The weights of adult males are normally distributed with a mean of 172 lb and a standard deviation of 29 lb.
1.Find the weight that divides the upper 2% of weights from the lower 98%.
Can anyone please explain how to do this,I have no idea from where to begin.
Thanks.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score that fits that proportion.
Insert values in above equation to find weight score.
To find the weight that divides the upper 2% from the lower 98%, you need to use the concept of the standard normal distribution.
Here are the steps to solve this problem:
Step 1: Convert the given weights to standard scores (z-scores). The formula for calculating the z-score is:
z = (x - μ) / σ
where:
x = the given weight
μ = the mean of the distribution (172 lb)
σ = the standard deviation of the distribution (29 lb)
Step 2: Find the z-score that corresponds to the upper 2% of the distribution. This can be done by finding the z-score associated with the area under the normal curve to the left of the desired upper percentile (2%). This is typically done using a standard normal distribution table, or you can use a calculator or software that provides this functionality.
Step 3: Once you have the z-score for the upper 2%, you can use it to find the weight that corresponds to that z-score. This can be done by rearranging the z-score formula:
x = z * σ + μ
where:
x = the weight that divides the upper 2% from the lower 98%
z = the z-score corresponding to the upper 2%
μ = the mean of the distribution (172 lb)
σ = the standard deviation of the distribution (29 lb)
By plugging in the values into the formula, you can find the weight that divides the upper 2% from the lower 98%.
I hope this explanation helps you understand how to approach and solve this problem.