1. The problem states that the mean score $\mu=32$ and the standard deviation $\sigma=5$. We are asked to find the z-score for a score $x=37$. 2. Recall the z-score formula is: $$z=\frac{x-\mu}{\sigma}$$ 3. Substitute the given values: $$z=\frac{37-32}{5}$$ 4. Calculate the numerator: $$37-32=5$$ 5. Divide by the standard deviation: $$z=\frac{5}{5}=1$$ 6. Therefore, the z-score corresponding to $x=37$ is $1$. This means the score 37 is 1 standard deviation above the mean.