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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: sisqo1111 on October 07, 2008, 06:30:56 pm

Title: Probability Question Help
Post by: sisqo1111 on October 07, 2008, 06:30:56 pm: Hi all would someone be able to help me with this question:

Q) At a qualifying meeting, the time taken for runners to complete 400 meters follows a normal distribution, with a mean time of 50 seconds and a standard deviation of 2 seconds. If the fastest 30% of runners qualify for the next meeting, how fast would you need to run to qualify?

Thanks
Title: Re: Probability Question Help
Post by: Collin Li on October 07, 2008, 06:37:10 pm: Let the top 30% of runners run faster than $a$ seconds (i.e: $\leq a$ ).

$\mbox{Pr}(X \leq a) = 0.30$

$a$ = invNorm(0.30,50,2) = 48.95 s

Therefore, you have to run faster than 48.95 seconds.

Slightly tricky because when we think about the "top 30%," we're usually taking the right-end of the distribution. Since this is a race, we want fewer seconds (rather than more), so we want the "bottom" end of the distribution (in times), which will give us the fastest runners.
Title: Re: Probability Question Help
Post by: username on October 07, 2008, 06:39:13 pm: Coblin, is there a way to do that without a calculator, or is that something we wouldn't really expect on exam 1?
Title: Re: Probability Question Help
Post by: sisqo1111 on October 07, 2008, 06:41:36 pm: thanks a heap
Title: Re: Probability Question Help
Post by: Collin Li on October 07, 2008, 06:42:23 pm: You can't do that on exam 1, unless they tell you something like:

Quote
Given $\mbox{Pr}(Z < -0.52) = 0.30$ to 2 decimal places

(or some other expression which you can manipulate to get that)

They used to give you tables of values of the standard normal distribution, but you don't get that anymore.
Title: Re: Probability Question Help
Post by: sisqo1111 on October 07, 2008, 06:51:51 pm: would it be okay if i just get some help on these other 2 questions as wel. I sort of have an idea but not sure lol.

Question 1) The probability that a full forward in Aussie Rules will kick a goal from the 50-meter line is 0.15. If the full forward has 10 kicks from outside the 50-meter line, find the probability that:
A)kick a goal every time
b) kick at least one goal
C) kick more than one goal given that he kicked at least one goal.

and the other question is

Question 2) The distribution of heights of navy officers was found to be normal with a mean (mu)=175cm and a standard deviation (sigma)=5cm. Determine:
A) The percentage of navy officers with heights between 170cm and 180 cm.
B) The percentage of officers with heights greater than 180 cm.
C) The approximate percentage of navy officers with heights greater than 185 cm.

Thanks a heap :)
Title: Re: Probability Question Help
Post by: Collin Li on October 07, 2008, 07:14:05 pm: Question 1

$n=10$ and $p=0.15$

a) $\mbox{Pr}(X = 10) = (0.15)^{10} = 5.7665\times 10^{-9}$

b) $\mbox{Pr}(X \geq 1) = 1 - \mbox{Pr}(X < 1) = 1 - \mbox{Pr}(X = 0) = 1 - (0.85)^{10} = 0.8031$

c) $\mbox{Pr}(X > 1 | X \geq 1) = \frac{\mbox{Pr}(X > 1)}{\mbox{Pr}(X \geq 1)} = \frac{1 - \mbox{Pr}(X \leq 1)}{0.8031} = \frac{1 - \mbox{Pr}(X = 1) - \mbox{Pr}(X = 0)}{0.8031}$

$= \frac{0.8031 - \mbox{Pr}(X = 1)}{0.8031}$

$= \frac{0.8031 - 10(0.15)^1(0.85)^9}{0.8031}$

$= \frac{0.4557}{0.8031}$

$= 0.5674$
Title: Re: Probability Question Help
Post by: Collin Li on October 07, 2008, 07:21:40 pm: Question 2

$\mu = 175$ and $\sigma = 5$

a) $\mbox{Pr}(170 < X < 180) = \mbox{Pr}(-1 < Z < 1) = 0.68$

b) $\mbox{Pr}(X > 180) = \mbox{Pr}(Z > 1) = \frac{1-0.68}{2} = 0.16$

c) $\mbox{Pr}(X > 185) = \mbox{Pr}(Z > 2) = \frac{1-0.95}{2} = 0.025$

All these questions can also be solved using the calculator by using:

$\mbox{Pr}(a < X < b)$ = normalcdf( $a,b,\mu , \sigma$ )

Hint: Use 1e99 and -1e99 for $\infty$ and $-\infty$ respectively.
Title: Re: Probability Question Help
Post by: sisqo1111 on October 07, 2008, 07:32:41 pm: thanks so much coblin