Question

When considering data with a mound-shaped distribution and a mean of $18$ and a variance of $4$, approximately what percentage of the data we would expect to be above $22$?

Answer 1

To find what percentage of the data:

Let $x$ be a random variable of mound-shaped distribution and a mean $\left(\text{μ}\right)$ of $18$ and a variance $\left({\sigma }^2\right)$ of $4$.

Using the standard deviation formula and calculate the standard deviation of $x$ in below:

$\begin{array}{rcl}\sigma & =& \sqrt{\text{var}\left(x\right)}\left[\text{Standard deviation formula}\right]\\ & \Rightarrow & \sigma =\sqrt{4}\left[\text{Simplifying}\right]\\ \therefore \sigma & =& 2\end{array}$

Since the data is $x=22$.

Using $z-$score formula and calculate the $z-$score below:

$\begin{array}{rcl}z& =& \frac{x-\mu }{\sigma }\left[z-\text{score formula}\right]\\ & \Rightarrow & z=\frac{22-18}{2}\left[\text{Substitute}x=22,\mu =18,\sigma =2\right]\\ & \Rightarrow & z=\frac{4}{2}\left[\text{Simplifying}\right]\\ \therefore z& =& 2\end{array}$

Using $z-$table, for $z=2,\text{P}\left(z<2\right)=0.9772$.

Calculate the required percentage in below:

$\begin{array}{rcl}\text{P}\left(x>22\right)& =& 1-\text{P}\left(x<22\right)\left[\text{Using}z-\text{table}\right]\\ & \Rightarrow & \text{P}\left(x>22\right)=1-\text{P}\left(z<2\right)\left[\text{Simplifying}\right]\\ & \Rightarrow & \text{P}\left(x>22\right)=1-0.9772\\ & \Rightarrow & \text{P}\left(x>22\right)=0.0228\\ \therefore \text{P}\left(x>22\right)& =& 2.28\%\end{array}$

Hence, the percentage of the data is $\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{\%}$.