96.0k views
0 votes
For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12? hint: use interval-based calculations because no critical region is involved.​

1 Answer

6 votes
The binomial distribution is given by,
P(X=x) =
(^(n)C_(x))p^(x) q^(n-x)
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) =
(^(100)C_(x))(0.2)^(x) (0.8)^(100-x)
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) =
(^(100)C_(0))(0.2)^(0) (0.8)^(100-0) + (^(100)C_(1))(0.2)^(1) (0.8)^(100-1) +
(^(100)C_(2))(0.2)^(2) (0.8)^(100-2) + (^(100)C_(3))(0.2)^(3) (0.8)^(100-3) +
(^(100)C_(4))(0.2)^(4) (0.8)^(100-4) + (^(100)C_(5))(0.2)^(5) (0.8)^(100-5) +
(^(100)C_(6))(0.2)^(6) (0.8)^(100-6) + (^(100)C_(7))(0.2)^(7) (0.8)^(100-7) +
(^(100)C_(8))(0.2)^(8) (0.8)^(100-8) + (^(100)C_(9))(0.2)^(9) (0.8)^(100-9) +
(^(100)C_(10))(0.2)^(10) (0.8)^(100-10) + (^(100)C_(11))(0.2)^(11) (0.8)^(100-11) +
(^(100)C_(12))(0.2)^(12) (0.8)^(100-12)


Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.
User Wamadahama
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.