Answer:
You need a z-score of at least 1.09 to get accepted.
Explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:

To get accepted into OSU, you need at least a 27. What is the z-score you need to get accepted?
We need to find Z when X = 27. So



You need a z-score of at least 1.09 to get accepted.