably be able to do this Q quite easily.

Hope that helps!

Edit: curry_in_a_hurry already did this but since I'd already typed it up...I'm putting it up anyway

[fredrick]

its just a normal distribution for first part use
normalcdf(81.80,E99,80.80,4.5) .
normalcdf(81.80,90.17,80.80,4.5)
normalcdf(90.17,E99,80.80,4.5).

pard b: (M-80.80)/4.5=invnorm(0.1)

and so on

edit:fixed part b