## Question 613:

1## Answer:

No answer provided yet.We need to find what score will achieve the 70^{th} percentile. The language is a bit confusing, but they need 70% of the students to score at 70%, which we would interpret as say 70 out of 100 questions. The sample size of 1000 just tells us this is a reasonably large sample and are OK in making inferences from the normal curve. They don't tell us anything about the normality of the data, so we must make that assumption.

Next thing we need to do is find the z-score which corresponds to 70% of the area under the curve. We can look this value up in a z-table or use the percentile to z-score calculator and enter 70%. Since we're interested in the 1-sided area only (since we don't want the top 15% and bottom 15% of scores) we select the One-Sided area. We get the z-score of .5244.

Now we need to find the score which is .524 standard deviations above the mean of 70. To do so we use the z-score formula which is (datapoint - mean)/sd. Lets fill in what we know.

- ( x – 70 ) /11.55 = .5244
- x - 70 = 6.05682
- x = 70+6.05682
- x = 76.05682

So the students need to achieve an average score of 76.057 to keep their funding. This assumes the average score for the whole test is 70 and a standard deviation of 11.55.