Chapter 6, Problem 6.1TI

What is the z-score of x, when x = 1 and X~N(12,3)?

To determine

The z-score of x.

Answer to Problem 6.1TI

x = 1 is 3.67 standard deviations below or to the left of the mean $\mu =12$.

Explanation of Solution

Given information:

x = 1 and $X\sim N(12,3)$

Formula used:

If X is the normally distributed random variable and $X\sim N(\mu ,\sigma )$, then the z-score is

$z=\frac{x-\mu }{\sigma }$

The z value tell u how many standard deviations the value of x is above (to the right) or below (to the left) the mean, $\mu$.

Calculation:

From the given information, we have the values,

$\begin{array}{l}x=1\\ \mu =12\\ \sigma =3\end{array}$

Therefore,

$\begin{array}{c}z=\frac{1-12}{3}\\ =-\frac{11}{3}\\ =-3.67\end{array}$

Hence, this means that x = 1 is 3.67 standard deviations below or to the left of the mean $\mu =12$.