45 A study found that the mean migration distance of a green

4-5 A study found that the mean migration distance of a green turtle was 2200 km and the standard deviation was 625 km. Assuming the migration distances are normally distributed0 find the probability that a randomly selected turtle migrates a distance of 4. Less than 1900 km 5. Between 2000 km and 2500 km

Solution

Z = X - mean/standard deviation

4)

P(X < 1900) = P( Z < 1900 - 2200/625)

= P ( Z < -0.48)

= 0.3156

5)

P(2000 < X < 2500)

= P ( X < 2500) - P( X < 2000)

= P ( Z < 2500 - 2200/625) - P( Z < 2000 - 2200/625)

= P ( Z < 0.48) - P ( Z < -0.32)

= 0.6844 - 0.3745

= 0.3099