Suppose that the average weight of students in a statistics class is 75.5 kilograms with a standard deviation of 10.3 kilograms.
A. 45% of the student weights are above ____ kilograms?
Z=.3264=x-75.5/10.3 Now what?
55% are below x
z = .1 ---> F(z) about .55
.1 = (x - 75.5)/10.3
x - 75.5 = 1.03
x = 76.5
Thanks!
You are welcome :)
To find the weight above which 45% of the student weights fall, you can use the z-score formula. The z-score formula is given by (x - μ) / σ, where x is the value you want to find, μ is the mean, and σ is the standard deviation.
Given that the mean weight is μ = 75.5 kilograms and the standard deviation is σ = 10.3 kilograms, you have the equation:
Z = (x - 75.5) / 10.3
You already calculated the z-score as Z = 0.3264. Now, you can solve for x by rearranging the equation:
0.3264 = (x - 75.5) / 10.3
To isolate x, multiply both sides of the equation by 10.3:
0.3264 * 10.3 = x - 75.5
3.36432 = x - 75.5
Add 75.5 to both sides of the equation:
3.36432 + 75.5 = x
78.86432 = x
So, the weight above which 45% of the student weights fall is approximately 78.86 kilograms.