The score on the LSAT are \approx imately normal with mean of 150.7 and standard deviation of 10.2. Queen’s School of Business in Kingston, Ontario requires a minimum LSAT score of 157 for admission. What is the z-score for a score of 157? Round to three de

The formula for z-score is: z = (x - ?) / ? where x is the score, ? is the mean, and ? is the standard deviation. In this case, we want to find the z-score for a score of 157 on the LSAT, with a mean of 150.7 and a standard deviation of 10.2. z = (157 - 150.7) / 10.2 z = 0.617 Rounding to three decimal places, the z-score for a score of 157 on the LSAT is 0.617. This means that a score of 157 is 0.617 standard deviations above the mean. Since the LSAT scores are \approx imately normally distributed, we can use the z-score to find the percentage of scores that fall above 157. We can use a standard normal distribution table or calculator to find that the percentage of scores above a z-score of 0.617 is about 26.18%. This means that about 26.18% of LSAT takers score higher than 157, which is the minimum score required for admission to Queen's School of Business in Kingston, Ontario.