Answer:
Probability that the sample contains at least five Roman Catholics = 0.995 .
Step-by-step explanation:
We are given that In Quebec, 90 percent of the population subscribes to the Roman Catholic religion.
The Binomial distribution probability is given by;
P(X = r) = [tex]\binom{n}{r}p^{r}(1-p)^{n-r}[/tex] for x = 0,1,2,3,.......
Here, n = number of trials which is 8 in our case
r = no. of success which is at least 5 in our case
p = probability of success which is probability of Roman Catholic of
0.90 in our case
So, P(X >= 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
= [tex]\binom{8}{5}0.9^{5}(1-0.9)^{8-5} + \binom{8}{6}0.9^{6}(1-0.9)^{8-6} + \binom{8}{7}0.9^{7}(1-0.9)^{8-7} + \binom{8}{8}0.9^{8}(1-0.9)^{8-8}[/tex]
= 56 * [tex]0.9^{5} * (0.1)^{3}[/tex] + 28 * [tex]0.9^{6} * (0.1)^{2}[/tex] + 8 * [tex]0.9^{7} * (0.1)^{1}[/tex] + 1 * [tex]0.9^{8}[/tex]
= 0.995
Therefore, probability that the sample contains at least five Roman Catholics is 0.995.
