91Ó°ÊÓ

The variable is normally distributed.

Assume that xis a random variable that follows a normal distribution with a mean $\mathrm{μ}$ and standard deviation$\mathrm{σ}$

z-score is defined as -

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

For 1 standard deviation above the mean, then the z-score isrole="math" localid="1660273418117" $x=\mathrm{μ}+\mathrm{σ}$

So,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{\mathrm{μ}+\mathrm{σ}-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{\mathrm{σ}}{\mathrm{σ}} \\ =1 \end{gathered}$$

So, z = 1

Thus, the z-score for 1 standard deviation above the mean is 1.

For 1 standard deviation below the mean, then the z-score is$x=\mathrm{μ}+\mathrm{σ}$

So,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{\mathrm{μ}-\mathrm{σ}-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{-\mathrm{σ}}{\mathrm{σ}} \\ =-1 \end{gathered}$$

So, z -1

Thus, the z-score for 1 standard deviation below the mean is -1

For equal to the mean, then the z-score is $x=\mathrm{μ}$

So,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{\mathrm{μ}-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0}{\mathrm{σ}} \\ =0 \end{gathered}$$

So, z = 0

Thus, the z-score for equal to the mean is 0.

d.

For 2.5 standard deviations below the mean, then the z-score is $x=\mathrm{μ}-2.5\mathrm{σ}$,

So,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{\mathrm{μ}-2.5\mathrm{σ}-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{-2.5\mathrm{σ}}{\mathrm{σ}} \\ =-2.5 \end{gathered}$$

So, z = 2.5

Thus, the z-score for 2.5 standard deviations below the mean is -2.5.

For 3 standard deviations above the mean, then the z-score is$x=\mathrm{μ}+3\mathrm{σ}$

So,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{\mathrm{μ}+3\mathrm{σ}-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{3\mathrm{σ}}{\mathrm{σ}} \\ =3 \end{gathered}$$

So, z = 3

Thus, the z-score for 3 standard deviations above the mean is 3.