Question

1. suppose the prevalence (p=.32) of type II diabetes follows a binomial distribution in a community. Researchers want to sample 15 households from the community for a new study investigating ethnographic ties to the chronic illness.

A. What is the cumulative probability of selecting 8 or more diabetics in the sample from the community?

B. What is the probability of selecting between 6 and 10 diabetics in the sample from the community?

Answer 1

a)

Here, n = 15, p = 0.32, (1 - p) = 0.68 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 <= 7) = (15C0 * 0.32^0 * 0.68^15) + (15C1 * 0.32^1 * 0.68^14) + (15C2 * 0.32^2 * 0.68^13) + (15C3 * 0.32^3 * 0.68^12) + (15C4 * 0.32^4 * 0.68^11) + (15C5 * 0.32^5 * 0.68^10) + (15C6 * 0.32^6 * 0.68^9) + (15C7 * 0.32^7 * 0.68^8)
P(X <= 7) = 0.0031 + 0.0217 + 0.0715 + 0.1457 + 0.2057 + 0.213 + 0.1671 + 0.1011
P(X <= 7) = 0.9289

P(X >= 8). = 1 - P(X < =7)
= 1 - 0.9289
= 0.0711

b)

Here, n = 15, p = 0.32, (1 - p) = 0.68, x1 = 6 and x2 = 10.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(6 <= X <= 10)
P(6 <= X <= 10) = (15C6 * 0.32^6 * 0.68^9) + (15C7 * 0.32^7 * 0.68^8) + (15C8 * 0.32^8 * 0.68^7) + (15C9 * 0.32^9 * 0.68^6) + (15C10 * 0.32^10 * 0.68^5)
P(6 <= X <= 10) = 0.1671 + 0.1011 + 0.0476 + 0.0174 + 0.0049
P(6 <= X <= 10) = 0.3381