9. A marine biologist was studying the size distribution of mussels in the inter
ID: 3074422 • Letter: 9
Question
9. A marine biologist was studying the size distribution of mussels in the intertidal zone. The mean length of mussels was 21.5 mm with a standard deviation of 8.0 mm. A mussel is classified as a "new recrui if it is less than 10.0 mm in length, a "first year mussel" if it is between 10.0 mm and 30.0 mm in length, and an "established musse" if it is above 30.0 mm in length. Assuming mussel length follows a normal distribution, answer the following questions (a) What is the percentile for mussel with a length of 25.0 mm? 67.00th percentile] (b) What mussel length corresponds to the 90th percentile? [31.74 mm] (c) What proportion of the mussel population are new recruits? What proportion of the population are first year mussels? What proportion of the population are established mussels? [0.0749, 0.7805, and 0.1446 for new recruit, first year, and established mussels, respectively]Explanation / Answer
Question 9
Here mean length of mussesls = 21. m
standard deviation = 8.0 m
(a) Here percentile for mussel with length of 25.0 mm. Here x is the lengh of mussels
Pr(x < 25.0) = NORMDIST (x < 25.0 ; 21.5 mm ; 8.0 mm)
Z = (25.0 - 21.5)/8 = 0.4375
Pr(x < 25.0) = NORMDIST (x < 25.0 ; 21.5 mm ; 8.0 mm) = Pr(Z < 0.4375) = 0.6691 or 0.67
(b) Here if 90th percentile is x0 then
Pr(x <x0) = 0.90
so here z value is
Z = 1.28155
(x0 - 21.5)/8 = 1.28155
x0 = 21.5 + 8 * 1.28155 = 31.75 mm
(c) Here the new recruits probabilty are = Pr(x <10 mm) = Pr(x < 10 mm ; 21.5 mm ; 8.0 mm)
Z = (10 - 21..5)/8 = -1.4375
Pr(x < 10 mm ; 21.5 mm ; 8.0 mm) = Pr(Z < -1.4375) = 0.0749
Pr(10 < x < 30 mm ; 21.5 mm ; 8.0 mm)
= Pr(x <30 mm ; 21.5 mm ; 8.0 mm) - Pr(x < 10 mm ; 21.5 mm ; 8.0 mm)
Z2 = (30 - 21.5) /8 = 1.0625
Z1 = -1.4375
Pr(x <30 mm ; 21.5 mm ; 8.0 mm) - Pr(x < 10 mm ; 21.5 mm ; 8.0 mm) = Pr(Z < 1.0625) - Pr(Z <-1.4375)
= 0.856 - 0.0749 = 0.7805
Pr(Established muscles) = 1 - Pr(New recruits) - Pr(Firs year mussel)
= 1 - 0.0749 - 0.7805 = 0.1446
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