Question

6.10n Green Sea Urchins. From the paper "Effects of Chronic Nitrate Exposure on Gonad Growth in Green Sea 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. (15 marks: 4 Marks For a., 3 Marks for b., 4 Marks Each For e. and d.) a. Find the percentage of adult green sea urchins with weights between 50 g and 60 g. b. Obtain the percentage of adult green sea urchins with weights above 40 g. c. Determine and interpret the 90th percentile for the weights. d. Find and interpret the 6th decile for the weights.

Answer 1

Let X be the weight of adult green sea urchin and it is given that X-N(u=52,0=17.2). (a) The percentage of adult green sea urchins with weights between 50 g and 60 g is computed as follows: 50-52 X-u 60-52 P150 - <- - <- 17.2 o 17.2 P[50<x<60)=P(107.92 874_697.32 = P(-0.12<Z<0.47] (2=454-N(0,1) = P[Z<0.47] - P[Z<-0.12] =NORM.S.DIST (0.47,TRUE) – NORM.S.DIST (-0.12,TRUE) (use the function in Excel) =0.6808-0.4522 =0.2286 Therefore, 22.86% of adult green sea urchins with weights between 50g and 60g.

The percentage of adult green sea urchins with weights above 40g is computed as follows: X-u 30-52 - > o 17.2 P[X> 40)=P(4,4 30732 =P[2>-1.28] z="SEAT- (0,1) =1- P[Z<-1.28] =1-NORM.S.DIST (-1.28, TRUE) (use the function in MS Excel) =1-0.1002 = 0.8997 Thus, 89.97% of adult green sea urchins with weights above 40g. (c)

The 90th percentile for the weights is computed as follows: P[X<x]=0.90 Ισ =0(1732 -0.90 3:1552 = 64(0.90) 17.2 r-52 Ï =NORM.S.INV (use the function in MS Excel) 17.2 -x-52 -1.2816 Ï 17.2 >x=52+(1.2816)17.2 x=74.0435 Therefore, the 90th percentile is 74.0435. This means that weight of 90% of the observations lie below 74.0435g . (d)

The 6th decile for weights is computed as follows: P[X<x]=0.6 > P[131172-06 =P[2<*72-06 = 172-9-(0.6) X-52 - 17.2 =*==NORM.S.INV (0.6) (use the function in MS Excel) 17.2 x-52 - ==0.2533 17.2 3x=52+(0.2533)17.2 x=56.3568 Therefore, the 6th decile is 56.3568). This means that weight of 60% of the observations lie below 56.3568g.