Shell lengths of sea turtles. refer to the aquatic biology (vol. 9, 2010) study of green sea turtles inhabiting the grand cayman south sound lagoon, exercise 2.85 (p. 97). researchers discovered that the curved carapace (shell) length of these turtles is approximately normally distributed with mean 55.7 centimeters and standard deviation 11.5 centimeters.
a. the minimum and maximum size limits for captured turtles in the legal marine turtle fishery are 40 cm and 60 cm, respectively. how likely are you to capture a green sea turtle that is considered illegal?
b. what maximum limit, l, should be set so that only 10% of the turtles captured have shell lengths greater than l?

   a) 44.46 %

b)  Z = 94.8

Step-by-step explanation:

We have a normal distribution with mean 55.7 and standard deviation 11.5 all in cms.

The problem stament ask for:

a)The minimum and maximum size limits for captured turtles in the legal marine turtle fishery are 40 cm and 60 cm, respectively. How likely are you to capture a green sea turtle that is considered illegal?

What is the probability of capture turtles in that range

P [ 40 ≤ z ≤ 60 ]

z = ( Z- μ) ÷ σ                     for Z = 40       z₁ = ( 40-55.7) ÷ 11.5 ⇒ z₁ = -1.365

                                          for Z = 60       z₂ = ( 60-55.7) ÷ 11.5 ⇒ z₂ = 0.3739

The area ( probability ) associated to:

1.-For z₁ we find value for  -1,36    ⇒ 0.0869

                                           -1.37    ⇒ 0.0853

We are looking for -1.365 half way between -1.36 and -1.37

Then z =  -1.365  from this point to the left tail  is 0.0861 (or the probability 8.61 %)  

And for z₂ =              value for 0.37    ⇒ 0.6443

                                                0.38    ⇒ 0.6480

using the same criteria as before the area  for z₂ is 0.6415 or 64.15 % (this area is from the left tail up to the point z₂ = 0.375 therefore to get the probability to capture turtles considered llegal is

P [ -1.365 ≤ z ≤ 0.6415 ] is  64.15 - 8.61 = 55.54 % and of capture turtles considered illegal is 1- 55.54 = 0,4446 or 44.46 %

b)

The poin of area (in z table) 0,1 is aprox 3.4

z = (Z-μ) ÷ 11.5   ⇒   (3.4 × 11.5) = Z - 55.7  ⇒  39.1 + 55.7   Z = 94.8

can you me with homework?

step-by-step explanation: