Question

UESTION 3 Researchers discovered 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 40cm and 60cm, respectively. How likely are you to capture a green sea turtle that is considered illegal?

Answer 1

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.

Let X = curved carapace (shell) length of turtles

SO, X ~ Normal($\mu=55.7,\sigma^{2} =12^{2}$)

The z-score probability distribution of normal distribution is given by;

                  Z =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean length = 55.7 cm

            $\sigma$ = 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 how likely is to capture a green sea turtle that is considered illegal.

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 $\leq$ 40 cm)

  P(X < 60 cm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{60-55.7}{12}$ ) = P(Z < 0.36) = 0.64058

  P(X $\leq$ 40 cm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{40-55.7}{12}$ ) = P(Z $\leq$ -1.31) = 1 - P(Z < 1.31)

                                                        = 1 - 0.90490 = 0.0951

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.

Therefore, P(40 cm < X < 60 cm) = 0.64058 - 0.0951 = 0.5455

So, Probability(Legal capturing of sea turtles) = 0.5455

Hence, Probability(illegal capturing of sea turtles) = 1 - 0.5455 = 0.4545 or 45.45%

Therefore, we are 45.45% likely to capture a green sea turtle that is considered illegal.