Answer:
We are 45.45% likely to capture a green sea turtle that is considered illegal.
Step-by-step explanation:
We are given that the curved carapace (shell) length of these turtles is approximately normally distributed with a mean of 55.7 centimeters and a standard deviation of 12 cm. The minimum and maximum size limits for captured sea turtles in the legal marine turtle fishery are 40 cm and 60 cm, respectively.
<em>Let X = curved carapace (shell) length of turtles</em>
SO, X ~ Normal()
The z-score probability distribution of normal distribution is given by;
Z = ~ N(0,1)
where, = population mean length = 55.7 cm
= standard deviation = 12 cm
Now, we are given the minimum and maximum size limits for captured sea turtles in the legal marine turtle fishery but we have to find that <u>how likely is to capture a green sea turtle that is considered illegal</u>.
As we know that; Probability(Legal capturing of sea turtles) = 1 - Probability(illegal capturing of sea turtles)
So, firstly we will find the probability for capturing sea turtles in the legal marine turtle fishery which is given by = P(40 cm < X < 60 cm)
P(40 cm < X < 60 cm) = P(X < 60 cm) - P(X 40 cm)
P(X < 60 cm) = P( <
) = P(Z < 0.36) = 0.64058
P(X 40 cm) = P(
) = P(Z
-1.31) = 1 - P(Z < 1.31)
= 1 - 0.90490 = 0.0951
<em>The above probabilities are calculated using z table by looking at the critical values of x = 0.36 and x = 1.31 which gives an probability area of 0.64058 and 0.9049 respectively.</em>
<em>Therefore, P(40 cm < X < 60 cm) = 0.64058 - 0.0951 = 0.5455</em>
So, Probability(Legal capturing of sea turtles) = 0.5455
<em>Hence, Probability(illegal capturing of sea turtles) = 1 - 0.5455 = 0.4545 or 45.45%</em>
Therefore, we are 45.45% likely to capture a green sea turtle that is considered illegal.