Answer :
To find the score that separates the top 59% from the bottom 41% on an English test with a mean of 37.3 and a standard deviation of 8, we identified the z-score for the 41st percentile and converted it to the actual test score, which is 35.5 when rounded to the nearest tenth.
The student is asking to find the score that separates the top 59% from the bottom 41% on an English test with a normal distribution, a mean of 37.3, and a standard deviation of 8. This is a question about percentiles and z-scores in statistics. To find the score corresponding to the top 59%, we need to identify the z-score that reflects the point where 41% of the data lies below since a normal distribution is symmetric around the mean.
We can use a z-table or a calculator to find that the z-score corresponding to the 41st percentile is approximately -0.22 (since tables usually give the area to the left of the z-score). Next, we apply the formula to convert the z-score to the actual test score:
X = \\(\\mu + z \\times \\sigma\\)
X = 37.3 + (-0.22 \\times 8) = 37.3 - 1.76 = 35.54. When rounded to the nearest tenth, the score is 35.5. Therefore, the score that separates the top 59% from the bottom 41% is 35.5.
