Question

Basis for the Range Rule of Thumb and the Empirical Rule. In Exercises 45–48, find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2.

About______ % of the area is between z = -2 and z = 2 (or within 2 standard deviation of the mean).

Short answer

The indicated area between -2 and 2 under the curve of the standard normal distribution is shown below.

95.44% of the area is between z = -2 and z = 2.

Step-by-step solution

Given information

The z-scores are -2 and 2.

Describe the distribution

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

Thus,

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

Draw the area under the curve of standard normal distribution

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.

Find the shaded area

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,

• the cumulative probability of -2 is obtained from the cell intersection for rows -2.0 and the column value 0.00, which is 0.0228, and

• the cumulative probability of 2 is obtained from the cell intersection for rows 2.0 and the column value 0.00, which is 0.9772.

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.