Suppose a college entrance exam has a mean of 1200 and a standard deviation of 100. An Ivy league college will only take applicants in the 90% percentile or higher. what is the lowest score they would need on the entrance exam?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.90) and its Z score. Insert data into equation and solve for score.

To determine the lowest score needed on the entrance exam to be in the 90th percentile or higher, we'll first need to find the z-score corresponding to the 90th percentile.

A z-score measures the number of standard deviations a particular value is from the mean. By looking up the z-score in a standard normal distribution table, we can find the corresponding raw score.

The formula to calculate the z-score is:
z = (x - mean) / standard deviation

In this case:
mean = 1200
standard deviation = 100
z = unknown (to be determined)
x = unknown (the lowest score needed)

Now, let's calculate the z-score:
z = (x - mean) / standard deviation

To find the z-score corresponding to the 90th percentile, we can use a standard normal distribution table or consult a z-score calculator. For simplicity, let's assume the z-score is 1.28, which corresponds to the 90th percentile.

We can rearrange the formula to solve for x:
x = (z * standard deviation) + mean

Plugging in the values:
x = (1.28 * 100) + 1200
x = 128 + 1200
x = 1328

Therefore, the lowest score needed on the entrance exam to be in the 90th percentile or higher is 1328.