Given a normal distribution with mean -30 and standard deviation 30 , use the table to find the area under the normal curve and above the interval [ $-60,-30$ ] on the horizontal axis.

The area is
(Type an integer or decimal rounded to four decimal places as needed.)

Step 1 ：First, we need to standardize the interval. The standardization formula is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value from the original distribution, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 2 ：For the lower limit of the interval, -60, the standardized value is \(Z = \frac{-60 - (-30)}{30} = -1\).

Step 3 ：For the upper limit of the interval, -30, the standardized value is \(Z = \frac{-30 - (-30)}{30} = 0\).

Step 4 ：Now, we need to find the area under the standard normal curve between -1 and 0. This is equivalent to finding the probability that a standard normal random variable falls in this interval.

Step 5 ：From the standard normal distribution table, we find that the area to the left of 0 is 0.5000 and the area to the left of -1 is 0.1587.

Step 6 ：Subtracting these two areas gives the area between -1 and 0, which is \(0.5000 - 0.1587 = 0.3413\).

Step 7 ：So, the area under the normal curve and above the interval [-60, -30] on the horizontal axis is \(\boxed{0.3413}\).