Assuming that the distribution is normal for weight relative to the ideal and 99% of the male participants

scored between (–53.68, 64.64), where did 95% of the values for weight relative to the ideal lie? Round your
answer to two decimal places.

you can find Z-table help here:

http://davidmlane.com/hyperstat/z_table.html

you know the mean is midway between these two extremes.

Use the tool to find how many STD's that is, then scale up to find the real score range.

To find where 95% of the values for weight relative to the ideal lie, we can use the concept of z-scores.

A z-score measures how many standard deviations a given value is from the mean of a distribution. In this case, assuming a normal distribution, we can use the z-score formula:

z = (x - mean) / standard deviation

Where x is the value we want to find the z-score for, mean is the mean of the distribution, and standard deviation is the standard deviation of the distribution.

Given that 99% of male participants scored between -53.68 and 64.64, we know that these values correspond to the z-scores -2.33 and 2.33 (since 99% of the values lie within 2.33 standard deviations from the mean in a normal distribution).

To find the mean and standard deviation of the distribution, we would need more information or data. Without this information, we cannot calculate the exact values for the mean and standard deviation.

However, we can still estimate where 95% of the values lie using the z-score. Since we know that 99% of the values are within 2.33 standard deviations from the mean, and a normal distribution is symmetric, we can assume that about 2.5% of the values lie beyond 2.33 standard deviations on each tail of the distribution.

So, to find where 95% of the values lie, we can subtract 2.5% from each tail:

Lower bound = -2.33 - 2.5% = -2.33 - 0.025 = -2.355 (approximately)
Upper bound = 2.33 + 2.5% = 2.33 + 0.025 = 2.355 (approximately)

Therefore, approximately 95% of the values for weight relative to the ideal lie between -2.355 and 2.355 (rounded to two decimal places).