Respuesta :
Answer: 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