Question

78% of all bald eagles survive their first year of life. If 43 bald eagles are randomly selected, find the probability that a. Exactly 33 of them survive their first year of life. b. At most 36 of them survive their first year of life. c. At least 33 of them survive their first year of life. d. Between 31 and 37 (including 31 and 37) of them survive their first year of life.

Answer 1

Mean = n * P = ( 43 * 0.78 ) = 33.54
Variance = n * P * Q = ( 43 * 0.78 * 0.22 ) = 7.3788
Standard deviation = = 2.7164

Part a)

P ( X = 33 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 33 - 0.5 < X < 33 + 0.5 ) = P ( 32.5 < X < 33.5 )

P ( 32.5 < X < 33.5 )
Standardizing the value

Z = ( 32.5 - 33.54 ) / 2.7164
Z = -0.38
Z = ( 33.5 - 33.54 ) / 2.7164
Z = -0.01
P ( -0.38 < Z < -0.01 )
P ( 32.5 < X < 33.5 ) = P ( Z < -0.01 ) - P ( Z < -0.38 )
P ( 32.5 < X < 33.5 ) = 0.4941 - 0.3509
P ( 32.5 < X < 33.5 ) = 0.1432

part b)

P ( X <= 36 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 36 + 0.5 ) = P ( X < 36.5 )

P ( X < 36.5 )
Standardizing the value

Z = ( 36.5 - 33.54 ) / 2.7164
Z = 1.09

P ( X < 36.5 ) = P ( Z < 1.09 )
P ( X < 36.5 ) = 0.8621

Part c)

P ( X >= 33 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 33 - 0.5 ) =P ( X > 32.5 )

P ( X > 32.5 ) = 1 - P ( X < 32.5 )
Standardizing the value

Z = ( 32.5 - 33.54 ) / 2.7164
Z = -0.38

P ( Z > -0.38 )
P ( X > 32.5 ) = 1 - P ( Z < -0.38 )
P ( X > 32.5 ) = 1 - 0.352
P ( X > 32.5 ) = 0.648

Part d)

P ( 31 <= X <= 37 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 31 - 0.5 < X < 37 + 0.5 ) = P ( 30.5 < X < 37.5 )

P ( 30.5 < X < 37.5 )
Standardizing the value

Z = ( 30.5 - 33.54 ) / 2.7164
Z = -1.12
Z = ( 37.5 - 33.54 ) / 2.7164
Z = 1.46
P ( -1.12 < Z < 1.46 )
P ( 30.5 < X < 37.5 ) = P ( Z < 1.46 ) - P ( Z < -1.12 )
P ( 30.5 < X < 37.5 ) = 0.9276 - 0.1315
P ( 30.5 < X < 37.5 ) = 0.796