91影视

The z-scores are -3 and 3.

Let Z be the random variable following the standard normal distribution.

Thus,

$$\begin{gathered} Z~N\left(渭,{蟽}^2\right) \\ ~N\left(0,1^2\right) \end{gathered}$$

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 鈥榸鈥� and 鈥楶(z)鈥�, respectively.
4. Shade the region between -3 and 3.

The shaded area of the graph indicates the probability that the z-score is between -3 and 3.

Due to a one-to-one correspondence between the area and probability, the area between -3 and 3 can be expressed as follows.

The probability between -3 and 3 is computed as

$P\left(-3<Z<3\right)=P\left(Z<3\right)-P\left(Z<-3\right)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

Referring to the standard normal table,

• the cumulative probability of -3 is obtained from the cell intersection for rows -3.0 and the column value 0.00, which is 0.0013, and
• the cumulative probability of 3 is obtained from the cell intersection for rows 3.0 and the column value 0.00, which is 0.9987.

Thus,

$$\begin{gathered} P\left(Z<3\right)=0.9987 \\ P\left(Z<-3\right)=0.0013 \end{gathered}$$

.

Substituting the values in equation (1),

$$\begin{gathered} P\left(-3<Z<3\right)=P\left(Z<3\right)-P\left(Z<-3\right) \\ =0.9987-0.0013 \\ =0.9974 \end{gathered}$$

Expressing the result as a percentage, about 99.74 % of the area is between z = -3 and z = 3.