The interval 80-96% is 2 standard deviations either side of the mean.

If you have learnt confidence intervals you would know that for a 95% confidence interval Z = 1.96, which is almost two standard deviations away from the mean. (For Z, mean is 0 and standard deviation is 1). So 95% would be the answer.

You could standardise the X value (the test scores) to the Z value and use a normal distribution table otherwise. To convert to Z, remember

Finding half of the interval:

Looking at a normal distribution table, the probability of this is 0.4772. Remember we just found the probability of the upper half of the interval in the question. So multiplying by 2 gives a probability very close to 0.95, i.e. 95%.

Hope you can understand.