ANSWER
0.2644
Step-by-step explanation
The tuition costs is a random variable X normally distributed with a mean of $23600 and a standard deviation of $6023.
We have to find the probability that a randomly selected college has an annual cost between $25000 and $30000. This is,

Using the standard normal distribution formula,

We have,

Now, we have to look up these z-values in a z-score table. These tables show the area under the standard normal curve to the left of each z-score - i.e. they show the probability that Z is less than that value. So, to find this probability we have to separate each interval and use a complement,
So we have,

These z-scores in a z-table are,
So the probability is,

Hence, the probability that a randomly selected college will have an annual cost between $25000 and $30000 is 0.2644.