Question

How do you find the $z$ score corresponding to the ${{27}^{th}}$ percentile?

Answer 1

Hint: In the given question, we have been asked to find the $z$ score corresponding to the ${{27}^{th}}$ percentile. From the concepts of $z$ distribution we know that the formula is given as $f\left( x \right)=\dfrac{1}{\sigma \sqrt{2\pi }}{{e}^{\dfrac{-1}{2}{{\left( \dfrac{x-\mu }{\sigma } \right)}^{2}}}}$ .The ${{27}^{th}}$ percentile corresponding z score will be given as $f\left( x \right)=27$ . For finding this let us assume that $z=\dfrac{x-\mu }{\sigma }$ and consider $\mu =0$ and $\sigma =1$ which corresponds to $z$ distribution.

Complete step-by-step answer:
Now considering from the given question, we have been asked to find the $z$ score corresponding to the ${{27}^{th}}$ percentile.
From the concepts of $z$ distribution we know that the formula is given as $f\left( x \right)=\dfrac{1}{\sigma \sqrt{2\pi }}{{e}^{\dfrac{-1}{2}{{\left( \dfrac{x-\mu }{\sigma } \right)}^{2}}}}$ .
The ${{27}^{th}}$ percentile corresponding z score will be given as $f\left( x \right)=27$ .
For finding this let us assume that $z=\dfrac{x-\mu }{\sigma }$ and consider $\mu =0$ and $\sigma =1$ which corresponds to $z$ distribution.
Now we can say that $f\left( z \right)=\dfrac{1}{\sqrt{2\pi }}{{e}^{\dfrac{-1}{2}{{\left( z \right)}^{2}}}}\Rightarrow 0.27$ .
By substituting $\pi =3.14$we will have $f\left( z \right)=\left( 0.4 \right){{e}^{\dfrac{-1}{2}{{\left( z \right)}^{2}}}}\Rightarrow 0.27$ .
By further simplifying this we will get $\Rightarrow {{e}^{\dfrac{-1}{2}{{z}^{2}}}}=\dfrac{0.27}{0.4}$ .
Now by simplifying this further we will have $\Rightarrow {{e}^{\dfrac{-1}{2}{{z}^{2}}}}=\dfrac{27}{4}$ .
Hence we have $\Rightarrow {{e}^{\dfrac{-1}{2}{{z}^{2}}}}=6.75$ .
Now by applying logarithm to base $e$ on both sides we will have $\Rightarrow {{\log }_{e}}{{e}^{\dfrac{-1}{2}{{z}^{2}}}}={{\log }_{e}}6.75$ .
Now by further simplifying this we will get $\Rightarrow \dfrac{-1}{2}{{z}^{2}}={{\log }_{e}}6.75$ since ${{\log }_{a}}{{a}^{x}}=x$ .
Now we can simply write this as $\Rightarrow {{z}^{2}}=-2{{\log }_{e}}6.75$ .
Now we can say that
$\begin{align}
  & \Rightarrow {{z}^{2}}=-2\left( 1.91 \right) \\
 & \Rightarrow {{z}^{2}}=-3.82 \\
\end{align}$
Hence we can conclude that the $z$ score corresponding to the ${{27}^{th}}$ percentile will be given as $-0.62$ approximately.

Note: In questions of this type we should carefully perform the calculations if we had made a mistake during calculation in between the steps it will lead us to end up having the wrong conclusion. So we should make sure twice if we are wrong or right.