In: Statistics and Probability
66% of all bald eagles survive their first year of life. If 35
bald eagles are randomly selected, find the probability that
a. Exactly 23 of them survive their first year of life.
b. At most 22 of them survive their first year of life.
c. At least 23 of them survive their first year of
life.
d. Between 21 and 26 (including 21 and 26) of them survive their
first year of life.
n = 35
p = 0.66
It is a binomial distribution.
P(X = x) = nCx * px * (1 - p)n - x
a) P(X = 23) = 35C23 * (0.66)^23 * (0.34)^12
b) P(X < 22) = 1 - P(X > 22)
= 1 - (P(X = 23) + P(X = 24) + P(X = 25) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30) + P(X = 31) + P(X = 32) +P(X = 33) + P(X = 34) + P(X = 35))
= 1 - (35C23 * (0.66)^23 * (0.34)^12 + 35C24 * (0.66)^24 * (0.34)^11 + 35C25 * (0.66)^25 * (0.34)^10 + 35C26 * (0.66)^26 * (0.34)^9 + 35C27 * (0.66)^27 * (0.34)^8 + 35C28 * (0.66)^28 * (0.34)^7 + 35C29 * (0.66)^29 * (0.34)^6 + 35C30 * (0.66)^30 * (0.34)^5 + 35C31 * (0.66)^31 * (0.34)^4 + 35C32 * (0.66)^32 * (0.34)^3 + 35C33 * (0.66)^33 * (0.34)^2 + 35C34 * (0.66)^34 * (0.34)^1 + 35C35 * (0.66)^35 * (0.34)^0)
= 1 - 0.5917 = 0.4083
c) P(X > 23) = P(X = 23) + P(X = 24) + P(X = 25) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30) + P(X = 31) + P(X = 32) +P(X = 33) + P(X = 34) + P(X = 35)
= 35C23 * (0.66)^23 * (0.34)^12 + 35C24 * (0.66)^24 * (0.34)^11 + 35C25 * (0.66)^25 * (0.34)^10 + 35C26 * (0.66)^26 * (0.34)^9 + 35C27 * (0.66)^27 * (0.34)^8 + 35C28 * (0.66)^28 * (0.34)^7 + 35C29 * (0.66)^29 * (0.34)^6 + 35C30 * (0.66)^30 * (0.34)^5 + 35C31 * (0.66)^31 * (0.34)^4 + 35C32 * (0.66)^32 * (0.34)^3 + 35C33 * (0.66)^33 * (0.34)^2 + 35C34 * (0.66)^34 * (0.34)^1 + 35C35 * (0.66)^35 * (0.34)^0 = 0.5917
d) P(21 < X < 26) = P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25) + P(X = 26)
= 35C21 * (0.66)^21 * (0.34)^14 + 35C22 * (0.66)^22 * (0.34)^13 + 35C23 * (0.66)^23 * (0.34)^12 + 35C24 * (0.66)^24 * (0.34)^11 + 35C25 * (0.66)^25 * (0.34)^10 + 35C26 * (0.66)^26 * (0.34)^9
= 0.7136