Answer:
The probability that exactly 97 of them finished the marathon is 0.221.
Step-by-step explanation:
We are given that in 2019, approximately 97.4% of all the runners who started the Boston Marathon were able to complete the 42.2 km race.
100 runners are chosen at random.
The above situation can be represented through the binomial distribution;
[tex]P(X = x) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x = 0,1,2,3,......[/tex]
where, n = number of trials(or samples) taken = 100 runners
r = number of success = exactly 97
p = probability of success which in our question is the probability
that runners finished the marathon, i.e; p = 0.974
So, X ~ Binom(n = 100, p = 0.974)
Now, the probability that exactly 97 of them finished the marathon is given by = P(X = 97)
P(X = 97) = [tex]\binom{100}{97} \times 0.974^{97} \times (1-0.974)^{100-97}[/tex]
= [tex]161700 \times 0.974^{97} \times 0.026^{3}[/tex]
= 0.221
Hence, the required probability is 0.221.
Using the binomial distribution, it is found that there is a 0.2207 = 22.07% probability that exactly 97 of them finished the marathon.
For each runner, there are only two possible outcomes, either they finished the marathon, or they did not. The probability of a runner finishing the marathon is independent of any other runner, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
The probability that exactly 97 of them finished the marathon is P(X = 97), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 97) = C_{100,97}.(0.974)^{97}.(0.026)^{3} = 0.2207[/tex]
There is a 0.2207 = 22.07% probability that exactly 97 of them finished the marathon.
A similar problem is given at https://brainly.com/question/24863377
