A. Z = (score-mean)/SD
Z = (50-52)/17.2 = -2/17.2 = -.12
Z = (60-52)/17.2 = 8/17.2 = .47
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 scores.
.0478 +.1808 = .2286 = 23%
B. Similar process.
C. Correct
The weight of adult green sea urchins are normally distributed with a mean 52g and standard deviation 17.2g?
A. Find the percentage of adult green sea urchins with weighs between 50g and 60g
My answer/work: normalcdf (50, 60, 52, 17.2) = .225, 23%
B. Obtain the percentage of adult green sea urchins with weights above 40g
My answer/work: normalcdf (40, e99, 52, 17.2) = .757, 76%
C. Determine and interpret the 90% percentile for the weights
My answer/work: invnorm: .9, 0,1) =1.28
z-52 + (1.28)*(17.2)
z= 74.04%
I'm a little unsure if I got C correct. As for the rest I'm wondering if I did them correctly. If not, could someone lean me to the right direction? Thank you :)
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