Suppose that the average weight of students in a statistics class is 75.5 kilograms with a standard deviation of 10.3 kilograms.
What percent of students have weights greater than 80.1 kilograms?
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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 related to the Z score.
To find the percentage of students with weights greater than 80.1 kilograms, we need to use the standard deviation and the mean of the weight distribution. We can relate a particular weight value to the average weight by calculating the z-score.
The z-score is given by the formula:
z = (x - μ) / σ
Where:
x is the weight value (80.1 kilograms),
μ is the mean weight,
σ is the standard deviation.
Plugging in the values we have:
z = (80.1 - 75.5) / 10.3
Calculating this value yields:
z ≈ 0.44660194
Now, we need to find the percentage of students with weights greater than 80.1 kilograms, which corresponds to finding the area under the standard normal distribution curve to the right of this z-score.
To find this percentage, we can use a standard normal distribution table or a statistical software. Let's say we use a standard normal distribution table and look up the z-score of 0.45 (the closest value) in the table.
Looking up the z-score 0.45 in the table, we find the corresponding area/probability to be approximately 0.6736.
However, we are interested in the students with weights greater than 80.1 kilograms, so we subtract this area value from 1 (since the total area under the normal distribution curve is 1) to get the percentage.
Percentage = 1 - 0.6736 ≈ 0.3264
Therefore, approximately 32.64% of students have weights greater than 80.1 kilograms in the statistics class.