A call center data set shows that in a sample of 30 individuals, 11 had prior call center experience. If we assume that the probability that any potential hire will also have experience with a probability of 11/30, what is the probability that among ten potential hires, more than half of them will have experience? Define the parameter(s) for this distribution based on the data.

P(X > 5) = 0.1164 to 4 d.p.

The parameters are defined in the explanation.

Step-by-step explanation:

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of potential hires = 10

x = Number of successes required = number of potential hires that have prior call centre experience = more than half; that is, x > 5

p = probability of success = probability that any potential hire will have experience = (11/30) = 0.367

q = probability of failure = probability that any potential hire will NOT have experience = 1 - p = 1 - 0.367 = 0.633

P(X > 5) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)

Inserting the parameters and computing the probabilities for each of those values of X,

P(X > 5) = 0.11641775484 = 0.1164 to 4 d.p.

Hope this Helps!!!

1.3 million hoops i believe

15 / 2 = 7.5

100 / 7.5 = 13.