91Ó°ÊÓ

The z-scores are -2 and 2.

Define Z as the random variable that follows the standard normal distribution.

Thus,

.$$\begin{gathered} Z~N\left(\mathrm{μ},{\mathrm{σ}}^2\right) \\ ~N\left(0,1^2\right) \end{gathered}$$

Steps to draw a normal curve:

1. Make a horizontal axis and a vertical axis.
2. Mark the points -3.5, -3.0, -2.0 up to 3 on the horizontal axis and points 0, 0.05, 0.10 up to 0.50 on the vertical axis.
3. Provide titles to the horizontal and vertical axes as ‘z’ and ‘P(z)’, respectively.
4. Shade the region between -2 and 2.

The shaded area of the graph indicates the probability that the z-score is between -2 and 2.

The area between -2 and 2 is computed as using probabilities, as the area has a one-to-one correspondence with the probability under the standard normal curve.

$P\left(-2<Z<2\right)=P\left(Z<2\right)-P\left(Z<-2\right)                  ...\left(1\right)$

Referring to the standard normal table,

Thus,

$$\begin{gathered} P\left(Z<2\right)=0.9772 \\ P\left(Z<-2\right)=0.0228 \end{gathered}$$

Substituting the values in equation (1),

$$\begin{gathered} P\left(-2<Z<2\right)=P\left(Z<2\right)-P\left(Z<-2\right) \\ =0.9772-0.0228 \\ =0.9544 \end{gathered}$$

Expressing the result as a percentage, about 95.44 % of the area is between z = -2 and z = 2.