The Random Variable X is normally distributed with mean 560 and standard deviation 20.

Find P(X<530) which I calculated as 0.0668.

However...

It is known that p(|X-560|<a)
Find the value of a

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

To find the value of a, we can use the properties of the normal distribution.

First, let's understand the given condition, p(|X-560|<a). This represents the probability that the absolute difference between X and 560 is less than a. In other words, it measures how close X is to the mean 560 within a certain distance a.

Using the properties of the normal distribution, we know that approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations.

Since we are interested in finding the value of a, which corresponds to the probability p(|X-560|<a), we need to find the standard deviation bound that represents that probability.

Let's consider the interval from 560-a to 560+a. This represents the range within which X falls if |X-560|<a. To find the probability, we want to find the proportion of the total area under the normal distribution curve that falls within this interval.

Using the properties of the normal distribution, we know that approximately 68% of the data falls within one standard deviation from the mean. Therefore, a reasonable estimate for a would be the value of one standard deviation.

In this case, the given standard deviation is 20. Hence, a ≈ 20.

Please note that this is an estimate, and the exact value of a can be calculated more precisely using statistical software or by referring to standard normal distribution tables.