You don't need the central limit theorem, what you need is the fact that

the mean of a sample of size N from a normal distribution N(mu, sigma^2)

is itself normal distributed ~N(mu, sigma^2/N).

So in this case the mean of the sample means is m5 = 87.0, and the standard deviation of the sample means is s5 =1.5/sqrt(5).

Now the interval 86.5 to 87.5 is the interval m5 - 0.75 s5 to m5 - 0.75 s5

or a +/-0.75 sd interval about the mean. The probability of this we look up

in a table of the standard normal distribution and is ~=0.4532

RonL