Question

71% of all bald eagles survive their first year of life. If 45 bald eagles are randomly selected, find the probability that

a. Exactly 32 of them survive their first year of life. ____
b. At most 35 of them survive their first year of life.____
c. At least 34 of them survive their first year of life.____
d. Between 29 and 36 (including 29 and 36) of them survive their first year of life.____

Answer 1

Mean = n * P = ( 45 * 0.71 ) = 31.95
Variance = n * P * Q = ( 45 * 0.71 * 0.29 ) = 9.2655
Standard deviation = = 3.0439

Part a)
P ( X = 32 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 32 - 0.5 < X < 32 + 0.5 ) = P ( 31.5 < X < 32.5 )

P ( 31.5 < X < 32.5 )
Standardizing the value

Z = ( 31.5 - 31.95 ) / 3.0439
Z = -0.15
Z = ( 32.5 - 31.95 ) / 3.0439
Z = 0.18
P ( -0.15 < Z < 0.18 )
P ( 31.5 < X < 32.5 ) = P ( Z < 0.18 ) - P ( Z < -0.15 )
P ( 31.5 < X < 32.5 ) = 0.5717 - 0.4412
P ( 31.5 < X < 32.5 ) = 0.1305

part b)

P ( X <= 35 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 35 + 0.5 ) = P ( X < 35.5 )

P ( X < 35.5 )
Standardizing the value

Z = ( 35.5 - 31.95 ) / 3.0439
Z = 1.17

P ( X < 35.5 ) = P ( Z < 1.17 )
P ( X < 35.5 ) = 0.879

Part c)

P ( X >= 34 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 34 - 0.5 ) =P ( X > 33.5 )

P ( X > 33.5 ) = 1 - P ( X < 33.5 )
Standardizing the value

Z = ( 33.5 - 31.95 ) / 3.0439
Z = 0.51

P ( Z > 0.51 )
P ( X > 33.5 ) = 1 - P ( Z < 0.51 )
P ( X > 33.5 ) = 1 - 0.695
P ( X > 33.5 ) = 0.305

Part d)

P ( 29 <= X <= 36 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 29 - 0.5 < X < 36 + 0.5 ) = P ( 28.5 < X < 36.5 )

P ( 28.5 < X < 36.5 )
Standardizing the value

Z = ( 28.5 - 31.95 ) / 3.0439
Z = -1.13
Z = ( 36.5 - 31.95 ) / 3.0439
Z = 1.49
P ( -1.13 < Z < 1.49 )
P ( 28.5 < X < 36.5 ) = P ( Z < 1.49 ) - P ( Z < -1.13 )
P ( 28.5 < X < 36.5 ) = 0.9325 - 0.1285
P ( 28.5 < X < 36.5 ) = 0.804