Question

LAB 6.2: Green Sea Urchins: From the paper "Effects of Chronic Nitrate Exposure on Gonad Growth in GreenSea Urchin Strongylocentrotus droebachienses" (Aquaculture, Vol. 242, No. 1-4, pp. 357-363) by S. Siikavuopio et al., we found that weights of adult green sea urchins are normally distributed with mean 52.0 g and standard deviation 17.2 g Answer the following questions completely, showing all calculations and stating a conclusion. Use MINITAB to find needed probabilities.(6 Marks: 2 Marks for each of a., b., and c.) Find the percentage of adult green sea urchins with weights between 50 g and 60 g. a. Obtain the percentage of adult green sea urchins with weights above 40 g b. Determine and interpret the 90th percentile for the weights C.

Answer 1

Let X denotes the weight of a randomly selected adult green sea urchins.

X ~ Normal(52, 17.2²)

a) The percentage of adult green sea urchins with weights between 50 g and 60 g

$= 100*P(50\leq X\leq 60)$

$= 100*[P(X\leq 60) - P(X\leq 50)]$

$= 100*(0.679076 - 0.453716)$

$= 22.536\%$

b) The percentage of adult green sea urchins with weights above 40 g

$= 100*P(X>40)$

$= 100*[1 - P(X\leq 40)]$

$= 100*(1 - 0.24269)$

$= 75.731\%$

c) 90^th percentile for the weights = 74.0427

..Minitab - Untitled - [Session] File Edit Data Calc Stat Graph Editor Tools Window Help Assistant Oxat 21-10-2019 23:05:50 – Welcome to Minitab, press F1 for help. ## part a) MTB > CDF 60; SUBC> Normal 52 17.2. Cumulative Distribution Function Normal with mean = 52 and standard deviation = 17.2 x 60 P ( x S x ) 0.679076 MTB > CDF 50; SUBC> Normal 52 17.2. Cumulative Distribution Function Normal with mean = 52 and standard deviation = 17.2 x P(XSx) 50 0.453716 MTB > let kl = 100* (0.679076 - 0.453716) MTB > print kl Data Display K1 22.5360

.: Minitab - Untitled - [Session] O File Edit Data Calc Stat Graph Editor Tools Window Help Assistant + N X QATOO ## part b) MTB > CDF 40; SUBC> Normal 52 17.2. Cumulative Distribution Function Normal with mean = 52 and standard deviation = 17.2 x 40 P ( x S x) 0.242690 MTB > let k2 = 100*(1 - 0.24269) MTB > print k2 Data Display K2 75.7310 ## part ) MTB > InvCDF 0.9; SUBC> Normal 52 17.2. Inverse Cumulative Distribution Function Normal with mean = 52 and standard deviation = 17.2 P ( X S x) 0.9 x 74.0427