does anyone know how the last line of the solution works? is it some sort of formula?
(Q17 tech active WACE 2020)
The first bit refers to this section screenshotted from the QCE syllabus.
We know that \( \overline{X}\) is approximately normal with mean \(\mu\) and standard deviation \( \frac{\sigma}{\sqrt{n}} \). Therefore \( \frac{\overline{X}-\mu}{\sigma/\sqrt{n}}\) is approximately standard normal. Note that:
a) This is a general rule. In general, for
any random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), if we know that \(X\) is normally distributed, then \( \frac{X-\mu}{\sigma}\) is standard normally distributed.
b) It's very similar to the third point where they use \(s\) instead of \(\sigma\). At a high school level, I'd suggest just taking for granted the approximate standard normality works with both \(\sigma\)
and with \(s\).
For the second bit, note that in theory that equal sign really should be an approximately equal sign. But let's not worry about that. The point is, first recall that \(\overline{X}\) denotes a sample mean for a larger sample size of \(2n\). And we've calculated that its corresponding standard deviation will therefore be \(9.0192\) (approximately). But also \(\overline{X}\) also has mean \(\mu\) (you should know why this is the case). Therefore here, \( \frac{\overline{X}-\mu}{9.0192} \) will be approximately standard normally distributed. Hence, it is tempting to start by dividing by \(9.0192\), and introducing \(Z\) as a placeholder for a standard normal random variable.
\[ P\left( |\overline{X}-\mu| <10\right) = P\left( \left|\frac{\overline{X}-\mu}{9.0192}\right| <\frac{10}{|9,0192|}\right) =P\left( |Z|<\frac{10}{9.0192} \right) . \]
We still have to remove the absolute value. First recall that the solution to the absolute value inequality \( |x| < a\) is \( -a < x < a\).
\[P\left( |Z|<\frac{10}{9.0192} \right) = P\left( -\frac{10}{9.0192} < Z < \frac{10}{9.0192} \right). \]
But we also know that the density of the standard normal distribution is a bell curve, and in particular
symmetric about
zero. The symmetry about zero allows you to conclude that \( P\left( -\frac{10}{9.0192} < Z < 0\right)\) and \( P\left( 0 < Z < \frac{10}{9.0192} \right)\) equal one another. Which is where that bit comes from.
\[ P\left( -\frac{10}{9.0192} < Z < \frac{10}{9.0192} \right) = 2 P\left( 0< Z < \frac{10}{9.0192} \right). \]
The final bit is just a calculator plug in. Can be computed by something along the lines of \( \verb|2*(normcdf(10/9.0192,mean=0,sd=1)-normcdf(0,mean=0,sd=1))| \)
Note: Use of symmetry is not required if your graphics calculator can just do \( \verb|normcdf(10/9.0192,mean=0,sd=1)-normcdf(-10/9.0192,mean=0,sd=1)| \) instead.