Question

70% of people who are 30 years old will be alive ten years from now. In the random sample of 15 people, calculate the probability that...

a) exactly 10 of them will be alive ten years from now

b) at least 9 will be alive ten years from now

c) at most 8 will be alive ten years from now

d) how many of them would you expect to be alive 10 years from now

e) let X be the variable denoting the number of people alive ten years from now. Calculate the standard deviation of X

Answer 1

a)

Here, n = 15, p = 0.7, (1 - p) = 0.3 and x = 10
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 10)
P(X = 10) = 15C10 * 0.7^10 * 0.3^5
P(X = 10) = 0.2061

b)

Here, n = 15, p = 0.7, (1 - p) = 0.3 and x = 9
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 8).
P(X <= 8) = (15C0 * 0.7^0 * 0.3^15) + (15C1 * 0.7^1 * 0.3^14) + (15C2 * 0.7^2 * 0.3^13) + (15C3 * 0.7^3 * 0.3^12) + (15C4 * 0.7^4 * 0.3^11) + (15C5 * 0.7^5 * 0.3^10) + (15C6 * 0.7^6 * 0.3^9) + (15C7 * 0.7^7 * 0.3^8) + (15C8 * 0.7^8 * 0.3^7)
P(X <= 8) = 0 + 0 + 0 + 0.0001 + 0.0006 + 0.003 + 0.0116 + 0.0348 + 0.0811
P(X <= 8) = 0.1312

P(x >= 9) = 1 - P(x< =8)
= 1 - 0.1312
= 0.8688

c)

Here, n = 15, p = 0.7, (1 - p) = 0.3 and x = 8
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 8).
P(X <= 8) = (15C0 * 0.7^0 * 0.3^15) + (15C1 * 0.7^1 * 0.3^14) + (15C2 * 0.7^2 * 0.3^13) + (15C3 * 0.7^3 * 0.3^12) + (15C4 * 0.7^4 * 0.3^11) + (15C5 * 0.7^5 * 0.3^10) + (15C6 * 0.7^6 * 0.3^9) + (15C7 * 0.7^7 * 0.3^8) + (15C8 * 0.7^8 * 0.3^7)
P(X <= 8) = 0 + 0 + 0 + 0.0001 + 0.0006 + 0.003 + 0.0116 + 0.0348 + 0.0811
P(X <= 8) = 0.1312

d)

mean = np = 0. 70 * 15 = 10.5

e)
std.dev = sqrt(npq)
=sqrt(15 * 0.7 * 0.3)
= 1.7748