Answer to Question #214016 in Statistics and Probability for jayden

Question

According to German bureau of statistics, the probability of a randomly selected 90 year old German male surviving at least another year is approximately 0.82. if a sample of 20 90 year old German males is chosen what is the probability that

1. 18
2. at least 18
3. at most 18 will survive for at least another year?

Answer 1

Let $X=$ the number of 90 year old German male surviving at least another year: $X\sim Bin(n, p).$

Given $n=20, p=0.82, q-1-p=1-0.82=0.18.$

1.

$$P(X=18)=\dbinom{20}{18}(0.82)^{18}(0.18)^{20-18}$$

$$\approx 0.17296090701$$

2.

$$P(X\geq18)=P(X=18)+P(X=19)+P(X=20)$$

$$=\dbinom{20}{18}(0.82)^{18}(0.18)^{20-18}+\dbinom{20}{19}(0.82)^{19}(0.18)^{20-19}$$

$$+\dbinom{20}{20}(0.82)^{20}(0.18)^{20-20}$$

$$\approx 0.27479318631$$

3.

$$P(X\leq18)=1-P(X=19)-P(X=20)$$

$$=1-\dbinom{20}{19}(0.82)^{19}(0.18)^{20-19}$$

$$-\dbinom{20}{20}(0.82)^{20}(0.18)^{20-20}$$

$$\approx0.8981677207$$