The weights of all one hundred (100) 9th graders at a school are measured, and it is found that the mean of all the measurements is 100 lbs., with a standard deviation of 15 lbs.   Explain how you would use this information to determine the percentage of students who weighed between 85 lbs and 115 lbs.  Make sure your explanation includes the use of z-scores.

Z = (score-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores.

To determine the percentage of students who weighed between 85 lbs and 115 lbs, we can use the concept of z-scores. A z-score measures how many standard deviations a given data point is away from the mean.

Here's how we can use the provided information to calculate the percentage:

1. Calculate the z-score for each weight within the range of 85 lbs to 115 lbs using the formula:

z = (x - mean) / standard deviation

For example, for 85 lbs:
z = (85 - 100) / 15 = -1

And for 115 lbs:
z = (115 - 100) / 15 = 1

2. Once we have the z-scores, we can consult a standard normal distribution table or use statistical software to find the corresponding percentage for each z-score.

For a z-score of -1, the standard normal distribution table will give us the percentage of students below that z-score. To find the percentage between -1 and 1, we need to subtract the percentage of students below -1 from the percentage below 1.

3. Look up the percentages corresponding to -1 and 1 in the standard normal distribution table. Let's assume the percentage below -1 is 0.1587 and the percentage below 1 is 0.8413.

The percentage of students between -1 and 1 is:
0.8413 - 0.1587 = 0.6826 (or 68.26%)

Therefore, approximately 68.26% of the ninth graders weigh between 85 lbs and 115 lbs based on the given mean and standard deviation.