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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Teen on November 06, 2012, 12:54:56 pm

Title: 2010 vcaa exam
Post by: Teen on November 06, 2012, 12:54:56 pm: Let x be a normally distributed random variable with mean of 5 and variance of 9 and let z be the random variable with the standard normal distribution

Find b such that that pr(x>7)=pr(z,b)
I am getting positive 1.5 but the answer is - 1.5 why ?

Thanks in advance
Title: Re: 2010 vcaa exam
Post by: BubbleWrapMan on November 06, 2012, 12:56:30 pm: Typo?
Title: Re: 2010 vcaa exam
Post by: Reckoner on November 06, 2012, 01:26:31 pm: Is this 5.b on exam 1?

The question is find b when $Pr(X>7)=Pr(Z<b)$

So using the formula
$<br />Z=\frac{X-\mu}{\sigma}<br />$
Where $\mu=5 \sigma=3 X=7$

We get $Z=\frac{2}{3}$

However, the the question asks for $Pr(Z<b)$ . The value we got was when $Pr(X>7)=Pr(Z>b)$ .

So through the symmetry of the normal distribution curve, $b=\frac{-2}{3}$

Here's the graph from the assessment report showing the symmetry of the areas of distribution.

(http://i.imgur.com/Fs8lp.jpg?1?2477)
Title: Re: 2010 vcaa exam
Post by: Teen on November 06, 2012, 02:05:48 pm: Thanks ;)